A problem from Hardy's_Pure Mathematics on Real variables about Linear equations in 3 variables Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Essay about the art and applications of differential equations?Quotient-lifting propertiesIs there a general algorithm to solve computable integral equation?Is there a codifferential for a covariant exterior derivative?Solvability of the equation $2a_1^2 = a_2^n + a_3^n + a_4^n$ when $n geq 5$ is prime?A monoid where every element has finitely many divisorsWhy does what I've written fail to define truth?existence of solution to set of non-linear equationsHelp me motivate a topic.When does a set of inequalities over series imply an inequality over variables?

Subalgebra of a group algebra

How much damage would a cupful of neutron star matter do to the Earth?

Why does the remaining Rebel fleet at the end of Rogue One seem dramatically larger than the one in A New Hope?

Central Vacuuming: Is it worth it, and how does it compare to normal vacuuming?

If Windows 7 doesn't support WSL, then what does Linux subsystem option mean?

Should there be a hyphen in the construction "IT affin"?

Project Euler #1 in C++

What is the appropriate index architecture when forced to implement IsDeleted (soft deletes)?

How to compare two different files line by line in unix?

How do I find out the mythology and history of my Fortress?

How would a mousetrap for use in space work?

Significance of Cersei's obsession with elephants?

Product of Mrówka space and one point compactification discrete space.

How to make a Field only accept Numbers in Magento 2

Take 2! Is this homebrew Lady of Pain warlock patron balanced?

Would it be possible to dictate a bech32 address as a list of English words?

What is "gratricide"?

Amount of permutations on an NxNxN Rubik's Cube

Did Mueller's report provide an evidentiary basis for the claim of Russian govt election interference via social media?

Importance of からだ in this sentence

Using audio cues to encourage good posture

Put R under double integral

Why do we bend a book to keep it straight?

How could we fake a moon landing now?



A problem from Hardy's_Pure Mathematics on Real variables about Linear equations in 3 variables



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Essay about the art and applications of differential equations?Quotient-lifting propertiesIs there a general algorithm to solve computable integral equation?Is there a codifferential for a covariant exterior derivative?Solvability of the equation $2a_1^2 = a_2^n + a_3^n + a_4^n$ when $n geq 5$ is prime?A monoid where every element has finitely many divisorsWhy does what I've written fail to define truth?existence of solution to set of non-linear equationsHelp me motivate a topic.When does a set of inequalities over series imply an inequality over variables?










2












$begingroup$


It may be my stupidity, yet I am unable to understand what the following problem asks

(Q.1 Miscellaneous Examples, Chapter-1: Real Variables).



It says:




What are the conditions that $ax+by+cz=0$,



  • (1) for all values of $x,y,z$ ;

  • (2) for all values of $x,y,z$ subject to $alpha x+beta y+gamma z=0$;

  • (3) for all values of $x,y,z$ subject to both $alpha x+beta y+gamma z=0$ and $Ax+By+Cz=0$?"



Initially I thought it was about linear diophantine equations in three variables, but then it is already given that the constant term is $0$ which means that it is a solvable diophantine equation (for all the coefficients divide $0$). So the problem is surely not about the solvability of the equation. Then, what it is about? What does the author mean by 'conditions'? Any suggestions are welcome.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    What are $a,b,c,x,y,z$???
    $endgroup$
    – amsmath
    Mar 23 at 5:20






  • 2




    $begingroup$
    But Diophantine equations only involve integers...
    $endgroup$
    – amsmath
    Mar 23 at 6:08






  • 1




    $begingroup$
    Very poorly written question, so not your fault.
    $endgroup$
    – user21820
    Mar 23 at 11:12







  • 1




    $begingroup$
    @user21820 It was perfectly clear and well written following the conventions used when Hardy wrote it. It's only "poorly written" in the sense that "Shakespeare is poorly written, because he couldn't even spell many of the words correctly according to a 21st-century dictionary".
    $endgroup$
    – alephzero
    Mar 23 at 14:17






  • 1




    $begingroup$
    @alephzero: Your claims are unjustified.
    $endgroup$
    – user21820
    Mar 23 at 14:57















2












$begingroup$


It may be my stupidity, yet I am unable to understand what the following problem asks

(Q.1 Miscellaneous Examples, Chapter-1: Real Variables).



It says:




What are the conditions that $ax+by+cz=0$,



  • (1) for all values of $x,y,z$ ;

  • (2) for all values of $x,y,z$ subject to $alpha x+beta y+gamma z=0$;

  • (3) for all values of $x,y,z$ subject to both $alpha x+beta y+gamma z=0$ and $Ax+By+Cz=0$?"



Initially I thought it was about linear diophantine equations in three variables, but then it is already given that the constant term is $0$ which means that it is a solvable diophantine equation (for all the coefficients divide $0$). So the problem is surely not about the solvability of the equation. Then, what it is about? What does the author mean by 'conditions'? Any suggestions are welcome.










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    What are $a,b,c,x,y,z$???
    $endgroup$
    – amsmath
    Mar 23 at 5:20






  • 2




    $begingroup$
    But Diophantine equations only involve integers...
    $endgroup$
    – amsmath
    Mar 23 at 6:08






  • 1




    $begingroup$
    Very poorly written question, so not your fault.
    $endgroup$
    – user21820
    Mar 23 at 11:12







  • 1




    $begingroup$
    @user21820 It was perfectly clear and well written following the conventions used when Hardy wrote it. It's only "poorly written" in the sense that "Shakespeare is poorly written, because he couldn't even spell many of the words correctly according to a 21st-century dictionary".
    $endgroup$
    – alephzero
    Mar 23 at 14:17






  • 1




    $begingroup$
    @alephzero: Your claims are unjustified.
    $endgroup$
    – user21820
    Mar 23 at 14:57













2












2








2





$begingroup$


It may be my stupidity, yet I am unable to understand what the following problem asks

(Q.1 Miscellaneous Examples, Chapter-1: Real Variables).



It says:




What are the conditions that $ax+by+cz=0$,



  • (1) for all values of $x,y,z$ ;

  • (2) for all values of $x,y,z$ subject to $alpha x+beta y+gamma z=0$;

  • (3) for all values of $x,y,z$ subject to both $alpha x+beta y+gamma z=0$ and $Ax+By+Cz=0$?"



Initially I thought it was about linear diophantine equations in three variables, but then it is already given that the constant term is $0$ which means that it is a solvable diophantine equation (for all the coefficients divide $0$). So the problem is surely not about the solvability of the equation. Then, what it is about? What does the author mean by 'conditions'? Any suggestions are welcome.










share|cite|improve this question











$endgroup$




It may be my stupidity, yet I am unable to understand what the following problem asks

(Q.1 Miscellaneous Examples, Chapter-1: Real Variables).



It says:




What are the conditions that $ax+by+cz=0$,



  • (1) for all values of $x,y,z$ ;

  • (2) for all values of $x,y,z$ subject to $alpha x+beta y+gamma z=0$;

  • (3) for all values of $x,y,z$ subject to both $alpha x+beta y+gamma z=0$ and $Ax+By+Cz=0$?"



Initially I thought it was about linear diophantine equations in three variables, but then it is already given that the constant term is $0$ which means that it is a solvable diophantine equation (for all the coefficients divide $0$). So the problem is surely not about the solvability of the equation. Then, what it is about? What does the author mean by 'conditions'? Any suggestions are welcome.







reference-request real-numbers linear-diophantine-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 23 at 6:32









MarianD

2,2761619




2,2761619










asked Mar 23 at 5:00









Awe Kumar JhaAwe Kumar Jha

638113




638113







  • 1




    $begingroup$
    What are $a,b,c,x,y,z$???
    $endgroup$
    – amsmath
    Mar 23 at 5:20






  • 2




    $begingroup$
    But Diophantine equations only involve integers...
    $endgroup$
    – amsmath
    Mar 23 at 6:08






  • 1




    $begingroup$
    Very poorly written question, so not your fault.
    $endgroup$
    – user21820
    Mar 23 at 11:12







  • 1




    $begingroup$
    @user21820 It was perfectly clear and well written following the conventions used when Hardy wrote it. It's only "poorly written" in the sense that "Shakespeare is poorly written, because he couldn't even spell many of the words correctly according to a 21st-century dictionary".
    $endgroup$
    – alephzero
    Mar 23 at 14:17






  • 1




    $begingroup$
    @alephzero: Your claims are unjustified.
    $endgroup$
    – user21820
    Mar 23 at 14:57












  • 1




    $begingroup$
    What are $a,b,c,x,y,z$???
    $endgroup$
    – amsmath
    Mar 23 at 5:20






  • 2




    $begingroup$
    But Diophantine equations only involve integers...
    $endgroup$
    – amsmath
    Mar 23 at 6:08






  • 1




    $begingroup$
    Very poorly written question, so not your fault.
    $endgroup$
    – user21820
    Mar 23 at 11:12







  • 1




    $begingroup$
    @user21820 It was perfectly clear and well written following the conventions used when Hardy wrote it. It's only "poorly written" in the sense that "Shakespeare is poorly written, because he couldn't even spell many of the words correctly according to a 21st-century dictionary".
    $endgroup$
    – alephzero
    Mar 23 at 14:17






  • 1




    $begingroup$
    @alephzero: Your claims are unjustified.
    $endgroup$
    – user21820
    Mar 23 at 14:57







1




1




$begingroup$
What are $a,b,c,x,y,z$???
$endgroup$
– amsmath
Mar 23 at 5:20




$begingroup$
What are $a,b,c,x,y,z$???
$endgroup$
– amsmath
Mar 23 at 5:20




2




2




$begingroup$
But Diophantine equations only involve integers...
$endgroup$
– amsmath
Mar 23 at 6:08




$begingroup$
But Diophantine equations only involve integers...
$endgroup$
– amsmath
Mar 23 at 6:08




1




1




$begingroup$
Very poorly written question, so not your fault.
$endgroup$
– user21820
Mar 23 at 11:12





$begingroup$
Very poorly written question, so not your fault.
$endgroup$
– user21820
Mar 23 at 11:12





1




1




$begingroup$
@user21820 It was perfectly clear and well written following the conventions used when Hardy wrote it. It's only "poorly written" in the sense that "Shakespeare is poorly written, because he couldn't even spell many of the words correctly according to a 21st-century dictionary".
$endgroup$
– alephzero
Mar 23 at 14:17




$begingroup$
@user21820 It was perfectly clear and well written following the conventions used when Hardy wrote it. It's only "poorly written" in the sense that "Shakespeare is poorly written, because he couldn't even spell many of the words correctly according to a 21st-century dictionary".
$endgroup$
– alephzero
Mar 23 at 14:17




1




1




$begingroup$
@alephzero: Your claims are unjustified.
$endgroup$
– user21820
Mar 23 at 14:57




$begingroup$
@alephzero: Your claims are unjustified.
$endgroup$
– user21820
Mar 23 at 14:57










2 Answers
2






active

oldest

votes


















6












$begingroup$

I think Hardy is asking for conditions on $(a,b,c)$, to be given in terms of $(A,B,C)$ and $(alpha,beta,gamma)$. The $(x,y,z)$ are dummy variables, note the "for all".



For (1), which $(a,b,c)$ are orthogonal to all vectors? Answer, clearly only the zero vector, so $(a,b,c)=(0,0,0)$.



For (2), which $(a,b,c)$ are orthogonal to all vectors which are orthogonal to $(alpha,beta,gamma)$? Answer, clearly those which are parallel to $(alpha,beta,gamma)$. So the condition is that for some scalar $lambda$, we must have $(a,b,c)=lambda (alpha, beta, gamma)$.



Case (3) is left to the reader; the condition is clearly that for some scalars $lambda, mu$ we must have that $(a,b,c)=lambda (alpha, beta, gamma)+mu (A,B,C)$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Geometrically, the solution to (1) is the origin. The solution to (2) is the line through the origin and $(alpha, beta, gamma)$. And the solution to (3) is the plane containing the origin, $(alpha, beta, gamma)$ and $(A,B,C)$.
    $endgroup$
    – Paul Sinclair
    Mar 23 at 13:57


















2












$begingroup$

Since I don't have the book available, nor do I recall ever reading it (at least not any recent version of it), I can't be sure what it is getting at as I don't have the context of knowing what was written in Chapter $1$. It should likely help give you an idea.



However, here is what it seems to me. As it's a book about real variables, I doubt the equations are supposed to be diophantine ones, i.e., where the variables can only be integers. Instead, it's to do with the degrees of freedom vs. how many linearly independent constraints you have. With just $1$ equation, you can generally pick any $2$ values and then determine the third, e.g., choose $y$ and $z$ to then get $x$ from



$$x = frac-by-cza tag1labeleq1$$



This is assuming that $a neq 0$.



With $2$ constraint equations, assuming they're linearly independent, you can solve each for $1$ variable to eliminate it, resulting in an equation with $2$ variables. You then have $1$ degree of freedom to choose $1$ of those remaining variables, with the other $2$ then being determined. For example, using your provided $2$ equations:



$$ax + by + cz = 0 tag2labeleq2$$
$$alpha x + beta y + gamma z = 0 tag3labeleq3$$



Multiply eqrefeq2 by $alpha$, multiply eqrefeq3 by $a$, and subtract the $2$ resulting equations to get



$$left(alpha b - abetaright)y + left(alpha c - agammaright)z = 0 ; Rightarrow ; y = left(fracagamma - alpha calpha b - abetaright)z tag4labeleq4$$



This is assuming that $alpha b - abeta neq 0$. Thus, for each $z$, you get a $y$ from eqrefeq4 and then an $x$ from eqrefeq1.



For $3$ linear constraint equations, assuming they're all linearly independent, you get just one set of values.



If any of these equations are linearly dependent, this results in an extra degree of freedom for each equation that is dependent.



Based on your understanding of chapter 1, please indicate if this doesn't match what you might expect.



Update: Based on the comment by ancientmathematician about a geometric argument, note that one equation gives a plane, having $2$ independent equations gives the intersection of $2$ planes, i.e., a line, and $3$ independent equations gives the intersection of $3$ planes, i.e., a point. Perhaps these are the "conditions" being referred to. However, based on how the question is specifically worded, I think a better fit would be what is described in ancientmathematician's answer.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    the 1st chapter is about rational and irrational numbers, irrationality, surds, linear arithmetic continuum, Dedekind theorem, Weierstrass theorem and limit points.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:31










  • $begingroup$
    @AweKumarJha Thanks for the info. Offhand, I don't see any direct connection between the problem and any of those topics, but linear arithmetic continuum seems like the closest one. Do you think what I wrote fits to at least some extent with what the problem is asking for?
    $endgroup$
    – John Omielan
    Mar 23 at 6:35







  • 1




    $begingroup$
    in fact what you wrote is the only thing that presently fits to a remarkable extent to the context, for I don't see other possibilities. Anyway, thanks for your support.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:41






  • 1




    $begingroup$
    In the 1921 3rd edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 31, and in the 1952 10th edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 32. Incidentally, I see nothing in Chapter I to suggest that this has anything to do with Diophantine equations. Indeed, about the only thing in this chapter that I can find even remotely involving Diophantine equations are some irrationality proofs by contradiction.
    $endgroup$
    – Dave L. Renfro
    Mar 23 at 7:16






  • 1




    $begingroup$
    I rather suspect that Hardy (and those for whom he wrote the book) would have given a geometric and not algebraic argument. I don't know why the problem is set here in the book, but of course it's the sort of thing needed when one comes to Lagrange multipliers.
    $endgroup$
    – ancientmathematician
    Mar 23 at 7:57











Your Answer








StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













draft saved

draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158966%2fa-problem-from-hardys-pure-mathematics-on-real-variables-about-linear-equations%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

I think Hardy is asking for conditions on $(a,b,c)$, to be given in terms of $(A,B,C)$ and $(alpha,beta,gamma)$. The $(x,y,z)$ are dummy variables, note the "for all".



For (1), which $(a,b,c)$ are orthogonal to all vectors? Answer, clearly only the zero vector, so $(a,b,c)=(0,0,0)$.



For (2), which $(a,b,c)$ are orthogonal to all vectors which are orthogonal to $(alpha,beta,gamma)$? Answer, clearly those which are parallel to $(alpha,beta,gamma)$. So the condition is that for some scalar $lambda$, we must have $(a,b,c)=lambda (alpha, beta, gamma)$.



Case (3) is left to the reader; the condition is clearly that for some scalars $lambda, mu$ we must have that $(a,b,c)=lambda (alpha, beta, gamma)+mu (A,B,C)$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Geometrically, the solution to (1) is the origin. The solution to (2) is the line through the origin and $(alpha, beta, gamma)$. And the solution to (3) is the plane containing the origin, $(alpha, beta, gamma)$ and $(A,B,C)$.
    $endgroup$
    – Paul Sinclair
    Mar 23 at 13:57















6












$begingroup$

I think Hardy is asking for conditions on $(a,b,c)$, to be given in terms of $(A,B,C)$ and $(alpha,beta,gamma)$. The $(x,y,z)$ are dummy variables, note the "for all".



For (1), which $(a,b,c)$ are orthogonal to all vectors? Answer, clearly only the zero vector, so $(a,b,c)=(0,0,0)$.



For (2), which $(a,b,c)$ are orthogonal to all vectors which are orthogonal to $(alpha,beta,gamma)$? Answer, clearly those which are parallel to $(alpha,beta,gamma)$. So the condition is that for some scalar $lambda$, we must have $(a,b,c)=lambda (alpha, beta, gamma)$.



Case (3) is left to the reader; the condition is clearly that for some scalars $lambda, mu$ we must have that $(a,b,c)=lambda (alpha, beta, gamma)+mu (A,B,C)$.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Geometrically, the solution to (1) is the origin. The solution to (2) is the line through the origin and $(alpha, beta, gamma)$. And the solution to (3) is the plane containing the origin, $(alpha, beta, gamma)$ and $(A,B,C)$.
    $endgroup$
    – Paul Sinclair
    Mar 23 at 13:57













6












6








6





$begingroup$

I think Hardy is asking for conditions on $(a,b,c)$, to be given in terms of $(A,B,C)$ and $(alpha,beta,gamma)$. The $(x,y,z)$ are dummy variables, note the "for all".



For (1), which $(a,b,c)$ are orthogonal to all vectors? Answer, clearly only the zero vector, so $(a,b,c)=(0,0,0)$.



For (2), which $(a,b,c)$ are orthogonal to all vectors which are orthogonal to $(alpha,beta,gamma)$? Answer, clearly those which are parallel to $(alpha,beta,gamma)$. So the condition is that for some scalar $lambda$, we must have $(a,b,c)=lambda (alpha, beta, gamma)$.



Case (3) is left to the reader; the condition is clearly that for some scalars $lambda, mu$ we must have that $(a,b,c)=lambda (alpha, beta, gamma)+mu (A,B,C)$.






share|cite|improve this answer









$endgroup$



I think Hardy is asking for conditions on $(a,b,c)$, to be given in terms of $(A,B,C)$ and $(alpha,beta,gamma)$. The $(x,y,z)$ are dummy variables, note the "for all".



For (1), which $(a,b,c)$ are orthogonal to all vectors? Answer, clearly only the zero vector, so $(a,b,c)=(0,0,0)$.



For (2), which $(a,b,c)$ are orthogonal to all vectors which are orthogonal to $(alpha,beta,gamma)$? Answer, clearly those which are parallel to $(alpha,beta,gamma)$. So the condition is that for some scalar $lambda$, we must have $(a,b,c)=lambda (alpha, beta, gamma)$.



Case (3) is left to the reader; the condition is clearly that for some scalars $lambda, mu$ we must have that $(a,b,c)=lambda (alpha, beta, gamma)+mu (A,B,C)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 23 at 8:07









ancientmathematicianancientmathematician

4,5471513




4,5471513











  • $begingroup$
    Geometrically, the solution to (1) is the origin. The solution to (2) is the line through the origin and $(alpha, beta, gamma)$. And the solution to (3) is the plane containing the origin, $(alpha, beta, gamma)$ and $(A,B,C)$.
    $endgroup$
    – Paul Sinclair
    Mar 23 at 13:57
















  • $begingroup$
    Geometrically, the solution to (1) is the origin. The solution to (2) is the line through the origin and $(alpha, beta, gamma)$. And the solution to (3) is the plane containing the origin, $(alpha, beta, gamma)$ and $(A,B,C)$.
    $endgroup$
    – Paul Sinclair
    Mar 23 at 13:57















$begingroup$
Geometrically, the solution to (1) is the origin. The solution to (2) is the line through the origin and $(alpha, beta, gamma)$. And the solution to (3) is the plane containing the origin, $(alpha, beta, gamma)$ and $(A,B,C)$.
$endgroup$
– Paul Sinclair
Mar 23 at 13:57




$begingroup$
Geometrically, the solution to (1) is the origin. The solution to (2) is the line through the origin and $(alpha, beta, gamma)$. And the solution to (3) is the plane containing the origin, $(alpha, beta, gamma)$ and $(A,B,C)$.
$endgroup$
– Paul Sinclair
Mar 23 at 13:57











2












$begingroup$

Since I don't have the book available, nor do I recall ever reading it (at least not any recent version of it), I can't be sure what it is getting at as I don't have the context of knowing what was written in Chapter $1$. It should likely help give you an idea.



However, here is what it seems to me. As it's a book about real variables, I doubt the equations are supposed to be diophantine ones, i.e., where the variables can only be integers. Instead, it's to do with the degrees of freedom vs. how many linearly independent constraints you have. With just $1$ equation, you can generally pick any $2$ values and then determine the third, e.g., choose $y$ and $z$ to then get $x$ from



$$x = frac-by-cza tag1labeleq1$$



This is assuming that $a neq 0$.



With $2$ constraint equations, assuming they're linearly independent, you can solve each for $1$ variable to eliminate it, resulting in an equation with $2$ variables. You then have $1$ degree of freedom to choose $1$ of those remaining variables, with the other $2$ then being determined. For example, using your provided $2$ equations:



$$ax + by + cz = 0 tag2labeleq2$$
$$alpha x + beta y + gamma z = 0 tag3labeleq3$$



Multiply eqrefeq2 by $alpha$, multiply eqrefeq3 by $a$, and subtract the $2$ resulting equations to get



$$left(alpha b - abetaright)y + left(alpha c - agammaright)z = 0 ; Rightarrow ; y = left(fracagamma - alpha calpha b - abetaright)z tag4labeleq4$$



This is assuming that $alpha b - abeta neq 0$. Thus, for each $z$, you get a $y$ from eqrefeq4 and then an $x$ from eqrefeq1.



For $3$ linear constraint equations, assuming they're all linearly independent, you get just one set of values.



If any of these equations are linearly dependent, this results in an extra degree of freedom for each equation that is dependent.



Based on your understanding of chapter 1, please indicate if this doesn't match what you might expect.



Update: Based on the comment by ancientmathematician about a geometric argument, note that one equation gives a plane, having $2$ independent equations gives the intersection of $2$ planes, i.e., a line, and $3$ independent equations gives the intersection of $3$ planes, i.e., a point. Perhaps these are the "conditions" being referred to. However, based on how the question is specifically worded, I think a better fit would be what is described in ancientmathematician's answer.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    the 1st chapter is about rational and irrational numbers, irrationality, surds, linear arithmetic continuum, Dedekind theorem, Weierstrass theorem and limit points.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:31










  • $begingroup$
    @AweKumarJha Thanks for the info. Offhand, I don't see any direct connection between the problem and any of those topics, but linear arithmetic continuum seems like the closest one. Do you think what I wrote fits to at least some extent with what the problem is asking for?
    $endgroup$
    – John Omielan
    Mar 23 at 6:35







  • 1




    $begingroup$
    in fact what you wrote is the only thing that presently fits to a remarkable extent to the context, for I don't see other possibilities. Anyway, thanks for your support.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:41






  • 1




    $begingroup$
    In the 1921 3rd edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 31, and in the 1952 10th edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 32. Incidentally, I see nothing in Chapter I to suggest that this has anything to do with Diophantine equations. Indeed, about the only thing in this chapter that I can find even remotely involving Diophantine equations are some irrationality proofs by contradiction.
    $endgroup$
    – Dave L. Renfro
    Mar 23 at 7:16






  • 1




    $begingroup$
    I rather suspect that Hardy (and those for whom he wrote the book) would have given a geometric and not algebraic argument. I don't know why the problem is set here in the book, but of course it's the sort of thing needed when one comes to Lagrange multipliers.
    $endgroup$
    – ancientmathematician
    Mar 23 at 7:57















2












$begingroup$

Since I don't have the book available, nor do I recall ever reading it (at least not any recent version of it), I can't be sure what it is getting at as I don't have the context of knowing what was written in Chapter $1$. It should likely help give you an idea.



However, here is what it seems to me. As it's a book about real variables, I doubt the equations are supposed to be diophantine ones, i.e., where the variables can only be integers. Instead, it's to do with the degrees of freedom vs. how many linearly independent constraints you have. With just $1$ equation, you can generally pick any $2$ values and then determine the third, e.g., choose $y$ and $z$ to then get $x$ from



$$x = frac-by-cza tag1labeleq1$$



This is assuming that $a neq 0$.



With $2$ constraint equations, assuming they're linearly independent, you can solve each for $1$ variable to eliminate it, resulting in an equation with $2$ variables. You then have $1$ degree of freedom to choose $1$ of those remaining variables, with the other $2$ then being determined. For example, using your provided $2$ equations:



$$ax + by + cz = 0 tag2labeleq2$$
$$alpha x + beta y + gamma z = 0 tag3labeleq3$$



Multiply eqrefeq2 by $alpha$, multiply eqrefeq3 by $a$, and subtract the $2$ resulting equations to get



$$left(alpha b - abetaright)y + left(alpha c - agammaright)z = 0 ; Rightarrow ; y = left(fracagamma - alpha calpha b - abetaright)z tag4labeleq4$$



This is assuming that $alpha b - abeta neq 0$. Thus, for each $z$, you get a $y$ from eqrefeq4 and then an $x$ from eqrefeq1.



For $3$ linear constraint equations, assuming they're all linearly independent, you get just one set of values.



If any of these equations are linearly dependent, this results in an extra degree of freedom for each equation that is dependent.



Based on your understanding of chapter 1, please indicate if this doesn't match what you might expect.



Update: Based on the comment by ancientmathematician about a geometric argument, note that one equation gives a plane, having $2$ independent equations gives the intersection of $2$ planes, i.e., a line, and $3$ independent equations gives the intersection of $3$ planes, i.e., a point. Perhaps these are the "conditions" being referred to. However, based on how the question is specifically worded, I think a better fit would be what is described in ancientmathematician's answer.






share|cite|improve this answer











$endgroup$








  • 1




    $begingroup$
    the 1st chapter is about rational and irrational numbers, irrationality, surds, linear arithmetic continuum, Dedekind theorem, Weierstrass theorem and limit points.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:31










  • $begingroup$
    @AweKumarJha Thanks for the info. Offhand, I don't see any direct connection between the problem and any of those topics, but linear arithmetic continuum seems like the closest one. Do you think what I wrote fits to at least some extent with what the problem is asking for?
    $endgroup$
    – John Omielan
    Mar 23 at 6:35







  • 1




    $begingroup$
    in fact what you wrote is the only thing that presently fits to a remarkable extent to the context, for I don't see other possibilities. Anyway, thanks for your support.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:41






  • 1




    $begingroup$
    In the 1921 3rd edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 31, and in the 1952 10th edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 32. Incidentally, I see nothing in Chapter I to suggest that this has anything to do with Diophantine equations. Indeed, about the only thing in this chapter that I can find even remotely involving Diophantine equations are some irrationality proofs by contradiction.
    $endgroup$
    – Dave L. Renfro
    Mar 23 at 7:16






  • 1




    $begingroup$
    I rather suspect that Hardy (and those for whom he wrote the book) would have given a geometric and not algebraic argument. I don't know why the problem is set here in the book, but of course it's the sort of thing needed when one comes to Lagrange multipliers.
    $endgroup$
    – ancientmathematician
    Mar 23 at 7:57













2












2








2





$begingroup$

Since I don't have the book available, nor do I recall ever reading it (at least not any recent version of it), I can't be sure what it is getting at as I don't have the context of knowing what was written in Chapter $1$. It should likely help give you an idea.



However, here is what it seems to me. As it's a book about real variables, I doubt the equations are supposed to be diophantine ones, i.e., where the variables can only be integers. Instead, it's to do with the degrees of freedom vs. how many linearly independent constraints you have. With just $1$ equation, you can generally pick any $2$ values and then determine the third, e.g., choose $y$ and $z$ to then get $x$ from



$$x = frac-by-cza tag1labeleq1$$



This is assuming that $a neq 0$.



With $2$ constraint equations, assuming they're linearly independent, you can solve each for $1$ variable to eliminate it, resulting in an equation with $2$ variables. You then have $1$ degree of freedom to choose $1$ of those remaining variables, with the other $2$ then being determined. For example, using your provided $2$ equations:



$$ax + by + cz = 0 tag2labeleq2$$
$$alpha x + beta y + gamma z = 0 tag3labeleq3$$



Multiply eqrefeq2 by $alpha$, multiply eqrefeq3 by $a$, and subtract the $2$ resulting equations to get



$$left(alpha b - abetaright)y + left(alpha c - agammaright)z = 0 ; Rightarrow ; y = left(fracagamma - alpha calpha b - abetaright)z tag4labeleq4$$



This is assuming that $alpha b - abeta neq 0$. Thus, for each $z$, you get a $y$ from eqrefeq4 and then an $x$ from eqrefeq1.



For $3$ linear constraint equations, assuming they're all linearly independent, you get just one set of values.



If any of these equations are linearly dependent, this results in an extra degree of freedom for each equation that is dependent.



Based on your understanding of chapter 1, please indicate if this doesn't match what you might expect.



Update: Based on the comment by ancientmathematician about a geometric argument, note that one equation gives a plane, having $2$ independent equations gives the intersection of $2$ planes, i.e., a line, and $3$ independent equations gives the intersection of $3$ planes, i.e., a point. Perhaps these are the "conditions" being referred to. However, based on how the question is specifically worded, I think a better fit would be what is described in ancientmathematician's answer.






share|cite|improve this answer











$endgroup$



Since I don't have the book available, nor do I recall ever reading it (at least not any recent version of it), I can't be sure what it is getting at as I don't have the context of knowing what was written in Chapter $1$. It should likely help give you an idea.



However, here is what it seems to me. As it's a book about real variables, I doubt the equations are supposed to be diophantine ones, i.e., where the variables can only be integers. Instead, it's to do with the degrees of freedom vs. how many linearly independent constraints you have. With just $1$ equation, you can generally pick any $2$ values and then determine the third, e.g., choose $y$ and $z$ to then get $x$ from



$$x = frac-by-cza tag1labeleq1$$



This is assuming that $a neq 0$.



With $2$ constraint equations, assuming they're linearly independent, you can solve each for $1$ variable to eliminate it, resulting in an equation with $2$ variables. You then have $1$ degree of freedom to choose $1$ of those remaining variables, with the other $2$ then being determined. For example, using your provided $2$ equations:



$$ax + by + cz = 0 tag2labeleq2$$
$$alpha x + beta y + gamma z = 0 tag3labeleq3$$



Multiply eqrefeq2 by $alpha$, multiply eqrefeq3 by $a$, and subtract the $2$ resulting equations to get



$$left(alpha b - abetaright)y + left(alpha c - agammaright)z = 0 ; Rightarrow ; y = left(fracagamma - alpha calpha b - abetaright)z tag4labeleq4$$



This is assuming that $alpha b - abeta neq 0$. Thus, for each $z$, you get a $y$ from eqrefeq4 and then an $x$ from eqrefeq1.



For $3$ linear constraint equations, assuming they're all linearly independent, you get just one set of values.



If any of these equations are linearly dependent, this results in an extra degree of freedom for each equation that is dependent.



Based on your understanding of chapter 1, please indicate if this doesn't match what you might expect.



Update: Based on the comment by ancientmathematician about a geometric argument, note that one equation gives a plane, having $2$ independent equations gives the intersection of $2$ planes, i.e., a line, and $3$ independent equations gives the intersection of $3$ planes, i.e., a point. Perhaps these are the "conditions" being referred to. However, based on how the question is specifically worded, I think a better fit would be what is described in ancientmathematician's answer.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 23 at 8:41

























answered Mar 23 at 6:09









John OmielanJohn Omielan

5,1992218




5,1992218







  • 1




    $begingroup$
    the 1st chapter is about rational and irrational numbers, irrationality, surds, linear arithmetic continuum, Dedekind theorem, Weierstrass theorem and limit points.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:31










  • $begingroup$
    @AweKumarJha Thanks for the info. Offhand, I don't see any direct connection between the problem and any of those topics, but linear arithmetic continuum seems like the closest one. Do you think what I wrote fits to at least some extent with what the problem is asking for?
    $endgroup$
    – John Omielan
    Mar 23 at 6:35







  • 1




    $begingroup$
    in fact what you wrote is the only thing that presently fits to a remarkable extent to the context, for I don't see other possibilities. Anyway, thanks for your support.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:41






  • 1




    $begingroup$
    In the 1921 3rd edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 31, and in the 1952 10th edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 32. Incidentally, I see nothing in Chapter I to suggest that this has anything to do with Diophantine equations. Indeed, about the only thing in this chapter that I can find even remotely involving Diophantine equations are some irrationality proofs by contradiction.
    $endgroup$
    – Dave L. Renfro
    Mar 23 at 7:16






  • 1




    $begingroup$
    I rather suspect that Hardy (and those for whom he wrote the book) would have given a geometric and not algebraic argument. I don't know why the problem is set here in the book, but of course it's the sort of thing needed when one comes to Lagrange multipliers.
    $endgroup$
    – ancientmathematician
    Mar 23 at 7:57












  • 1




    $begingroup$
    the 1st chapter is about rational and irrational numbers, irrationality, surds, linear arithmetic continuum, Dedekind theorem, Weierstrass theorem and limit points.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:31










  • $begingroup$
    @AweKumarJha Thanks for the info. Offhand, I don't see any direct connection between the problem and any of those topics, but linear arithmetic continuum seems like the closest one. Do you think what I wrote fits to at least some extent with what the problem is asking for?
    $endgroup$
    – John Omielan
    Mar 23 at 6:35







  • 1




    $begingroup$
    in fact what you wrote is the only thing that presently fits to a remarkable extent to the context, for I don't see other possibilities. Anyway, thanks for your support.
    $endgroup$
    – Awe Kumar Jha
    Mar 23 at 6:41






  • 1




    $begingroup$
    In the 1921 3rd edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 31, and in the 1952 10th edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 32. Incidentally, I see nothing in Chapter I to suggest that this has anything to do with Diophantine equations. Indeed, about the only thing in this chapter that I can find even remotely involving Diophantine equations are some irrationality proofs by contradiction.
    $endgroup$
    – Dave L. Renfro
    Mar 23 at 7:16






  • 1




    $begingroup$
    I rather suspect that Hardy (and those for whom he wrote the book) would have given a geometric and not algebraic argument. I don't know why the problem is set here in the book, but of course it's the sort of thing needed when one comes to Lagrange multipliers.
    $endgroup$
    – ancientmathematician
    Mar 23 at 7:57







1




1




$begingroup$
the 1st chapter is about rational and irrational numbers, irrationality, surds, linear arithmetic continuum, Dedekind theorem, Weierstrass theorem and limit points.
$endgroup$
– Awe Kumar Jha
Mar 23 at 6:31




$begingroup$
the 1st chapter is about rational and irrational numbers, irrationality, surds, linear arithmetic continuum, Dedekind theorem, Weierstrass theorem and limit points.
$endgroup$
– Awe Kumar Jha
Mar 23 at 6:31












$begingroup$
@AweKumarJha Thanks for the info. Offhand, I don't see any direct connection between the problem and any of those topics, but linear arithmetic continuum seems like the closest one. Do you think what I wrote fits to at least some extent with what the problem is asking for?
$endgroup$
– John Omielan
Mar 23 at 6:35





$begingroup$
@AweKumarJha Thanks for the info. Offhand, I don't see any direct connection between the problem and any of those topics, but linear arithmetic continuum seems like the closest one. Do you think what I wrote fits to at least some extent with what the problem is asking for?
$endgroup$
– John Omielan
Mar 23 at 6:35





1




1




$begingroup$
in fact what you wrote is the only thing that presently fits to a remarkable extent to the context, for I don't see other possibilities. Anyway, thanks for your support.
$endgroup$
– Awe Kumar Jha
Mar 23 at 6:41




$begingroup$
in fact what you wrote is the only thing that presently fits to a remarkable extent to the context, for I don't see other possibilities. Anyway, thanks for your support.
$endgroup$
– Awe Kumar Jha
Mar 23 at 6:41




1




1




$begingroup$
In the 1921 3rd edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 31, and in the 1952 10th edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 32. Incidentally, I see nothing in Chapter I to suggest that this has anything to do with Diophantine equations. Indeed, about the only thing in this chapter that I can find even remotely involving Diophantine equations are some irrationality proofs by contradiction.
$endgroup$
– Dave L. Renfro
Mar 23 at 7:16




$begingroup$
In the 1921 3rd edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 31, and in the 1952 10th edition it's the first of the Miscellaneous Examples on Chapter I and is on p. 32. Incidentally, I see nothing in Chapter I to suggest that this has anything to do with Diophantine equations. Indeed, about the only thing in this chapter that I can find even remotely involving Diophantine equations are some irrationality proofs by contradiction.
$endgroup$
– Dave L. Renfro
Mar 23 at 7:16




1




1




$begingroup$
I rather suspect that Hardy (and those for whom he wrote the book) would have given a geometric and not algebraic argument. I don't know why the problem is set here in the book, but of course it's the sort of thing needed when one comes to Lagrange multipliers.
$endgroup$
– ancientmathematician
Mar 23 at 7:57




$begingroup$
I rather suspect that Hardy (and those for whom he wrote the book) would have given a geometric and not algebraic argument. I don't know why the problem is set here in the book, but of course it's the sort of thing needed when one comes to Lagrange multipliers.
$endgroup$
– ancientmathematician
Mar 23 at 7:57

















draft saved

draft discarded
















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid


  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.

Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158966%2fa-problem-from-hardys-pure-mathematics-on-real-variables-about-linear-equations%23new-answer', 'question_page');

);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

Slayer Innehåll Historia | Stil, komposition och lyrik | Bandets betydelse och framgångar | Sidoprojekt och samarbeten | Kontroverser | Medlemmar | Utmärkelser och nomineringar | Turnéer och festivaler | Diskografi | Referenser | Externa länkar | Navigeringsmenywww.slayer.net”Metal Massacre vol. 1””Metal Massacre vol. 3””Metal Massacre Volume III””Show No Mercy””Haunting the Chapel””Live Undead””Hell Awaits””Reign in Blood””Reign in Blood””Gold & Platinum – Reign in Blood””Golden Gods Awards Winners”originalet”Kerrang! Hall Of Fame””Slayer Looks Back On 37-Year Career In New Video Series: Part Two””South of Heaven””Gold & Platinum – South of Heaven””Seasons in the Abyss””Gold & Platinum - Seasons in the Abyss””Divine Intervention””Divine Intervention - Release group by Slayer””Gold & Platinum - Divine Intervention””Live Intrusion””Undisputed Attitude””Abolish Government/Superficial Love””Release “Slatanic Slaughter: A Tribute to Slayer” by Various Artists””Diabolus in Musica””Soundtrack to the Apocalypse””God Hates Us All””Systematic - Relationships””War at the Warfield””Gold & Platinum - War at the Warfield””Soundtrack to the Apocalypse””Gold & Platinum - Still Reigning””Metallica, Slayer, Iron Mauden Among Winners At Metal Hammer Awards””Eternal Pyre””Eternal Pyre - Slayer release group””Eternal Pyre””Metal Storm Awards 2006””Kerrang! Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029