1
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • $\begingroup$ $\sqrt 8=2\sqrt 2$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $3$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ I am not fast enough. Those have been answered. congrats. $\endgroup$ – Siong Thye Goh May 26 at 8:31
2
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

|cite|improve this answer
$\endgroup$
1
$\begingroup$

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.)

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.