EAPS 11100 Lecture Notes - Lecture 17: Rational Number

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is irrational
2
(Proof by contradiction)
Suppose is a rational number. Then, we can say that =
where
a
,
b
are
22b
a
whole numbers,
b
is not zero, and a and b are both not even .
We additionally assume that this is simplified to lowest terms, since that can
b
a
obviously be done with any fraction. Notice that in order for
a/b
to be in simplest
terms, both of
a
and
b
cannot be even. One or both must be odd. Otherwise, we
could simplify further.
b
a
From the equality = , we can write 2 =
a
2/
b
2, or
a
2 = 2 ·
b
2. So the square of
2b
a
a
is an even number since it is 2 times a number.
From this we know that
a
itself is also an even number, because it can't be odd; if
a
itself was odd, then
a
·
a
would be odd too. Odd number times odd number is always
odd.
If
a
itself is an even number, then
a
is 2 times another whole number. So,
a
= 2k
where k is this other number. Now here is the proof by contradiction
If we substitute
a
= 2k into the original equation 2 =
a
2/b2, this is what we get:
2
=
(2k)2/b2
2
=
4k2/b2
2*b2
=
4k2
b2
=
2k2
This means that
b
2 is even, from which follows again that
b
itself is even, thus
contradicting the premise
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Document Summary

2 is irrational a b where a . 2 is not zero, and a and b are both not even . (proof by contradiction) We additionally assume that this obviously be done with any fraction. Notice that in order for terms, both of could simplify. From the equality is an even number since it is 2 times a number. , we can write 2 = a and further. a/b . 2 a b a b a b is simplified to lowest terms, since that can to be in simplest cannot be even. So the square of a itself is also an even number, because it can"t be odd; if would be odd too. From this we know that a itself was odd, then a a odd. If where k is this other number. = 2k into the original equation 2 = itself is an even number, then is 2 times another whole number. This means that b contradicting the premise.

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