Request help on proof of irrationality of $\sqrt {8}$.

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.

• $\sqrt 8=2\sqrt 2$ may simplify things. – David Mitra May 17 at 9:14
• Use $\sqrt 8 = 2\sqrt 2$ and $\sqrt 2$ is irrational. – Wuestenfux May 17 at 9:14
• @DavidMitra will that help with linear combination $p-2q$ directly. – jiten May 17 at 9:16
• Why $p\cdot\frac{p-2q}{q}$ cannot be an integer? – CY Aries May 17 at 9:17
• 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. – CY Aries May 17 at 9:30

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 $$3$$.

• 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 $n\rightarrow \infty$. – jiten May 17 at 14:06
• 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 $1$. – jiten May 17 at 14:44
• 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 $f'(n)\approx\frac{\Delta y}{\Delta n}= \lim_{\Delta n \to0 } (\frac{f(n+\Delta n)-f(n)}{(n+\Delta n)-n})_{n=1}^{\infty}$ $= \lim_{h\to0 }(\frac{f(n+h)-f(n)}{h})_{n=1}^{\infty}$ $=\lim_{h\to0 }(\frac{\frac {n+h}{n+h+1}-\frac n{n+1}}{h})_{n=1}^{\infty}$ $=\lim_{h\to0 }(\frac{\frac {(n^2+n+nh+h) -(n^2+nh+n)}{(n^2+n+nh+h+n+1)}}{h})_{n=1}^{\infty}$ $=(\frac {1}{n^2+2n+1})_{n=1}^{\infty}$ – jiten May 17 at 15:04
• Have made a question at : math.stackexchange.com/q/3230261/424260, for the last comments. – jiten May 18 at 5:31
• I am not fast enough. Those have been answered. congrats. – Siong Thye Goh May 26 at 8:31

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.

• 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 $p-2q$. . – jiten May 17 at 9:17
• 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. $n\lt i \lt n+1$. – jiten May 17 at 9:27

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

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.)

• 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 $(2,3)$ – jiten May 18 at 2:54
• 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. – CY Aries May 18 at 5:02
• 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. – CY Aries May 18 at 5:05