Ttyjukl

Dilemma of explaining to interviewer that he is the reason for declining second interview

What are these green text/line displays shown during the livestream of Crew Dragon's approach to dock with the ISS?

Do any poskim exempt 13-20-year-olds from Mussaf?

Meaning of すきっとした

Crystal compensation for temp and voltage

Avoiding morning and evening handshakes

What's the purpose of these copper coils with resitors inside them in A Yamaha RX-V396RDS amplifier?

Metadata API deployments are failing in Spring '19

Has the Isbell–Freyd criterion ever been used to check that a category is concretisable?

Finding ratio of the area of triangles

Do commercial flights continue with an engine out?

How to properly claim credit for peer review?

Why do neural networks need so many training examples to perform?

If all harmonics are generated by plucking, how does a guitar string produce a pure frequency sound?

What's the rationale behind the objections to these measures against human trafficking?

ip vs ifconfig commands pros and cons

Why is commutativity optional in multiplication for rings?

raspberry pi change directory (cd) command not working with USB drive

Using AWS Fargate as web server

What is better: yes / no radio, or simple checkbox?

How can I make an algorithm in C++ for finding variations of a set without repetition (i.e. n elements, choose k)?

What is the purpose of easy combat scenarios that don't need resource expenditure?

What does a violin mute do?

How do I add a variable to this curl command?

Proof by Induction - New to proofs

Dominoes and induction, or how does induction work?Proving that $frac{phi^{400}+1}{phi^{200}}$ is an integer.Generating induction proofs from graphs/integralsMathematical induction proof; $g_k=3g_{k-1} - 2g_{k-2}$induction proof for kleene starTips on constructing a proof by induction.Induction proofs for subsets of integersproof using the mathematical inductionProof by Induction involving divisibilityInduction Proof:Inequality proof by induction(?)Proof by induction?

2

$begingroup$

totally new to proofs and found this challenge problem and struggling a bit. Any help would be appreciated!

There are some real numbers $x$ such that $x+frac{1}{x}$ is an
integer. For example, $2+sqrt{3}+frac{1}{2+sqrt{3}}=4$,
$1+frac{1}{1}=2$, and $2sqrt{6}-5+frac{1}{2sqrt{6}-5}=-10$.

Prove for all $xinmathbb{R}$ that if $x+frac{1}{x}$ is an integer,
then $x^n +frac{1}{x^n}$ also is an integer for all $ninmathbb{N}$.

induction

share|cite|improve this question

asked 3 hours ago

RobinRobin

283

New contributor

Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I recommend this answer as a good start
  $endgroup$
  – Ross Millikan
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/936479/…
  $endgroup$
  – lab bhattacharjee
  3 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

totally new to proofs and found this challenge problem and struggling a bit. Any help would be appreciated!

There are some real numbers $x$ such that $x+frac{1}{x}$ is an
integer. For example, $2+sqrt{3}+frac{1}{2+sqrt{3}}=4$,
$1+frac{1}{1}=2$, and $2sqrt{6}-5+frac{1}{2sqrt{6}-5}=-10$.

Prove for all $xinmathbb{R}$ that if $x+frac{1}{x}$ is an integer,
then $x^n +frac{1}{x^n}$ also is an integer for all $ninmathbb{N}$.

induction

share|cite|improve this question

asked 3 hours ago

RobinRobin

283

New contributor

Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I recommend this answer as a good start
  $endgroup$
  – Ross Millikan
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  math.stackexchange.com/questions/936479/…
  $endgroup$
  – lab bhattacharjee
  3 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

totally new to proofs and found this challenge problem and struggling a bit. Any help would be appreciated!

There are some real numbers $x$ such that $x+frac{1}{x}$ is an
integer. For example, $2+sqrt{3}+frac{1}{2+sqrt{3}}=4$,
$1+frac{1}{1}=2$, and $2sqrt{6}-5+frac{1}{2sqrt{6}-5}=-10$.

Prove for all $xinmathbb{R}$ that if $x+frac{1}{x}$ is an integer,
then $x^n +frac{1}{x^n}$ also is an integer for all $ninmathbb{N}$.

induction

share|cite|improve this question

asked 3 hours ago

RobinRobin

283

New contributor

Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked 3 hours ago

RobinRobin

283

New contributor

Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked 3 hours ago

RobinRobin

283

New contributor

Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

RobinRobin

283

New contributor

Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

RobinRobin

283

1 Answer
1

active

oldest

votes

9

$begingroup$

HINT: Note that for $ngeq1$ you have
$$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$

For more details, hover over the the block below:

The equation above can be rewritten to get
$$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

26.5k33997

$endgroup$

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
});


}
});

Robin is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3134247%2fproof-by-induction-new-to-proofs%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

9

$begingroup$

HINT: Note that for $ngeq1$ you have
$$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$

For more details, hover over the the block below:

The equation above can be rewritten to get
$$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

26.5k33997

$endgroup$

add a comment | 

9

$begingroup$

HINT: Note that for $ngeq1$ you have
$$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$

For more details, hover over the the block below:

The equation above can be rewritten to get
$$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

26.5k33997

$endgroup$

add a comment | 

$begingroup$

HINT: Note that for $ngeq1$ you have
$$left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)=left(x^{n+1}+frac{1}{x^{n+1}}right)+left(x^{n-1}+frac{1}{x^{n-1}}right).$$

For more details, hover over the the block below:

The equation above can be rewritten to get
$$x^{n+1}+frac{1}{x^{n+1}}=left(x^n+frac{1}{x^n}right)left(x+frac{1}{x}right)-left(x^{n-1}+frac{1}{x^{n-1}}right).$$
If the three terms in parentheses on the right hand side are integers, then so is the left hand side. Now to use induction, all you need is that $x^n+frac{1}{x^n}$ is an integer for $n=0$ and $n=1$.

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

26.5k33997

$endgroup$

share|cite|improve this answer

answered 3 hours ago

ServaesServaes

26.5k33997

answered 3 hours ago

ServaesServaes

26.5k33997

answered 3 hours ago

ServaesServaes

26.5k33997