How do you prove that the square root of 14 is irrational?

2 Answers
Oct 10, 2015

Answer:

A rational number is expressed by ratio of integers.

Explanation:

The only square roots that are rational numbers are those who are perfect squares. #sqrt16# for example is a rational number because it equals 4 and 4 is an integer. #sqrt14#=3.74, which is not an integer and therefore is an irrational number.

Oct 10, 2015

Answer:

Use proof by contradiction...

Explanation:

Suppose #sqrt(14)# is rational.

Then #sqrt(14) = p/q# for some positive integers #p, q# with #q != 0#.

Without loss of generality, we can suppose that #p# and #q# are the smallest such pair of integers.

#(p/q)^2 = 14#

So:

#p^2 = 14 q^2#

In particular, #p^2# is even.

If #p^2# is even, then #p# must be even too, so #p = 2k# for some positive integer #k#.

So:

#14 q^2 = (2k)^2 = 4 k^2#

Dividing both sides by #2#, we get:

#7 q^2 = 2 k^2#

So #k^2# must be divisible by #7#. So #k# must be divisible by #7# too. So #k = 7m# for some positive integer #m#.

#7 q^2 = 2 (7m)^2 = 7*14m^2#

Divide both sides by #7# to find:

#q^2 = 14 m^2#

So #14 = q^2/m^2 = (q/m)^2#

So #sqrt(14) = q/m#

Now #m < q < p#, contradicting our supposition that #p# and #q# are the smallest pair of positive integers such that #sqrt(14) = p/q#.

So our supposition is false and therefore our hypothesis that #sqrt(14)# is rational is also false.