Proof regarding the square root of a natural number $m>1$ and a prime number $p$, when $p \nmid m$

Question

I am self studying from "A logical introduction to proof" by Daniel W.Cunningham,one of those textbooks that unfortunately do not contain answers to end-of-chapter exercises.Could someone help me with the following problem?

Let p be a prime number and m>1 be a natural number.Prove that if $p\nmid m$ then $\sqrt{mp}$ is irrational.

My attempt:

Let p be a prime number and m $\in \Bbb{N} $ where m>1 (i.e.m>=2).The proof proceeds by regular mathematical induction.

The base step invovles proving that (p $\nmid 2$) $\Rightarrow$ ($\sqrt{2p} $ is irrational).By contraposition,we need to prove that ($\sqrt{2p} $ is rational) $\Rightarrow$ ($p\mid 2$).By defintion $\sqrt{2p}={a\over b} $ for some integers a and b where b$\neq 0 $ and gcd(a,b)=1.Therefore ${2p=}{a^2\over b^2} $ and hence ${2b^2 p=a^2} $ by squaring both sides and subsequent multplication.

Furthermore, ${p\mid a^2}$ and by implication ${p\mid a}$ which by definition suggests ${a=pk}$ for some k $\in \Bbb{Z}$. Consequently ${2b^2 p=p^2k^2 }$ then ${2b^2 = pk^2}$ hence ${p \mid 2b^2}$ which by Euclid's Lemma suggests either (i) ${p \mid 2}$ or (ii) ${p \mid b^2}$ .

Now in the case where (i) ${p\mid 2}$,the conclusion to the base step has been verified.

Consider the case that (ii) ${p\mid b^2}$,then by the corrolary to Euclid's Lemma, ${p\mid b}$ as well.With ${p\mid b}$ and ${p\mid a}$,we have a prime number p that divides both a and b when by assumption gcd(a,b)=1.This implies that ${p\mid b}$ is false and hence that ${p\mid b^2}$ is false by contraposition,leaving us with case(i) as verified above.

Does this prove the base step?

Regarding the inductive step i do not know how to link the induction conclusion to the induction hypothesis...it is as if the conclusion can be arrived at without even considering the hypothesis in the inductive step,which to my understanding means the 'dominos have not been aligned' in the typical induction-dominos imagery.

Any help would be great. Thanks!

Answer 1

It's not a good idea to try induction to prove this because induction will prove it for ALL $m$, but this is false if $p \mid m$.

Your proof works more directly for $m$:

Suppose $\sqrt{pm}=\frac{a}{b}$ for some coprime positive integers $a$ and $b$. Then $b^2 p m = a^2$. Then $p \mid a^2$. As $p$ is prime we have $p\mid a$, that is $a=p k$ for some integer $k$. Then $b^2 p m = p^2 k^2$, so $b^2 m = pk^2$. This means $p\mid b^2m$. Again, as $p$ is prime this implies $p\mid b$ or $p\mid m$, but the hypothesis tell us $p \not\mid m$, so we must have $p\mid b$. So we now have $p\mid a$ and $p\mid b$, a contradiction to the fact that $a$ and $b$ are coprime.

Answer 2

You actually don’t need to recur, the proof is standard, and almost the same as the proof that $\sqrt2$ is irrationnal.

Let’s assume there is a $x = \frac{a}{b}$ so that $x^2 = mp$ and $pgcd(a,b) =1$

Then $\frac{a^2}{b^2} = mp$

$a^2 = mp b^2$

then

either p is a prime factor of b, which means it is not one of a (since $pgcd(a,b) = 1$) and therefore $a^2 = m p b^2$ is impossible

or p is not a prime factor of b. since $a^2 = mp b^2$, we have $p | a^2$ so $p | a$ ($p|xy$ implies $p|x$ or $p|y$ when $p$ is prime, by euclide’s theorem. We just set $x = y = a$)

then $p^2 | a^2$ then $p^2 | m p b^2$, that is $p | mb^2$ which contradicts with our hypothesis