NCERT Class 10 Math Chapter 1 Important Questions Answer – Real Numbers

Class 10
Chapter  Real Numbers
Subject Math
Category Important Question Answer

Class 10 Math Chapter 1 Important Question Answer

Q1. Find HCF of 510 and 92. Most Most Important

Ans – The prime factorisation of 510 and 92 gives :

510 = 2×3×5×17

92 = 22 ×23

Therefore, the HCF of these two integers is 2.

Q2. Prove that \displaystyle \sqrt{5} is an irrational number.  Most Important

Ans – Let us assume, to the contrary, that \displaystyle \sqrt{5}  is rational.

That is, we can find integers a and b(≠0) such that \displaystyle \sqrt{5} = \displaystyle \frac{a}{b}

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b \displaystyle \sqrt{5}= a.

Squaring both sides, and rearranging, we get 5b2 = a2 .

Therefore, a2 is divisible by 5, so a is also divisible by 5.

So, we can write a = 5c for some integer c.

Substituting for a, we get 5b2 = 25c2 , that is b2 = 5c2 .

This means that b2 is divisible by 5, and so b is also divisible by 5.

Therefore, a and b have at least 5 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that \displaystyle \sqrt{5} is rational.

So, we can conclude that \displaystyle \sqrt{5} is irrational.

Q3. Prove that \displaystyle \sqrt{3} is an irrational number. Most Important

Ans – Let us assume, to the contrary, that \displaystyle \sqrt{3} is rational.

That is, we can find integers a and b(≠0) such that \displaystyle \sqrt{3} = \displaystyle \frac{a}{b}

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b \displaystyle \sqrt{3}= a.

Squaring both sides, and rearranging, we get 3b2 = a2 .

Therefore, a2 is divisible by 3, so a is also divisible by 3.

So, we can write a = 3c for some integer c.

Substituting for a, we get 3b2 = 9c2 , that is b2 = 3c2 .

This means that b2 is divisble by 3, and so b is also divisible by 3.

Therefore, a and b have at least 3 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that \displaystyle \sqrt{3} is rational.

So, we can conclude that \displaystyle \sqrt{3} is irrational.

Q4. Prove that \displaystyle \sqrt{2} is an irrational number. Most Important

Ans – Let us assume, to the contrary, that \displaystyle \sqrt{2} is rational.

That is, we can find integers a and b(≠0) such that \displaystyle \sqrt{2} = \displaystyle \frac{a}{b} .

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.

So, b \displaystyle \sqrt{2}= a.

Squaring both sides, and rearranging, we get 2b2 = a2 .

Therefore, a2 is divisible by 2, so a is also divisible by 2.

So, we can write a = 2c for some integer c.

Substituting for a, we get 2b2 = 4c2 , that is b2 = 2c2 .

This means that b2 is divisble by 2, and so b is also divisible by 2.

Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that \displaystyle \sqrt{2} is rational.

So, we can conclude that \displaystyle \sqrt{2} is irrational.

Q5. Prove that \displaystyle 3\sqrt{2} is an irrational number. Most Important

Ans – Let us assume, to the contrary, that \displaystyle 3\sqrt{2} is rational.

That is, we can find coprime a and b (≠0) such that \displaystyle 3\sqrt{2}=\frac{a}{b}

Rearranging, we get \displaystyle \sqrt{2}=\frac{a}{{3b}}.

Since 3, a and b are integers, \displaystyle \frac{a}{{3b}} is rational, and so \displaystyle \sqrt{2} is rational.

But this contradicts the fact that \displaystyle \sqrt{2} is irrational.

So, we can conclude that \displaystyle 3\sqrt{2} is irrational.

Q6. Show that \displaystyle 5-3\sqrt{2} is an irrational number.

Ans – Let us assume, to the contrary, that \displaystyle 5-3\sqrt{2} is rational.

That is, we can find coprime a and b (b≠0) such that \displaystyle 5-3\sqrt{2}=\frac{a}{b}.

Therefore, \displaystyle 5-\frac{a}{b}=3\sqrt{2}.

Rearranging this equation, we get \displaystyle \sqrt{2}=\frac{5}{3}-\frac{a}{{3b}}.

Since a and b are integers we get \displaystyle \frac{5}{3}-\frac{a}{{3b}} is rational, and so \displaystyle \sqrt{2} is rational.

But this contradicts the fact that \displaystyle \sqrt{2} is irrational.

This contradiction has arisen because of our incorrect assumption that \displaystyle 5-3\sqrt{2} is rational.

So, we conclude that \displaystyle 5-3\sqrt{2} is irrational.

Q7. Show that 7+ √5 is an irrational number.

Ans – Let us assume, to the contrary, that \displaystyle 7+\sqrt{5} is rational.

That is, we can find coprime a and b (b≠0) such that \displaystyle 7+\sqrt{5}=\frac{a}{b}.

Therefore, \displaystyle 7-\frac{a}{b}=\sqrt{5}.

Rearranging this equation, we get \displaystyle \sqrt{5}=7-\frac{a}{b}.

Since a and b are integers we get \displaystyle 7-\frac{a}{b} is rational, and so \displaystyle \sqrt{5} is rational.

But this contradicts the fact that \displaystyle \sqrt{5} is irrational.

This contradiction has arisen because of our incorrect assumption that \displaystyle 7+\sqrt{5} is rational.

So, we conclude that \displaystyle 7+\sqrt{5} is irrational.

Q8. \displaystyle 3+2\sqrt{5} is a rational number or irrational number.

Ans – \displaystyle 3+2\sqrt{5} is an irrational number.

As we know that the sum of a rational and an irrational number is irrational.

Q9. Express 3825 as a product of prime factors: Most Important

Ans – Prime factorisation of 3825 is

3825 = 3 × 3 × 5 × 5 × 17

= \displaystyle {{3}^{2}}\times {{5}^{2}}\times 17

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