epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Special Kinds of Theorems 45
and
^
aix-\-biy = c\
V^i + ba2)x H- (fci + hb2)y = ci + hc2
where ax^ai.bubi.cu and C2 are real numbers, and b^Q.
The pair of values (xo,>'o) is a solution of iSi if and only if it is a
solution of S2.
USE OF COUNTEREXAMPLES
An example can be very useful when trying to make a point or explain
the result obtained in a proof, but it cannot be used as a proof of the fact
that a statement is true.
Let us see what might happen if we used and accepted examples as proofs.
We could make the claim that if a and b are any two real numbers, then
{a^bf=a^
+ b^.
When asked to support our claim, we can produce a multitude of pairs of
numbers that satisfy this equality. For example, consider a = 0 and b = \\
(^ + i)2 = (0+1)2 = 1
a^ + b^ = 0^ + \^ = \.
Thus, it is true that {a + bf= a^ + fc^.
Consider some more examples, such as: a = 0 and fc = —1; a——A and
i = 0; a = 7X and b = 0\ and so on. For all these pairs, the equality
{a + b)^= a^ + b^ holds true. But, it is possible to notice that in all the
pairs hsted above at least one number is equal to zero. The claim states that
the equality holds true for any two real numbers, not just for some special
pairs. What happens if we consider a=\ and fo = 2?
{a + bf = 2>^ = 9
a^-^b^ = 1^ + 2^ = 5.
Therefore, the equaUty is false. In spite of all the examples that seem to
support it, we have found an example that contradicts it, a counterexample,
A counterexample is an acceptable proof of the fact that the statement
"If A, then B" is false because it shows that B can be and is false while A