Question

prove
$\sqrt{3}$
is a irrational number

Answer 1

Hey User

Here's your answer friend,

==> Let √3 be a rational number.

==> Therefore,

==> √3 = a/b...............{ Where a and b are coprime numbers and b ≠ 0 }

==> Squaring on both sides we get,

==> [√3]² = [a/b]²

==> 3 = a²/b²

==> a² = 3b²..................(1)

==> 3|a²

==> 3|a...............(2)

and therefore,

===> a = 3c...... (a = bc)

==> Squaring on both sides we get,

==> a² = (3c)²

==> a² = 9c²

==> 3b² = 9c² ............(from 1)

==> b² = 3c²

==> 3|b²

==> 3|b..................(3)

From (2) and (3) we get,

==> 3 divides both a and b.

==> But a and b are coprime numbers.

==> Therefore our assumption proved wrong.

√3 is an irrational number.

HOPE IT HELPS YOU :)

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