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There is a proof in my math textbook about the fact that $\root \of 2$ is irrational.

Proof:

Lets assume that $\root \of 2$ is rational.Then there will exist $2$ coprime natural numbers $p$ , $q > 1$ such that , $$ \root \of 2 = \frac{p}{q} \to 2 = \frac{p^2}{q^2} \to 2q = \frac{p^2}{q}$$

Obviously , $2q$ is an integer but $\frac{p^2}{q}$ is not a integer , because $p$ and $q$ are natural numbers , coprime and $q > 1$. So,

$$2q \neq \frac{p^2}{q} \to \root \of 2 \neq \frac{p}{q}$$

So , $\root \of 2$ is an irrational number.$\square$

But the confusion to me is , it seems like I can use this argument to show that $\root \of 4$ is an irrational number.

Proof:

Lets assume that $\root \of 4$ is rational.Then there will exist $2$ coprime natural numbers $p$ , $q > 1$ such that , $$ \root \of 4 = \frac{p}{q} \to 4 = \frac{p^2}{q^2} \to 4q = \frac{p^2}{q}$$

Obviously , $4q$ is an integer but $\frac{p^2}{q}$ is not a integer , because $p$ and $q$ are natural numbers , coprime and $q > 1$. So,

$$4q \neq \frac{p^2}{q} \to \root \of 4 \neq \frac{p}{q}$$

So , $\root \of 4$ is an irrational number.$\square$

Can someone tell me what is wrong with this proof?

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    $\begingroup$ $q$ could be $1$ $\endgroup$
    – NL1992
    May 16, 2021 at 10:11
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    $\begingroup$ If $p=2, q=1$ then $4q = \frac{p^2}{q}$ and $\root \of 4 = \frac{p}{q}$ $\endgroup$
    – Henry
    May 16, 2021 at 10:14
  • $\begingroup$ Similar: math.stackexchange.com/q/2270591/42969 $\endgroup$
    – Martin R
    May 16, 2021 at 10:16
  • $\begingroup$ @MartinR I would not say it is a duplicate , because the proof is kind of different than your linked post. $\endgroup$ May 16, 2021 at 10:17
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    $\begingroup$ @vitamind It is not the same proof. $\endgroup$ May 16, 2021 at 10:29

4 Answers 4

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You begin your proof saying that $\sqrt4=\frac pq$ with $p$ and $q$ natural coprimes greater than $1$. That works in the case of $\sqrt2$, since $\sqrt2$ is not a natural number, and then, yes, if it could be written as $\frac pq$ with $p$ and $q$ natural coprime numbers, then both of them would have to be greater than $1$. But $\sqrt4$ is a natural number. And the assumption that $\sqrt4$ can be written as $\frac pq$ with $p$ and $q$ coprime and $p,q>1$ is false.

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    $\begingroup$ @JoséCarlosSantos how does one prove that $\root \of 4$ is a natural number but $\root \of 2$ is not. $\endgroup$ May 16, 2021 at 10:21
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    $\begingroup$ You have $1^2=1<2$ and, if $n$ is a natural number greater than $1$, then $n\geqslant2$, and therefore $n^2\geqslant4$. So, there is no $n\in\Bbb N$ such that $n^2=2$. And $\sqrt4=2\in\Bbb N$. $\endgroup$ May 16, 2021 at 10:23
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If $p,\,q$ are coprime natural numbers, $q|p^2$ iff $q=1$. The condition $p/q=\sqrt{2}$ contradicts $q=1$ because $\sqrt{2}$ isn't an integer, but $p/q=\sqrt{4}$ doesn't because $\sqrt{4}$ is. So while $q>1$ finishes the first proof, it can't be deduced in the second. In particular, we need only check $1^2<2<2^2=4$ to see $\sqrt{4}$, but not $\sqrt{2}$, is natural.

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You seem to be trying to understand the Proof by contradiction method used for the irrationality of numbers.


You can do the right steps as follows:

Let $\gcd(m,n)=1$, then you have

$$\sqrt 4=\frac mn$$

$$4=\frac {m^2}{n^2}$$

$$m^2=4n^2$$

This implies, $m=2k$. Hence,

$$4k^2=4n^2$$

$$k=n$$

This means, $$m=2k=2n$$

Finally,

$$\sqrt 4=\frac mn=\frac{2n}{n}=2.$$


Proof - verification

Lets assume that $\root \of 4$ is rational.Then there will exist $2$ coprime natural numbers $p$ , $q > 1$ such that , $$ \root \of 4 = \frac{p}{q} \to 4 = \frac{p^2}{q^2} \to 4q = \frac{p^2}{q}$$

Obviously , $4q$ is an integer but $\frac{p^2}{q}$ is not a integer , because $p$ and $q$ are natural numbers , coprime and $q > 1$.

Your error strarts here:

You claim that $p,q> 1$. This claim is wrong. The correct claim should be as follows:

$$\gcd(p,q)=1.$$

Obviously , $4q$ is an integer but $\frac{p^2}{q}$ is not a integer.

This claim is also wrong. Because, $q=1$ is a counter-example.

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    $\begingroup$ @MorganRodgers It seems the OP has now added something to the answer $\endgroup$ May 16, 2021 at 11:12
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    $\begingroup$ @AlbusDumbledore Thank you for your reviewing. $\endgroup$ May 16, 2021 at 11:32
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The restriction that $p,q$ should be $>1$ is unwarranted in either proofs, unnecessary in the first one and false for the second one. It should have been stated for what it was: that $p,q\in\Bbb Z$ coprime numbers such that $q\ge1$. Then the proof for $\sqrt 2$ goes more or less $$[2q^2=p^2]\Rightarrow [2\mid p]\Rightarrow [q^2=2\left(p/2\right)^2]\Rightarrow [2\mid q]$$

But then $2\mid\operatorname{gcd}(p,q)=1$.

Your second attempt at a proof relies on a false theorem, therefore the rest is hardly worth worrying about.

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