Prove that \sqrt{7} is irrational

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Question

Prove that

\sqrt{7}

is irrational

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Let us assume that

\sqrt{7}

is rational
It can be written as

\sqrt{7} = \frac{a}{b}

where

a

and

b

are integers in the lowest terms and

b \neq, 0

a

and

b

are integers in the lowest terms, it means that donot share any common factors.
So, our equation can be written as

7 = \frac{a^{2}}{b^{2}}

(by squaring both sides)

=&gt;7b^2=a^2

Since, both sides are equal, if

7

divides the L.H.S., then it must also divide R.H.S. So we can rewrite '

a

' as

7 k

, where

k

is some integer and

k \neq 0

=&gt;7b^2=(7k)^2

=&gt;7b^2=49k^2

=&gt;b^2=7k^2

Similarly,

b

is also divisible by

7

Thus,

a

and

b

are divisible by

7

But,

a

and

b

are in the lowest terms and donot share any common factors.

Thus, we have a contradiction which means our assumption is wrong.
Therefore,

\sqrt{7}

is irrational