AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 1 Real Numbers Ex 1.4 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 1st Lesson Real Numbers Exercise 1.4

### 10th Class Maths 1st Lesson Real Numbers Ex 1.4 Textbook Questions and Answers

Question 1.

Prove that the following are irrational,

i) \(\frac{1}{\sqrt{2}}\)

ii) √3 + √5

iii) 6 + √2

iv) √5

v) 3 + 2√5

Answer:

i) \(\frac{1}{\sqrt{2}}\)

On the contrary, suppose that is a \(\frac{1}{\sqrt{2}}\) rational number;

then \(\frac{1}{\sqrt{2}}\) is of the form where \(\frac{p}{q}\) and q are integers.

∴ \(\frac{1}{\sqrt{2}}\) = \(\frac{p}{q}\)

⇒ \(\frac{\sqrt{2}}{1}\) = \(\frac{q}{p}\)

(i.e.,) √2 is a rational number and it is a contradiction. This contradiction arised due to our supposition that \(\frac{1}{\sqrt{2}}\) is a rational number.

Hence \(\frac{1}{\sqrt{2}}\) is an irrational number.

ii) Suppose √3 + √5 is not an irrational number.

Then √3 + √5 must be a rational number.

√3 + √5 = \(\frac{p}{q}\), q ≠ 0 and p, q ∈ Z

Squaring on both sides

but √15 is an irrational number.

\(\frac{p^{2}-8 q^{2}}{2 q^{2}}\) is a rational number

(p^{2} – 8q^{2}, 2q^{2} ∈ Z, 2q^{2} ≠ 0)

but an irrational number can’t be equal to a rational number, so our supposition that √3 + √5 is not an irrational number is false.

∴ √3 + √5 is an irrational number.

iii) 6 + √2

To prove: 6 + √2 is an irrational number.

Let us suppose that 6 + √2 is a rational number.

∴ 6 + √2 = \(\frac{p}{q}\), q ≠ 0

⇒ √2 = \(\frac{p}{q}\) – 6

⇒ √2 = Difference of two rational numbers

⇒ √2 = rational number But this contradicts the fact that √2 is an irrational number.

∴ Our supposition is wrong.

Hence the given statement is true.

6 + √2 is an irrational number.

iv) √5

To prove: √5 is an irrational number.

On the contrary, let us assume that √5 is a rational number.

∴ √5 = \(\frac{p}{q}\), q ≠ 0

If p, q have a common factor, on cancelling the common factor let it be

reduces to \(\frac{a}{b}\) where a, b are co-primes.

Now √5 = \(\frac{a}{b}\), where HCF (a, b) = 1

Squaring on both sides we get

⇒ (√5)^{2} = \(\left(\frac{a}{b}\right)^{2}\)

⇒ 5 = \(\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}\)

⇒ 5b^{2} = a^{2}

⇒ 5 divides a^{2} and thereby 5 divides 5

Now, take a = 5c

then, a^{2} = 25c^{2}

i.e., 5b^{2} = 25c^{2}

⇒ b^{2} = 5c^{2}

⇒ 5 divides b^{2} and thereby b.

⇒ 5 divides both b and a.

This contradicts that a and b are co-primes.

This contradiction arised due to our assumption that √5 is a rational number.

Hence our assumption is wrong and the given statement is true, i.e., √5 is an irrational number,

v) 3 + 2√5

To Prove: 3 + 2√5 is an irrational.

On the contrary, let us assume that 3 + 2√5 is a rational number.

Here p, q being integers we can say that \(\frac{p-3q}{2q}\) is a rational number.

This contradicts the fact that √5 is an irrational number. This is due to our assumption “3 + 2√5 is a rational number”.

Hence our assumption is wrong.

∴ 3 + 2√5 is an irrational number.

Question 2.

Prove that √p + √q is an irrational, where p, q are primes.

Answer:

Given that p, q are primes.

Hence fp and fq are irrationals.

[∵ p, q have no factors other than 1 ∵ they are primes.]

Now √p + √q = sum of two irrational numbers = an irrational number

Hence proved.