false proof of $\root \of 4$ is irrational.

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?

• $q$ could be $1$ May 16, 2021 at 10:11
• If $p=2, q=1$ then $4q = \frac{p^2}{q}$ and $\root \of 4 = \frac{p}{q}$ May 16, 2021 at 10:14
• May 16, 2021 at 10:16
• @MartinR I would not say it is a duplicate , because the proof is kind of different than your linked post. May 16, 2021 at 10:17
• @vitamind It is not the same proof. May 16, 2021 at 10:29

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.

• @JoséCarlosSantos how does one prove that $\root \of 4$ is a natural number but $\root \of 2$ is not. May 16, 2021 at 10:21
• 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 $\sqrt4=2\in\Bbb N$. May 16, 2021 at 10:23

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.

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 $$q > 1$$.

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.

• @MorganRodgers It seems the OP has now added something to the answer May 16, 2021 at 11:12
• @AlbusDumbledore Thank you for your reviewing. May 16, 2021 at 11:32

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.