# HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

## AP Board Class 10 Maths Chapter 1 Real Numbers Ex 1.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 1 Real Numbers Ex 1.4 Book Answers AP Board Class 10 Maths Chapter 1 Real Numbers Ex 1.4 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 1 Real Numbers Ex 1.4 Book Answers

## Andhra Pradesh State Board Class 10th Maths Chapter 1 Real Numbers Ex 1.4 Books Solutions

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## AP Board Class 10th Maths Chapter 1 Real Numbers Ex 1.4 Textbooks Solutions with Answer PDF Download

Find below the list of all AP Board Class 10th Maths Chapter 1 Real Numbers Ex 1.4 Textbook Solutions for PDF’s for you to download and prepare for the upcoming exams:

Question 1.
Prove that the following are irrational,
i) 12√
ii) √3 + √5
iii) 6 + √2
iv) √5
v) 3 + 2√5
i) 12√
On the contrary, suppose that is a 12√ rational number;
then 12√ is of the form where 𝑝𝑞 and q are integers.
∴ 12√ = 𝑝𝑞
⇒ 2√1 = 𝑞𝑝
(i.e.,) √2 is a rational number and it is a contradiction. This contradiction arised due to our supposition that 12√ is a rational number.
Hence 12√ is an irrational number.

ii) Suppose √3 + √5 is not an irrational number.
Then √3 + √5 must be a rational number.
√3 + √5 = 𝑝𝑞, q ≠ 0 and p, q ∈ Z
Squaring on both sides
but √15 is an irrational number.
𝑝2−8𝑞22𝑞2 is a rational number
(p2 – 8q2, 2q2 ∈ Z, 2q2 ≠ 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 = 𝑝𝑞, q ≠ 0
⇒ √2 = 𝑝𝑞 – 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 = 𝑝𝑞, q ≠ 0
If p, q have a common factor, on cancelling the common factor let it be
reduces to 𝑎𝑏 where a, b are co-primes.
Now √5 = 𝑎𝑏, where HCF (a, b) = 1
Squaring on both sides we get
⇒ (√5)2 = (𝑎𝑏)2
⇒ 5 = a2 b2
⇒ 5b2 = a2
⇒ 5 divides a2 and thereby 5 divides 5
Now, take a = 5c
then, a2 = 25c2
i.e., 5b2 = 25c2
⇒ b2 = 5c2
⇒ 5 divides b2 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 𝑝−3𝑞2𝑞 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.
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.

## Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.4 Textbooks for Exam Preparations

Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.4 Textbook Solutions can be of great help in your Andhra Pradesh Board Class 10th Maths Chapter 1 Real Numbers Ex 1.4 exam preparation. The AP Board STD 10th Maths Chapter 1 Real Numbers Ex 1.4 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 10th Maths Chapter 1 Real Numbers Ex 1.4 Books State Board syllabus with maximum efficiency.

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Yes you can get Andhra Pradesh Board Text Book PDF for all classes using the links provided in the above article.

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