$$\sqrt{4}=\frac{p}{q}, p \wedge q \in \mathbb{Z}$$
$$4=\frac{p^2}{q^2}$$
$$p^2=q^2 \cdot 4$$
$4$ is a multiple of $p^2$ but not $p$, why?
Why do you say is 4 not a multiple of p? It is if q = 1.
By the way, this is the first step in a short proof that the square root of any integer must either be irrational or an integer, but never a fraction (i.e. if the square root of an integer = p/q, then q must equal 1).
Well, that's what my book says.
"In this case, the fact that p^2 is a multiple of 4 does not imply p is also a multiple of 4. Thus, our proof breaks down at this point."
And, well, the proof's got to fail somewhere, though, since 4 is rational (not implying it should be here necessarily).
OK, I see. If p^2 is a multiple of a number n, that does not mean that p is also a multiple of n. It may be, but not necessarily. For example if p = 6 then p^2 is a multiple of 4, whereas p is not. On the other hand p^2 is a multiple of 3, and so is p.
What I was trying to point out is that for any proper fraction p/q the numerator p and the denominator q can not share any prime factors, which means if $\displaystyle p^2 = n q^2$ it must be the case that q = 1.
Okay, this is what should have been mentioned: https://en.wikipedia.org/wiki/Euclid%27s_lemma
If $\displaystyle x^2$ has p as a factor, where p is a prime number, the x must have p as a factor. but if p is a square, $\displaystyle p= q^2$ then p can be "distributed" between the two copies of x in the square- it might be q, not p, that is a factor of x.