Question

use the method of contradiction to show that root 3 is an rational number

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Answer 1

Answer:

Suppose for the sake of contradiction that √3 is rational.

We know that rational numbers are those numbers which can be expressed in the form p/q, where p and q are integers and q is not equal to 0

Hence,

√3 = p/q

where p and q are integers with no factor in common.

Squaring both sides,

(√3)2 = (p/q)2

3 =p2/q2 -- (1)

That is, since p2 =3q2, which is multiple of 3, mean p itself must be a multiple of 3 such as p=3n.

Now we have that p

2 =(3n)2

=9n2 ---- (2)

From (1) and (2),

9n2 =3q2

=>3n2 =q2

This means, q is also a multiple of 3, contradicting the fact that p and q had no common factors.

Hence,

√3 is an irrational number.

HOPE IT HELPS