Methods of Proof
Methods of Proof
To prove $p \to q$: assume $p$ is true, then use definitions and theorems to show $q$ is true.
Instead of $p \to q$, prove $\lnot q \to \lnot p$ (logically equivalent).
Assume $\lnot p$. Derive a contradiction. Conclude $p$ must be true.
To prove $P(n)$ for all $n \geq n_0$:
- Base case: Show $P(n_0)$
- Inductive step: Assume $P(k)$, prove $P(k+1)$
Prove: If $n$ is even, then $n^2$ is even.
$n = 2k$ for some integer $k$. Then $n^2 = 4k^2 = 2(2k^2)$, which is even. ∎
Prove: $\sqrt{2}$ is irrational.
Assume $\sqrt{2} = a/b$ in lowest terms. Then $2b^2 = a^2$, so $a$ is even, $a = 2c$, so $2b^2 = 4c^2$, $b^2 = 2c^2$, so $b$ is even. Contradiction: not in lowest terms. ∎
Prove: $1+2+\cdots+n = \frac{n(n+1)}{2}$.
Base: $n=1$: $1 = 1(2)/2$ ✓. Step: Assume true for $k$. Then $1+\cdots+k+(k+1) = \frac{k(k+1)}{2}+(k+1) = \frac{(k+1)(k+2)}{2}$. ∎
Practice Problems
Show Answer Key
1. $2a + 2b = 2(a+b)$, even ∎
2. Contrapositive: $n$ even → $n^2$ even. $n=2k$, $n^2=4k^2=2(2k^2)$ ∎
3. If $N$ largest, then $N+1 > N$, contradiction ∎
4. Base: $1=1^2$. Step: $k^2+(2k+1)=(k+1)^2$ ∎
5. Proof by contrapositive
6. $b = am$, $c = bn = amn$, so $a \mid c$ ∎
7. Without it, the chain of implications has no starting point
8. $2$ is prime and even
9. Every non-empty subset of $\mathbb{N}$ has a least element
10. The assumption that $P(k)$ is true
11. $n^2+n = n(n+1)$; consecutive integers, one is even ∎
12. No — it only shows one case, not all