Question #c74dd

1 Answer
Nov 3, 2016

Answer:

#sqrt(2)/2# can not be rationalized.

(see below)

Explanation:

Defintion
Rationalization is the expression of a number in as a ratio (or "fraction") of the form: #a/b# where #a# and #b# are integers and have no common factors greater than #1#.

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#color(black)(sqrt(2)/2)# is an irrational number.

Proof:
Suppose that #sqrt(2)/2# were a rational number.
Then, by definition of rational numbers, we could write
#color(white)("XXX")sqrt(2)/2=a/b#
where #a# and #b# were integer values with no common factors.

This would mean
#color(white)("XXX")sqrt(2)b=2a#
Squaring both sides:
#color(white)("XXX")2b^2=4a^2#

#color(white)("XXX")b^2=2a^2#

#color(white)("XXX")rarr b^2# must be even
and since squared odd numbers are odd
#color(white)("XXX")rarr b# must be even.

So we could replace #b# with #2c# for some integer #c#
#color(white)("XXX")sqrt(2)/2=a/(2c)#

#color(white)("XXX")cancel2sqrt(2)c=cancel2a#

#color(white)("XXX")cancel2c^2=cancel4^2a^2#

#color(white)("XXX")rarr a^2# is even

#color(white)("XXX")rarr a# is even.

Therefore both #a# and #b# would need to be even;
that is both #a# and #b# would have a common factor of #2#

...but this is contrary to the initial declaration that #a# and #b# have no common factors.