Proof by Induction - New to proofs
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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
asked 1 hour ago
RobinRobin
261
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add a comment |
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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
asked 1 hour ago
RobinRobin
261
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.
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I recommend this answer as a good start
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– Ross Millikan
1 hour ago
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math.stackexchange.com/questions/936479/…
$endgroup$
– lab bhattacharjee
1 hour 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
asked 1 hour ago
RobinRobin
261
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.
$endgroup$
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
induction
asked 1 hour ago
RobinRobin
261
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 1 hour ago
RobinRobin
261
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 1 hour ago
RobinRobin
261
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 1 hour ago
RobinRobin
261
asked 1 hour ago
RobinRobin
261
261
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.
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.
Robin is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
I recommend this answer as a good start
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
math.stackexchange.com/questions/936479/…
$endgroup$
– lab bhattacharjee
1 hour ago
add a comment |
$begingroup$
I recommend this answer as a good start
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
math.stackexchange.com/questions/936479/…
$endgroup$
– lab bhattacharjee
1 hour ago
$begingroup$
I recommend this answer as a good start
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
I recommend this answer as a good start
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
math.stackexchange.com/questions/936479/…
$endgroup$
– lab bhattacharjee
1 hour ago
$begingroup$
math.stackexchange.com/questions/936479/…
$endgroup$
– lab bhattacharjee
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $n=1$.
answered 1 hour ago
ServaesServaes
26.5k33997
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $n=1$.
answered 1 hour 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 $n=1$.
answered 1 hour 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 $n=1$.
answered 1 hour ago
ServaesServaes
26.5k33997
$endgroup$
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 $n=1$.
answered 1 hour ago
ServaesServaes
26.5k33997
answered 1 hour ago
ServaesServaes
26.5k33997
answered 1 hour ago
ServaesServaes
26.5k33997
answered 1 hour ago
ServaesServaes
26.5k33997
26.5k33997
add a comment |
add a comment |
Robin is a new contributor. Be nice, and check out our Code of Conduct.
Robin is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I recommend this answer as a good start
$endgroup$
– Ross Millikan
1 hour ago
$begingroup$
math.stackexchange.com/questions/936479/…
$endgroup$
– lab bhattacharjee
1 hour ago