Question

prove that 3/2√5 is irrational

Answer 1

Let 3/2root5 be rational number
Multiply 3/2root5 by 2/3
( product of rational is always rational)
therefore root5 is rational
this conteadicts the fact that root5 is irrational
so our assumption is wrong 3/2root5 is irrational

Answer 2

$\dfrac{3}{2\sqrt{5}}$ is an irrational number.

Explanation:

Rational number: If a number can be defined as p/q where, p and q are integers and q≠0, then it is called a rational number.

Irrational number: If a number can not be defined as p/q where, p and q are integers and q≠0, then it is called an irrational number.

The given number is

$\dfrac{3}{2\sqrt{5}}$

We need to prove that 3/2√5 is irrational.

Rationalize the denominator.

$\dfrac{3}{2\sqrt{5}}\times \dfrac{\sqrt{5}}{\sqrt{5}}$

$\dfrac{3\sqrt{5}}{2(5)}$

$\dfrac{3\sqrt{5}}{10}$

$0.3\sqrt{5}}$

Note: Prove $\sqrt{5}$ is an irrational number.

Let $\sqrt{5}$ is a rational number. So, it can written in the form of p/q where p and q are distinct integers.

$\sqrt{5}}=\frac{p}{q}\Rightarrow 5q^2=p^2$

It means 5 is the factor p² ⇒ 5 is factor of p.

That is, p=5a

⇒$5q^2=(5a)^2$

⇒$5q^2=25a^2$

⇒$q^2=5a^2$

It means 5 is the factor q² ⇒ 5 is factor of q.

It conclude that p and q are not distinct. Which is a contradiction.

Hence $\sqrt{5}}$ in an irrational number.

Here, 0.3 is a rational number and $\sqrt{5}}$ in an irrational number.

$0.3\sqrt{5}}$ is an irrational number because we know that product of a rational number and an irrational number is always an irrational number.

Hence $\dfrac{3}{2\sqrt{5}}$ is an irrational number.

#Learn more

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