Proof by contradiction, the classic proof that the square root of 2 is irrational, and how to extend it to other roots and to sums.
You cannot check every fraction $\dfrac{p}{q}$ to see that none equals $\sqrt{2}$. Instead we use proof by contradiction:
Lemma. If $a$ is an integer and $a^2$ is even, then $a$ is even.
Proof. Suppose $a$ is odd, $a = 2k + 1$. Then $a^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$ is odd. So an odd number has an odd square. Equivalently, if the square is even the number cannot be odd. $\square$
Theorem. $\sqrt{2}$ is not rational.
Proof. Suppose $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{a}{b}$ with integers $a, b > 0$ that have no common factor (we cancel any common factor first). Squaring, $2 = \dfrac{a^2}{b^2}$, so $$a^2 = 2b^2.$$ Thus $a^2$ is even, so $a$ is even by the lemma. Write $a = 2k$. Then $4k^2 = 2b^2$, so $b^2 = 2k^2$ is even, and $b$ is even too.
Now $a$ and $b$ are both even, so they have the common factor 2. This contradicts "no common factor". Hence $\sqrt{2}$ is irrational. $\square$
The same idea shows $\sqrt{p}$ is irrational for every prime $p$, using "if the prime $p$ divides $a^2$ then $p$ divides $a$".
Example: Prove that $\sqrt{5}$ is irrational. Suppose $\sqrt{5} = \dfrac{a}{b}$ with integers $a, b > 0$ with no common factor. Then $a^2 = 5b^2$, so 5 divides $a^2$. If 5 did not divide $a$, then $a = 5k + r$ with $r \in {1, 2, 3, 4}$ and $a^2 = 25k^2 + 10kr + r^2$ would leave the same remainder as $r^2$ on division by 5. For $r = 1, 2, 3, 4$ we have $r^2 = 1, 4, 9, 16$, which leave remainders $1, 4, 4, 1$, never $0$. So 5 divides $a$: $a = 5k$. Then $25k^2 = 5b^2$, so $b^2 = 5k^2$ and 5 divides $b$ by the same argument. So 5 is a common factor of $a$ and $b$, a contradiction. Hence $\sqrt{5}$ is irrational.
Once $\sqrt{2}$ is known to be irrational, a short contradiction handles many other numbers.
Rule A. If $x$ is irrational and $r$ is rational, then $r + x$ is irrational. Reason: if $r + x = s$ were rational, then $x = s - r$ would be rational.
Rule B. If $x$ is irrational and $r \neq 0$ is rational, then $rx$ is irrational. Reason: if $rx = s$ were rational then $x = \dfrac{s}{r}$ would be rational. (The condition $r \neq 0$ matters: $0 \cdot \sqrt{2} = 0$.)
Example: Prove that $\dfrac{1}{1 + \sqrt{2}}$ is irrational. Multiply top and bottom by $\sqrt{2} - 1$: $\dfrac{1}{1 + \sqrt{2}} = \dfrac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \dfrac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1$. Suppose $\sqrt{2} - 1 = r$ with $r$ rational. Then $\sqrt{2} = r + 1$ is rational, which contradicts the irrationality of $\sqrt{2}$. So $\dfrac{1}{1 + \sqrt{2}}$ is irrational.
Example: Prove that $\sqrt{2} + \sqrt{3}$ is irrational (you may use that $\sqrt{2}$ is irrational). Suppose $\sqrt{2} + \sqrt{3} = r$ with $r$ rational; $r > 0$. Then $\sqrt{3} = r - \sqrt{2}$. Squaring, $3 = r^2 - 2r\sqrt{2} + 2$, so $2r\sqrt{2} = r^2 - 1$ and $\sqrt{2} = \dfrac{r^2 - 1}{2r}$. The right side is rational (because $r \neq 0$), but $\sqrt{2}$ is irrational. This is a contradiction, so $\sqrt{2} + \sqrt{3}$ is irrational.
Exam tip: Start every such proof with the exact words "Suppose, for a contradiction, that ... is rational" and end with the exact fact that is contradicted. In the $\sqrt{2}$ proof, say clearly that $\dfrac{a}{b}$ is in lowest terms; marks are given for that assumption and for the final contradiction "both even".