Question

1. Classify the following numbers as rational or irrational

$\sf \: [ a ] \: (2 - \sqrt{5} ) \\ \sf \: [ b ] \: (3 + \sqrt{23} ) - \sqrt{23} \\ \sf \: [ c ]\frac{2 \sqrt{7} }{7 \sqrt{7} } ] \\ \sf \: [ d ] \frac{\ \textless \ br /\ \textgreater \ 1}{ \sqrt{2} } \\ \sf \: [ e ]2\pi$
Answer need full explanation .

Explain each sub - question .
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Answer 1

Question :-

• To classify the following as rational or irrational numbers

• (2 - √5)

• (3 + √23) - √23

• $\sf \dfrac{2\sqrt{7} }{7\sqrt{7}}$

• $\sf \dfrac{1}{\sqrt{2} }$

• 2π

Answer :-

Rational numbers :-

Rational numbers are those which can be expressed in the form of $\dfrac{p}{q}$, where

p and q are co - primes and q ≠ 0.

Decimals which come under rational numbers are of 2 types :-

• Terminating decimals
• Non terminating and repeating/recurring decimals

Terminating decimals :-

Terminating decimals are those which end after a few digits.

Example :- 0.123467

Non terminating and repeating decimals :-

Non terminating decimals are never ending. Non terminating repeating decimals are those decimals which are never ending but the same digits keep occurring.

Example :- 0.9999999...

Example :- 0.12341234...

Irrational numbers :-

Numbers which cannot be expressed in the form $\dfrac{p}{q}$ where p and q are co - primes and q ≠ 0.

Non terminating non repeating decimals come under this category as these cannot be represented in the fraction form.

Non terminating non repeating decimals :-

Those which are never ending and the digits are non repetitive.

Solution :-

1. (2 - √5)

(2 - √5) is irrational.

Let us see why.

Let us assume  (2 - √5) to be a rational number which can be expressed in the form $\dfrac{p}{q}$, where p and q are co - primes and q ≠ 0.

$\implies \sf (2 - \sqrt{5}) = \dfrac{p}{q}$

Transposing 2,

$\implies \sf \sqrt{5} = \dfrac{p}{q} - 2$

By taking LCM = q,

$\implies \sf \sqrt{5} = \dfrac{p - 2q}{q}$

Here, LHS ≠ RHS as an irrational number cannot be equal to a rational number.

Hence, our assumption that  (2 - √5) is rational is wrong.

Therefore (2 - √5) has to be irrational.

2. (3 + √23) - √23

This number is rational.

Proof :-

By removing the brackets,

⇒ 3 + √23 - √23

Subtracting,

⇒ 3

3 is a rational number.

3. $\sf \dfrac{2\sqrt{7} }{7\sqrt{7} }$

(3) is a rational number.

Proof :-

In this fraction, since the terms are multiplied, by cancelling the common factor √7, we get :-

$\implies \sf \dfrac{2}{7}$

The fraction is expressed in the form $\dfrac{p}{q}$, and both the numerator and denominator are co - primes. q ≠ 0.

4. $\sf \dfrac{1}{\sqrt{2} }$

It is an irrational number.

Proof :-

The denominator satisfices q ≠ 0, but 1 and √2 are not co - prime as √2 is irrational.

5. 2π

2π is irrational.

• We know that π is irrational.

In order to prove 2π is irrational too, let us use the Contradiction method

Proof :-

Let 2π be a rational number which can be expressed in a fraction form and the numerator and denominator are co - primes, and q ≠ 0.

$\implies \sf 2\pi = \dfrac{p}{q}$

Transposing 2,

$\implies \sf \pi = \dfrac{p}{q} - 2$

Taking LCM = q,

$\implies\sf \pi = \dfrac{p - 2q}{q}$

As we already know that π is irrational, an irrational number cannot be equal to a rational number. Hence this contradicts the fact that 2π is rational.

Therefore 2π has to be irrational.