Use division method of contradiction to show that √3 and √5 are irrational numbers. Also find the value of √15×√3×√5
Suppose for the sake of contradicton that √3 is rational.
We know that rational numbers are those numbers which can be expressed in the form pq, where p and q are integers and q≠0
Hence, √3=pq where p and q are integers with no factor in common.
Squaring both sides,
3=p2q2
=p2=3q2 -- (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 p2=(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.
Suppose for the sake of contradicton that √5 is rational.
We know that rational numbers are those numbers which can be expressed in the form pq, where p and q are integers and q≠0
Hence, √5=pq where p and q are integers with no factor in common.
Squaring both sides,
5=p2q2
=p2=5q2 -- (1)
That is, since
p2=5q2 , which is multiple of 3, mean p itself must be a multiple of 5 such as p=5n.
Now we have that p2=(5n)2=25n2 ---- (2)
From (1) and (2),
25n2=5q2
=>5n2=q2
This means, q is also a multiple of 5, contradicting the fact that p and q had no common factors.
Hence, √5 is an irrational number.
The correct answer for √15×√3×√5=√15×15=15