A proof by deduction starts from definitions and
known facts, then uses a chain of logical, algebraic steps to reach the
result — and it must hold for all cases, not just a few examples you happen
to try.
The trick is to use algebra to stand for the general case. Instead of testing
particular numbers, you write a symbol that represents every number of that kind:
\text{even} = 2n, \qquad \text{odd} = 2m + 1, \qquad \text{consecutive} = n,\ n+1
Here n and m are whole numbers. Because
2n covers every even number at once, an argument about
2n proves the statement for them all.
Worked example
Claim: the sum of two odd numbers is always even.
Let the two odd numbers be 2m + 1 and
2n + 1, where m and
n are whole numbers. Add them and gather the terms:
(2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1)
Now m + n + 1 is a whole number, so the sum is
2 \times (\text{a whole number}) — which is exactly what it means to
be even. Since m and n
stood for any whole numbers, this proves the claim for every pair of odd numbers.
To prove a statement by deduction:
- start from definitions and known results;
-
use algebra for the general case — even
= 2n, odd = 2n + 1, consecutive
= n,\ n+1;
- proceed by valid algebraic and logical steps to the conclusion;
-
remember that testing examples is not a proof — a handful of cases never
covers them all.