Here's the old (circa 450 BCE) proof of the incommensurability of the diagonal and the side of the square.  The proof uses modern notation and concepts (e.g., fractions), but structurally it's quite similar to the original proof.
Now we need some terminology:

• Two quantities A and B are commensurable when there is a quantity Q such that Q is contained in A and B respectively n and m times. For example, a foot and a yard are commensurable because there is a quantity, the inch, which is contained in them respectively 12 and 36 times. If two quantities A and B are commensurable, then their ratio A/B can be expressed as the ratio n/m, in this case, 12/36.
• A number that can be expressed as a ratio n/m where both n and m are natural numbers (i.e., 1, 2, 3, 4....) is a rational number.
• A ratio is expressed in the smallest possible numbers when there is no number (different from 1)by which both the nominator and the denominator can be divided. In our example, both the nominator (12) and the denominator (36) can  be divided by 12, thus obtaining 1/3, which expresses the ratio between a foot and a yard in the smallest possible numbers.
• Two quantities which are not commensurable are incommensurable.

Given the square ABCD (see figure), consider the side AB and the diagonal AC.  We are going to prove that AC and AB are incommensurable. This is an
astonishing result. To undertand why, we must realize that in geometry there is no limit to the smallness of the parts into which a line can be divided. So, given two
line segments a and b, one would think that one could always find a very small line segment q such that q is in a and b n and m times respectively. This theorem
shows that this intuition is wrong.
We shall proceed by reductio. So, we shall assume that AC and AB are commensurable, derive a contradiction (in our case that the same number b is both odd
and even) and infer the negation of what we assumed. This will prove that AC and AB are incommensurable.

The Proof:

1. Suppose AC and AB are commensurable; let a/b be their ratio expressed in the smallest possible numbers.  Then, (AC/AB) = (a/b). (I)
2. Since AC > AB, it follows that a > b. Consequently, a > 1.
3. Now from (I) we get: (AC²/AB²) = (a²/b²) (II).
4. By the Pythagorean Theorem, AC² =AB²+BC², and since AB=BC, we can infer AC²=2AB². (III)
5. If we plug (III) into (II), we get (2AB²/AB²)=(a²/b²),that is, 2=(a²/b²).
Hence, multiplying both members of the equation by b²,
2b²=a². (IV)
6. So, a² is even, and consequently a is even as well. (For if the square of a number is even, the number is even as well). Since a/b is expressed in the smallest
possible numbers, b must be odd, otherwise both a and b could be divided by 2 and a/b would not be expressed in the smallest possible numbers.
7. Since a is even, there must be a number n such that a=2n. From this and (IV) (by substituting 2n for a in (IV)) we can infer 4n²=2b², that is, dividing both
members of the equation by 2, 2n²=b².
8. Hence, b² is even, and consequently b is even as well.
9. So, b is both even and odd, which is impossible.
10.Hence our assumption that AC is commensurable with AB must be false.
11.Therefore, AC and AB are incommensurable.

So, what is the value of AC/AB?  The Pythagorean Theorem tells us that it is the square root of 2.  But we have just seen that AC/AB is not a rational number, and consequently the square root of 2 is not a rational number either.