In typical proofs of the irrationality of $\sqrt{2}$, I have seen the following logic:
If $p^2$ is divisible by $2$, then $p$ is divisible by $2$.
Perhaps I am being over-analytical, but how do we know this to be true? IE. do we require a proof of this implication, or is it simply fact?
The quickest proof of that fact is to note that every whole number $n$ is either even or odd.
If $n$ is even, $n=2k$ for some whole number $k$: $n^2 = 4k^2 = 2(2k^2)$ is even.
If $n$ is odd, $n=2k+1$ for some whole number $k$: $n^2 = (2k+1)^2 = 4k^2 +4k + 1 = 2(2k^2 +2k) +1$ is odd.
Therefore the square of a whole number is even if and only if that number is even.
According to Euclid's lemma if $p$ is a prime and $a,b$ are integers, then $p \mid ab$ implies that $p \mid a$ or $p \mid b$.
Since $2$ is prime and it divides $p^2 = p\cdot p$ then either $2 \mid p$ or $2 \mid p$. In either case, $2 \mid p$.
A simple and (maybe) less technical way to state it could be to draw the following table: $$ \begin{array}{|c|c|} \hline \times&\text{even}&\text{odd}\\ \hline \text{even}&\text{even}&\text{even}\\ \hline \text{odd}&\text{even}&\text{odd}\\ \hline \end{array} $$
As an aside, I personally prefer the proof that no non-integer fraction has non-integer powers by exploiting unique prime factorization: $$ \left(\frac{p_1\cdot p_2\cdots p_n}{q_1\cdot q_2\cdot q_m}\right)^s=\frac{p_1^s\cdot p_2^s\cdots p_n^s}{q_1^s\cdot q_2^s\cdot q_m^s} $$ since if you could not cancel any of the prime factors $p_1,p_2,...,p_n,q_1,q_2,...,q_m$ before, you still cannot do it when they appear with multiplicity $s$. This proves that among integers, only perfect squares, cubes etc. have integer square roots, cube roots, etc.
Using modular arithmetic,
$$p^2\equiv p\mod2$$ because
$$p^2-p=p(p-1)\equiv0\mod2$$ as one of $p$ and $p-1$ is even.
It is a simple fact that you can take as an axiom, or you can have it proven easily enough.
Suppose $p$ is even, which means that $p = 2k$, where $k$ is also an integer. Then $p^2 = (2k)^2 = 4k^2$.
But if $p$ is odd, then $p = 2k + 1$, and $p^2 = (2k + 1)^2 = 4k^2 + 4k + 1$.
So if $p$ is divisible by $2$, then so is $p^2$; but if $p$ is not divisible by $2$, then neither is $p^2$.
Some mathematicians distinguish between theorems, lemmas, and corollaries.
Though they are all statements that require proofs, the difference between them is their intention.
Theorems are considered to be important. The appearance of the word theorem is the equivalent of drawing a circle around the subsequent statement and pointing at it with arrows.
Corollaries are (almost) immediate consequences of a theorem.
Lemmas are a way of breaking up the proof of a theorem into smaller pieces. They are generally smaller points that will be useful in proving a consequent theorem.
Sometimes these distinctions are ignored, sometimes they are carried to extremes. Zorn's lemma for example has every right to be called a theorem. It is of major importance. But it is used to prove theorems. So it is properly a lemma.
I have always felt that he Fundamental Theorem Of Arithmetic should also be called a lemma.
The Fundamental Theorem Of Arithmetic. Every integer greater than $1$ is either a prime number or is the product of prime numbers, and this product is unique, up to the order of the factors.
For example, $180 = 2 \times 2 \times 3 \times 3 \times 5 = 2^2 \times 3^2 \times 5$ , and this is the only way that $180$ can be expressed as a product of prime numbers.
Lemma. Let $p$ be a prime number and let $n$ be a positive integer. If $p$ divides $n^2$, then $p$ divides $n$.
Proof. If $n=1$, then $n^2 = 1$ and $1$ has no prime factors. So $n$ must be greater than $1$. Hence $n = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_m^{a_m}$ where, for some positive integer $m$, $\; p_1, p_2, \dots, p_m$ are distinct prime numbers and $\; a_1, a_2, \dots, a_m$ are non negative integers. Hence $n^2 = p_1^{2a_1} \times p_2^{2a_2} \times \cdots \times p_m^{2a_m}$. Since $p$ is a prime number, if $p$ divides $n^2$, then the fundamental theorem of arithmetic implies that $p$ must be one of $\; p_1, p_2, \dots, p_m$. It follows that $p$ divides $n$.
Note that it is crucial that $p$ be a prime number. For example, $18$ divides $6^2$ but $18$ does not divide $6$.
Theorem Let $p$ be a prime number. Then $\sqrt p$ is an irrational number.
Proof by contradiction. If $\sqrt p$ were rational, then we could say there exists positive integers, $m$ and $n$, such that $$ \sqrt p = \dfrac mn$$ and $m$ and $n$ have no common factors (that is, the fraction $\dfrac mn$ has been reduced to lowest terms).
It follows that $m^2 = p n^2$. Since $p$ divides $p n^2$, then $p$ divides $m^2$. Since $p$ is a prime number, then $p$ divides $m$. So $m = px$ for some integer $x$. Then \begin{align} m^2 &= p n^2\\ (p x)^2 &= p n^2 \\ p^2 x^2 &= p n^2 \\ p x^2 &= n^2 \end{align}
Since $p$ divides $p x^2$, then $p$ divides $n^2$. Since $p$ is a prime number, then $p$ divides $n$.
But then $p$ divides both $m$ and $n$, which contradicts our assumption that $m$ and $n$ had no common factors. Hence the theorem is proved.
Since $2$ is a prime number, we have shown that
Corollary. $\sqrt 2$ is an irrational number.
$\, n\,$ odd $\Rightarrow n^2$ odd $ $ by $(1\!+\!2i)^2 = 1\!+\!2(2i\!+\!2i^2).\, $ So $\,n^2$ even $\,\Rightarrow n\,$ even by contraposition.
My favorite proof of the irrationality of $\sqrt{2}$ invokes the Fundamental Theorem of Arithmetic at this point.
The number $p$ has some unique factorization as primes, say $p = p_1^{n_1} \cdot p_2^{n_2} \cdot p_3^{n_3}...$; so $p^2 = p \cdot p = (p_1^{n_1} \cdot p_2^{n_2} \cdot p_3^{n_3}...)(p_1^{n_1} \cdot p_2^{n_2} \cdot p_3^{n_3}...)$. The only way that $p^2$ can be divisible by 2 is for one of those prime factors to be 2; and the only way for that to happen is for one of the factors of $p$ to be 2, because $p^2$ is composed of exactly the same factors.
I'll give a tip of the hat that this is frequently the most troublesome point for my students when this is taught; we're usually so focused on understanding the proof-by-contradiction structure that this likely-new observation about prime factors gets short shrift.
There's a much simpler answer that doesn't require reaching out to other theorems - just intuition.
Suppose $p^2$ is divisible by 2, but $p$ is not. Then, $p$ is a product of only odd numbers. However, if this were the case, then $p^2$ would also be the product of only odd numbers. This can't be true because $p^2$ is divisible by 2, so $p$ must be divisible by 2 by contradiction.
$p^2$ is even iff $p$ is even. So if $2$ divides $p^2,$ it must divide $p.$
We know that any natural number can be expressed as a product of primes. Thus, p can be expressed as product of primes and so p².
Let p = a x b x c x d x ........x n
Where a, b, c, d,....... all are primes.
By hypothesis 2 divides p²,
That means there is a 2 hiding in p²
So, we can write p² = 2q, where q is a natural number.
So, q = (p²/2)
Since q is a natural number, p² must be even.
So, we may conclude that p must be even.
[If p is odd, p² is odd.]
Thus, 2 divides p.
$p^2$ can't have any factors that $p$ lacks.
$p$ is the product of its factors, each raised to an appropriate degree.
$p = f_1^{x_1} * f_2^{x_2}... f_n^{x_n}$
So $p^2$ must be the product of that product with itself.
$p = (f_1^{x_1} * f_2^{x_2}... f_n^{x_n}) * (f_1^{x_1} * f_2^{x_2}... f_n^{x_n})$
The set of factors is identical. Only the exponents have changed.
$p = f_1^{2x_1} * f_2^{2x_2}... f_n^{2x_n}$
So any factor appearing in one list must appear in the other, which implies the hypothesis above.
