Mixing
Proving $sqrt{2013^{2016}+2014^{2016}}$ is irrational
$begingroup$
Prove $sqrt{2013^{2016}+2014^{2016}}$ is irrational.
I've looked at the proofs for proving $sqrt{mathstrut 2}$, $sqrt{mathstrut 3}$, and$sqrt{mathstrut 15}$ irrational. Using proof by contradiction, assuming this number is rational and in the form of $frac{m}{n}$ where m&n are relatively prime. I understand that for the $sqrt{mathstrut 2}$ proof you essentially get that they're both even and thus not relatively prime, and $sqrt{mathstrut 3}$ proof involves rewriting m and n as 2a+1 and 2b+1 and getting LHS odd and RHS even. For the $sqrt{mathstrut 15}$ you get the factors of 3 and 5 being odd on one side and even on the other.
I'm having difficulty using any of these techniques to prove the problem is irrational. Is there a way I'm supposed to rewrite 2013 and 2014 in order to simplify this problem into one of the above mention proofs?
Any thoughts or suggestions would be extremely helpful. Thank you!
proof-explanation irrational-numbers
asked Jan 17 at 11:56
Abe CainAbe Cain
133
$endgroup$
add a comment |
$begingroup$
Prove $sqrt{2013^{2016}+2014^{2016}}$ is irrational.
I've looked at the proofs for proving $sqrt{mathstrut 2}$, $sqrt{mathstrut 3}$, and$sqrt{mathstrut 15}$ irrational. Using proof by contradiction, assuming this number is rational and in the form of $frac{m}{n}$ where m&n are relatively prime. I understand that for the $sqrt{mathstrut 2}$ proof you essentially get that they're both even and thus not relatively prime, and $sqrt{mathstrut 3}$ proof involves rewriting m and n as 2a+1 and 2b+1 and getting LHS odd and RHS even. For the $sqrt{mathstrut 15}$ you get the factors of 3 and 5 being odd on one side and even on the other.
I'm having difficulty using any of these techniques to prove the problem is irrational. Is there a way I'm supposed to rewrite 2013 and 2014 in order to simplify this problem into one of the above mention proofs?
Any thoughts or suggestions would be extremely helpful. Thank you!
proof-explanation irrational-numbers
asked Jan 17 at 11:56
Abe CainAbe Cain
133
$endgroup$
add a comment |
$begingroup$
Prove $sqrt{2013^{2016}+2014^{2016}}$ is irrational.
I've looked at the proofs for proving $sqrt{mathstrut 2}$, $sqrt{mathstrut 3}$, and$sqrt{mathstrut 15}$ irrational. Using proof by contradiction, assuming this number is rational and in the form of $frac{m}{n}$ where m&n are relatively prime. I understand that for the $sqrt{mathstrut 2}$ proof you essentially get that they're both even and thus not relatively prime, and $sqrt{mathstrut 3}$ proof involves rewriting m and n as 2a+1 and 2b+1 and getting LHS odd and RHS even. For the $sqrt{mathstrut 15}$ you get the factors of 3 and 5 being odd on one side and even on the other.
I'm having difficulty using any of these techniques to prove the problem is irrational. Is there a way I'm supposed to rewrite 2013 and 2014 in order to simplify this problem into one of the above mention proofs?
Any thoughts or suggestions would be extremely helpful. Thank you!
proof-explanation irrational-numbers
asked Jan 17 at 11:56
Abe CainAbe Cain
133
$endgroup$
Prove $sqrt{2013^{2016}+2014^{2016}}$ is irrational.
I've looked at the proofs for proving $sqrt{mathstrut 2}$, $sqrt{mathstrut 3}$, and$sqrt{mathstrut 15}$ irrational. Using proof by contradiction, assuming this number is rational and in the form of $frac{m}{n}$ where m&n are relatively prime. I understand that for the $sqrt{mathstrut 2}$ proof you essentially get that they're both even and thus not relatively prime, and $sqrt{mathstrut 3}$ proof involves rewriting m and n as 2a+1 and 2b+1 and getting LHS odd and RHS even. For the $sqrt{mathstrut 15}$ you get the factors of 3 and 5 being odd on one side and even on the other.
I'm having difficulty using any of these techniques to prove the problem is irrational. Is there a way I'm supposed to rewrite 2013 and 2014 in order to simplify this problem into one of the above mention proofs?
Any thoughts or suggestions would be extremely helpful. Thank you!
proof-explanation irrational-numbers
proof-explanation irrational-numbers
asked Jan 17 at 11:56
Abe CainAbe Cain
133
asked Jan 17 at 11:56
Abe CainAbe Cain
133
asked Jan 17 at 11:56
Abe CainAbe Cain
133
asked Jan 17 at 11:56
Abe CainAbe Cain
133
asked Jan 17 at 11:56
Abe CainAbe Cain
133
133
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
What are the possible last digits in base 10 for square numbers?
What is the last digit of the radicand? (bit inside the root)
Let me know if you need further help.
answered Jan 17 at 12:00
Ben CrossleyBen Crossley
904418
$endgroup$
$begingroup$
What about rest from division by $5$ instead? May be less troublesome
$endgroup$
– Jakobian
Jan 17 at 12:02
$begingroup$
Can this number be divided by 5?
$endgroup$
– Ben Crossley
Jan 17 at 12:04
$begingroup$
Instead of division by $10$, let's do division by $5$. Then finding possible rests of squares will be easier
$endgroup$
– Jakobian
Jan 17 at 12:04
$begingroup$
Ahh, I see. Yes that would work, seems sensible to stay in the base it's written in to me :)
$endgroup$
– Ben Crossley
Jan 17 at 12:05
2
$begingroup$
Edit - Thank you for the start - so only 1, 4, 9, 6 and 5. And since you would get last digits of 1 and 6 for 2013 and 2014, your last digit would be 7 in the radicand, and since 7 is not one of the 5 numbers above, it cannot be a perfect square. Thanks!
$endgroup$
– Abe Cain
Jan 17 at 12:13
|
show 1 more comment
$begingroup$
$$ 2013^{2016}+2014^{2016}equiv 2pmod{5} $$
(by Fermat's little theorem) implies that $ 2013^{2016}+2014^{2016}$ is not a square, hence its square root is irrational.
answered Jan 17 at 12:10
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
Or, for a different approach, a sum of two perfect fourth powers is never a perfect square unless they're $0$ and $1$. Reference - this is a theorem that Fermat actually proved.
And, of course, the square root of any positive integer that isn't a perfect square is irrational.
answered Jan 17 at 12:13
jmerryjmerry
10.2k1225
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076889%2fproving-sqrt2013201620142016-is-irrational%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What are the possible last digits in base 10 for square numbers?
What is the last digit of the radicand? (bit inside the root)
Let me know if you need further help.
answered Jan 17 at 12:00
Ben CrossleyBen Crossley
904418
$endgroup$
$begingroup$
What about rest from division by $5$ instead? May be less troublesome
$endgroup$
– Jakobian
Jan 17 at 12:02
$begingroup$
Can this number be divided by 5?
$endgroup$
– Ben Crossley
Jan 17 at 12:04
$begingroup$
Instead of division by $10$, let's do division by $5$. Then finding possible rests of squares will be easier
$endgroup$
– Jakobian
Jan 17 at 12:04
$begingroup$
Ahh, I see. Yes that would work, seems sensible to stay in the base it's written in to me :)
$endgroup$
– Ben Crossley
Jan 17 at 12:05
2
$begingroup$
Edit - Thank you for the start - so only 1, 4, 9, 6 and 5. And since you would get last digits of 1 and 6 for 2013 and 2014, your last digit would be 7 in the radicand, and since 7 is not one of the 5 numbers above, it cannot be a perfect square. Thanks!
$endgroup$
– Abe Cain
Jan 17 at 12:13
|
show 1 more comment
$begingroup$
What are the possible last digits in base 10 for square numbers?
What is the last digit of the radicand? (bit inside the root)
Let me know if you need further help.
answered Jan 17 at 12:00
Ben CrossleyBen Crossley
904418
$endgroup$
$begingroup$
What about rest from division by $5$ instead? May be less troublesome
$endgroup$
– Jakobian
Jan 17 at 12:02
$begingroup$
Can this number be divided by 5?
$endgroup$
– Ben Crossley
Jan 17 at 12:04
$begingroup$
Instead of division by $10$, let's do division by $5$. Then finding possible rests of squares will be easier
$endgroup$
– Jakobian
Jan 17 at 12:04
$begingroup$
Ahh, I see. Yes that would work, seems sensible to stay in the base it's written in to me :)
$endgroup$
– Ben Crossley
Jan 17 at 12:05
2
$begingroup$
Edit - Thank you for the start - so only 1, 4, 9, 6 and 5. And since you would get last digits of 1 and 6 for 2013 and 2014, your last digit would be 7 in the radicand, and since 7 is not one of the 5 numbers above, it cannot be a perfect square. Thanks!
$endgroup$
– Abe Cain
Jan 17 at 12:13
|
show 1 more comment
$begingroup$
What are the possible last digits in base 10 for square numbers?
What is the last digit of the radicand? (bit inside the root)
Let me know if you need further help.
answered Jan 17 at 12:00
Ben CrossleyBen Crossley
904418
$endgroup$
What are the possible last digits in base 10 for square numbers?
What is the last digit of the radicand? (bit inside the root)
Let me know if you need further help.
answered Jan 17 at 12:00
Ben CrossleyBen Crossley
904418
answered Jan 17 at 12:00
Ben CrossleyBen Crossley
904418
answered Jan 17 at 12:00
Ben CrossleyBen Crossley
904418
answered Jan 17 at 12:00
Ben CrossleyBen Crossley
904418
904418
$begingroup$
What about rest from division by $5$ instead? May be less troublesome
$endgroup$
– Jakobian
Jan 17 at 12:02
$begingroup$
Can this number be divided by 5?
$endgroup$
– Ben Crossley
Jan 17 at 12:04
$begingroup$
Instead of division by $10$, let's do division by $5$. Then finding possible rests of squares will be easier
$endgroup$
– Jakobian
Jan 17 at 12:04
$begingroup$
Ahh, I see. Yes that would work, seems sensible to stay in the base it's written in to me :)
$endgroup$
– Ben Crossley
Jan 17 at 12:05
2
$begingroup$
Edit - Thank you for the start - so only 1, 4, 9, 6 and 5. And since you would get last digits of 1 and 6 for 2013 and 2014, your last digit would be 7 in the radicand, and since 7 is not one of the 5 numbers above, it cannot be a perfect square. Thanks!
$endgroup$
– Abe Cain
Jan 17 at 12:13
|
show 1 more comment
$begingroup$
What about rest from division by $5$ instead? May be less troublesome
$endgroup$
– Jakobian
Jan 17 at 12:02
$begingroup$
Can this number be divided by 5?
$endgroup$
– Ben Crossley
Jan 17 at 12:04
$begingroup$
Instead of division by $10$, let's do division by $5$. Then finding possible rests of squares will be easier
$endgroup$
– Jakobian
Jan 17 at 12:04
$begingroup$
Ahh, I see. Yes that would work, seems sensible to stay in the base it's written in to me :)
$endgroup$
– Ben Crossley
Jan 17 at 12:05
2
$begingroup$
Edit - Thank you for the start - so only 1, 4, 9, 6 and 5. And since you would get last digits of 1 and 6 for 2013 and 2014, your last digit would be 7 in the radicand, and since 7 is not one of the 5 numbers above, it cannot be a perfect square. Thanks!
$endgroup$
– Abe Cain
Jan 17 at 12:13
$begingroup$
What about rest from division by $5$ instead? May be less troublesome
$endgroup$
– Jakobian
Jan 17 at 12:02
$begingroup$
What about rest from division by $5$ instead? May be less troublesome
$endgroup$
– Jakobian
Jan 17 at 12:02
$begingroup$
Can this number be divided by 5?
$endgroup$
– Ben Crossley
Jan 17 at 12:04
$begingroup$
Can this number be divided by 5?
$endgroup$
– Ben Crossley
Jan 17 at 12:04
$begingroup$
Instead of division by $10$, let's do division by $5$. Then finding possible rests of squares will be easier
$endgroup$
– Jakobian
Jan 17 at 12:04
$begingroup$
Instead of division by $10$, let's do division by $5$. Then finding possible rests of squares will be easier
$endgroup$
– Jakobian
Jan 17 at 12:04
$begingroup$
Ahh, I see. Yes that would work, seems sensible to stay in the base it's written in to me :)
$endgroup$
– Ben Crossley
Jan 17 at 12:05
$begingroup$
Ahh, I see. Yes that would work, seems sensible to stay in the base it's written in to me :)
$endgroup$
– Ben Crossley
Jan 17 at 12:05
2
2
$begingroup$
Edit - Thank you for the start - so only 1, 4, 9, 6 and 5. And since you would get last digits of 1 and 6 for 2013 and 2014, your last digit would be 7 in the radicand, and since 7 is not one of the 5 numbers above, it cannot be a perfect square. Thanks!
$endgroup$
– Abe Cain
Jan 17 at 12:13
$begingroup$
Edit - Thank you for the start - so only 1, 4, 9, 6 and 5. And since you would get last digits of 1 and 6 for 2013 and 2014, your last digit would be 7 in the radicand, and since 7 is not one of the 5 numbers above, it cannot be a perfect square. Thanks!
$endgroup$
– Abe Cain
Jan 17 at 12:13
|
show 1 more comment
$begingroup$
$$ 2013^{2016}+2014^{2016}equiv 2pmod{5} $$
(by Fermat's little theorem) implies that $ 2013^{2016}+2014^{2016}$ is not a square, hence its square root is irrational.
answered Jan 17 at 12:10
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
$$ 2013^{2016}+2014^{2016}equiv 2pmod{5} $$
(by Fermat's little theorem) implies that $ 2013^{2016}+2014^{2016}$ is not a square, hence its square root is irrational.
answered Jan 17 at 12:10
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
add a comment |
$begingroup$
$$ 2013^{2016}+2014^{2016}equiv 2pmod{5} $$
(by Fermat's little theorem) implies that $ 2013^{2016}+2014^{2016}$ is not a square, hence its square root is irrational.
answered Jan 17 at 12:10
Jack D'AurizioJack D'Aurizio
290k33282664
$endgroup$
$$ 2013^{2016}+2014^{2016}equiv 2pmod{5} $$
(by Fermat's little theorem) implies that $ 2013^{2016}+2014^{2016}$ is not a square, hence its square root is irrational.
answered Jan 17 at 12:10
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 17 at 12:10
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 17 at 12:10
Jack D'AurizioJack D'Aurizio
290k33282664
answered Jan 17 at 12:10
Jack D'AurizioJack D'Aurizio
290k33282664
290k33282664
add a comment |
add a comment |
$begingroup$
Or, for a different approach, a sum of two perfect fourth powers is never a perfect square unless they're $0$ and $1$. Reference - this is a theorem that Fermat actually proved.
And, of course, the square root of any positive integer that isn't a perfect square is irrational.
answered Jan 17 at 12:13
jmerryjmerry
10.2k1225
$endgroup$
add a comment |
$begingroup$
Or, for a different approach, a sum of two perfect fourth powers is never a perfect square unless they're $0$ and $1$. Reference - this is a theorem that Fermat actually proved.
And, of course, the square root of any positive integer that isn't a perfect square is irrational.
answered Jan 17 at 12:13
jmerryjmerry
10.2k1225
$endgroup$
add a comment |
$begingroup$
Or, for a different approach, a sum of two perfect fourth powers is never a perfect square unless they're $0$ and $1$. Reference - this is a theorem that Fermat actually proved.
And, of course, the square root of any positive integer that isn't a perfect square is irrational.
answered Jan 17 at 12:13
jmerryjmerry
10.2k1225
$endgroup$
Or, for a different approach, a sum of two perfect fourth powers is never a perfect square unless they're $0$ and $1$. Reference - this is a theorem that Fermat actually proved.
And, of course, the square root of any positive integer that isn't a perfect square is irrational.
answered Jan 17 at 12:13
jmerryjmerry
10.2k1225
answered Jan 17 at 12:13
jmerryjmerry
10.2k1225
answered Jan 17 at 12:13
jmerryjmerry
10.2k1225
answered Jan 17 at 12:13
jmerryjmerry
10.2k1225
10.2k1225
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3076889%2fproving-sqrt2013201620142016-is-irrational%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
