Answer:
To prove:
[tex]\frac{2\sqrt{3}}{5}[/tex] is irrational number.
Proof:
Let us first assume [tex]\frac{2\sqrt{3}}{5}[/tex] is a rational number.
So, it can be written as:
[tex]\frac{2\sqrt{3}}{5}=\frac{a}{b}[/tex] where [tex]b\neq0[/tex]
Multiplying both sides by 5.
[tex]5\times \frac{2\sqrt{3}}{5}=5\times \frac{a}{b}[/tex]
[tex]2\sqrt{3}=\frac{5a}{b}[/tex]
Dividing both sides by 2.
[tex]\frac{2\sqrt{3}}{2}=\frac{5a}{2b}[/tex]
[tex]\sqrt{3}=\frac{5a}{2b}[/tex]
We see that [tex]\frac{5a}{2b}[/tex] is a rational number as it is in the [tex]\frac{p}{q}[/tex] form but we know [tex]\sqrt3[/tex] is irrational which makes the equation false making it contradicting with our assumption.
Thus our assumption is wrong.
Hence [tex]\frac{2\sqrt{3}}{5}[/tex] is an irrational number.
