Question

Use division method of contradiction to show that $\sqrt{3}$ and $\sqrt{5}$ are irrational numbers. Also find the value of $\sqrt{15}\times \sqrt{3}\times \sqrt{5}$

Answer 1

Suppose for the sake of contradicton that $\sqrt{3}$ is rational.
We know that rational numbers are those numbers which can be expressed in the form $\frac{p}{q}$, where p and q are integers and $q\ne 0$
Hence, $\sqrt{3}=\frac{p}{q}$ where p and q are integers with no factor in common.
Squaring both sides,
$3=\frac{p^2}{q^2}$
$=p^2=3q^2$ -- (1)
That is, since $p^2=3q^2$ , which is multiple of 3, mean p itself must be a multiple of 3 such as $p=3n$.
Now we have that $p^2=({3n)}^2=9n^2$ ---- (2)
From (1) and (2),
$9n^2=3q^2$
$=>3n^2=q^2$
This means, q is also a multiple of 3, contradicting the fact that p and q had no common factors.
Hence, $\sqrt{3}$ is an irrational number.
Suppose for the sake of contradicton that $\sqrt{5}$ is rational.
We know that rational numbers are those numbers which can be expressed in the form $\frac{p}{q}$, where p and q are integers and $q\ne 0$
Hence, $\sqrt{5}=\frac{p}{q}$ where p and q are integers with no factor in common.
Squaring both sides,
$5=\frac{p^2}{q^2}$
$=p^2=5q^2$ -- (1)

That is, since $p^2=5q^2$ , which is multiple of 3, mean p itself must be a multiple of 5 such as $p=5n$.
Now we have that $p^2=({5n)}^2=25n^2$ ---- (2)
From (1) and (2),

$25n^2=5q^2$
$=>5n^2=q^2$
This means, q is also a multiple of 5, contradicting the fact that p and q had no common factors.
Hence, $\sqrt{5}$ is an irrational number.

The correct answer for $\sqrt{15}\times \sqrt{3}\times \sqrt{5}=\sqrt{15\times 15}=15$