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Prove that $\sqrt{8}$ is an irrational number.

As know that $2\lt \sqrt{8} \lt 3$, so if assume (to attempt proof by contradiction) that $\sqrt{8} = \frac pq$, where $p,q$ are coprime integers; then $2\lt \frac pq \lt 3$.

Attempt 1:
Subtracting $2$ from all terms, get : $0\lt \frac {p-2q}q \lt 1$.

This means $p-2q$ is an integer with no common terms with $q$. Also $q$ has no common terms with $p$.
But, $p-2q$ is a linear combination of $p,q$.

Am unable to use any property of linear combination of co-prime integers to directly prove by contradiction. Request help by this approach.

Attempt 2:
$2\lt \frac pq \lt 3 \implies 2q \lt p \lt 3q \implies 0\lt p-2q \lt q$. So, $\frac{p-2q}{q}$ is not an integer value & must be $\lt 1$.

As $\sqrt{8}=\frac pq$ is assumed to be a rational, so its product with $p-2q$ is also rational. But, this product cannot be an integer, as $\frac pq (p-2q)= p.\frac {p-2q}q$.

But $\frac pq(p-2q) = \frac {p^2}{q} -2p = 8q -2p$ which is a linear combination of integers. This contradicts the earlier statement.

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  • $\begingroup$ $\sqrt 8=2\sqrt 2$ may simplify things. $\endgroup$ – David Mitra May 17 at 9:14
  • $\begingroup$ Use $\sqrt 8 = 2\sqrt 2$ and $\sqrt 2$ is irrational. $\endgroup$ – Wuestenfux May 17 at 9:14
  • $\begingroup$ @DavidMitra will that help with linear combination $p-2q$ directly. $\endgroup$ – jiten May 17 at 9:16
  • $\begingroup$ Why $p\cdot\frac{p-2q}{q}$ cannot be an integer? $\endgroup$ – CY Aries May 17 at 9:17
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    $\begingroup$ How about $p=10$ and $q=4$? $\frac{p-2q}{q}=\frac{1}{2}$ and $p\cdot \frac{1}{2}=5$. Are you using the fact that $p$ and $q$ is coprime? $\frac{p-2q}{q}<1$ is not enough. $\endgroup$ – CY Aries May 17 at 9:30
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Suppose $\sqrt8 = \frac{p}q$ where $gcd(p,q)=1$.

then we can find integer $x,y$, such that $$px+qy =1\tag{1}$$

$$8q^2=p^2$$

Hence $p$ is an even number, $p=2k$, $$2q^2=k^2$$

Hence $k$ must also be an even number. $k=2l$.

$$2q^2=(2l)^2$$

$$q^2=2l^2$$

Hence $q$ must be an even number.

Since $p$ and $q$ are both even. $px+qy$ must be even. They cannot be equal to $1$.

Remark:

Once you lose track of the property of $\sqrt8$. Your proof shouldn't work. After all, we know that there are rational numbers between $2$ and $3$.

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  • $\begingroup$ Need this for next post. Request to show how $lim (\frac {1}{n^2+2n+1})_{n=1}^{\infty} =1$, as unable to evaluate $\frac {1}{\infty^2+2.\infty+1}$ at values of the domain tending to infinity. I know that limit of a sequence is its value for large values of domain, which here is continuous & unbounded; hence at $n\rightarrow \infty$. $\endgroup$ – jiten May 17 at 14:06
  • $\begingroup$ I got this idea that the only way can be $lim(\frac {1}{n^2+2.n+1} )_{n=1}^{\infty} = \frac {\frac{1}{n^2}}{1+2.\frac 1n+\frac 1{n^2}} = \frac 01$. But, this is still $0$, not $1$. $\endgroup$ – jiten May 17 at 14:44
  • $\begingroup$ Please help, can we chat. Else the problem arose in Q.3 of my next intended post as: > Q.3. Identify limit of Seq. 3 using calculus. Let, $f(n) = (\frac n{n+1})_{n=1}^{\infty}$, find $f'(n)\approx\frac{\Delta y}{\Delta n}= \lim_{\Delta n \to0 } (\frac{f(n+\Delta n)-f(n)}{(n+\Delta n)-n})_{n=1}^{\infty}$ $= \lim_{h\to0 }(\frac{f(n+h)-f(n)}{h})_{n=1}^{\infty}$ $=\lim_{h\to0 }(\frac{\frac {n+h}{n+h+1}-\frac n{n+1}}{h})_{n=1}^{\infty}$ $=\lim_{h\to0 }(\frac{\frac {(n^2+n+nh+h) -(n^2+nh+n)}{(n^2+n+nh+h+n+1)}}{h})_{n=1}^{\infty}$ $=(\frac {1}{n^2+2n+1})_{n=1}^{\infty}$ $\endgroup$ – jiten May 17 at 15:04
  • $\begingroup$ Have made a question at : math.stackexchange.com/q/3230261/424260, for the last comments. $\endgroup$ – jiten May 18 at 5:31
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    $\begingroup$ I am not fast enough. Those have been answered. congrats. $\endgroup$ – Siong Thye Goh May 26 at 8:31
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Why don't you just use $2\sqrt{2} = \sqrt{8}$? Once you do this just prove by contradiction that if $\sqrt{8}$ is rational, so must $\sqrt{2}$ be.

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  • $\begingroup$ Then the whole approach is changed, isn't it. I will attempt, but if I want with my current approach, then how to develop over approach by directly taking the linear combination $p-2q$. . $\endgroup$ – jiten May 17 at 9:17
  • $\begingroup$ Also, it can be a generic proof , i.e. where any integer factor of $^n\sqrt{p}$ is not known, like here for $n = 2, p =8$ with integer factor of $2$. This generic approach will work there too, as can always find two integers $(n,n+1)$surrounding the given irrational, let $i$, s.t. $n\lt i \lt n+1$. $\endgroup$ – jiten May 17 at 9:27
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There are faster methods, but this is essentially the same as showing square-roots of prime numbers are irrational

  • Suppose $\sqrt{8}$ can be written as $\frac pq$ in lowest terms,

  • So $p^2=8q^2$, so $p^2$ is a multiple of $8$, which requires $p$ to be a multiple of $2$, so write $p=2a$ for some integer $a$

  • Then $(2a)^2=8q^2$ and $4a^2=8q^2$ and $a^2=2q^2$, so $a^2$ is a multiple of $2$, which requires $a$ to be a multiple of $2$, so write $a=2b$ for some integer $b$

  • Then $(2b)^2=2q^2$ and $4b^2=2q^2$ and $2b^2=q^2$, so $q^2$ is a multiple of $2$, which requires $q$ to be a multiple of $2$

  • But then $p$ and $q$ are both divisible by $2$, contrary to the supposition that $\frac pq$ is in lowest terms

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At first glance, I think the proof will not work as the following statement hold for all real numbers, rational or irrational, between $2$ and $3$. I can't see how to use the linear combination of $p$, $q$.

$2\lt \frac pq \lt 3 \implies 2q \lt p \lt 3q \implies 0\lt p-2q \lt q$. So, $\frac{p-2q}{q}$ is not an integer value & must be $\lt 1$.

Then we have

As $\sqrt{8}=\frac pq$ is assumed to be a rational, so its product with $p-2q$ is also rational. But, this product cannot be an integer, as $\frac pq (p-2q)= p\cdot\frac {p-2q}q$.

I can't see why $\frac pq (p-2q)$ cannot be an integer just because it is equal to $ p\cdot\frac {p-2q}q$. But I think the conclusion is correct, as $\gcd(p,q)=1$ implies that

$$\gcd(p(p-2q),q)=\gcd(p(p-2q)+2p(q),q)=\gcd(p^2,q)=1$$

So, $\frac pq (p-2q)$ is not an integer. This holds for any rational number between $2$ and $3$ when it is written in the lowest term.

Then jiten makes a beautiful argument and complete the proof.

But $\frac pq(p-2q) = \frac {p^2}{q} -2p = 8q -2p$ which is a linear combination of integers. This contradicts the earlier statement.

Here note that $\frac{p^2}{q}=\left(\frac{p}{q}\right)^2q=8q$.

The second attempt is a nice proof, although there is one argument I can't really follow (but the conclusion is correct.)

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  • $\begingroup$ I am sorry, but could not understand how $$\gcd(p(p-2q),q)=\gcd(p(p-2q)+2p(q),q)=\gcd(p^2,q)=1$$ implies that $\frac pq (p-2q)$ is not an integer. Request some more elaboration. For me, it is even tougher is to approach the next line, i.e. 'This holds for any rational number $2$ and $3$ when it is written in the lowest term." I.e., how the $gcd$ property leads to application only in the interval $(2,3)$ $\endgroup$ – jiten May 18 at 2:54
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    $\begingroup$ As $\frac{p-2q}{q}$ lies between $2$ and $3$, it is not an integer. Therefore $q$ is not $1$. $\gcd(p(p-2q),q)=1$ means that $\frac{p(p-2q)}{q}$ is already in the lowest term. So $q\ne1$ implies that $\frac{p(p-2q)}{q}$ is not an integer. $\endgroup$ – CY Aries May 18 at 5:02
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    $\begingroup$ Also, it holds for all rational number in $(2,3)$, but not only in $(2,3)$. Just because that $\sqrt{8}\in(2,3)$, I mention this range only. It actually holds when $q\ne1$ and $\frac pq$ is in the lowest term. $\endgroup$ – CY Aries May 18 at 5:05

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