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PROOF BY CONTRADICTION

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PROOF BY CONTRADICTION

Let r be a proposition.

A proof of r by contradiction consists of

proving that not(r) implies a contradiction,

thus concluding that not(r) is false,

which implies that r is true.

In particular if r is

if p then q

then not(r) is logically equivalent

to p AND (not(q)).

This can be verified by constructing

a truth table.

So a true proposition

if p then q

may be proved by contradiction as follows:

Assume that p is true and q is false,

and show that this assumption implies

a contradiction.

One way of proving

that the assumption that

p is true and q is false implies

a contradiction is proving that this

assumption implies that p is false.

Since the same assumption also

implies that p is true, we conclude that

the assumption implies that p is true and p

is false, which is a contradiction.

EXAMPLE:

Prove that the sum of an even integer

and a non-even integer is non-even.

(Note: a non-even integer is an integer

that is not even.)

We have to prove that for every even integer

a and every non-even integer b, a+b

is non-even.

This is the same as proving that

For all integers a,b, if [a is even and

b is non-even] then [a+b is non-even].

We prove that by contradiction.

Assume that

[a is even and b is non-even],

and that [a+b is even]. So for some

integers m,n, a=2m and a+b=2n.

Since b=(a+b)-a, b=2n-2m=2(n-m).

We conclude that b is even. This leads

to a contradiction, since we assumed that

b is non-even.