﻿ Class 10 NCERT Math Solution
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Exercise - 1.3

Question-1 :- Prove that √5 is irrational.

Solution :-
``` Let √5 is a rational number. Therefore, we can find two integers a, b (b ≠ 0) such that √5 = a/b
Let a and b have a common factor other than 1. Then we can divide them by the common factor,
and assume that a and b are co-prime.
a = b√5
squaring both sides
a² = 5b²
Therefore, a² is divisible by 5 and it can be said that a is divisible by 5. Let a = 5k, where k is an integer
(5k)² = 5b²
b² = 5k²
This means that b² is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
Hence, √5 cannot be expressed as p/q or it can be said that √5 is irrational.
```

Question-2 :-  Prove that 3+2√5 is irrational.

Solution :-
``` Let 3+2√5 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
3+2√5 = a/b
2√5 = (a/b) - 3
√5 = 1/2(a/b - 3)
Since a and b are integers, 1/2(a/b - 3) will also be rational and therefore,√5 is rational.
This contradicts the fact that √5 is irrational. Hence, our assumption that 3+2√5 is rational is false.
Therefore, 3+2√5 is irrational.
```

Question-3 :- Prove that the following are irrationals:
(i) 1/√2,  (ii) 7√5, (iii) 6+√2

Solution :-
```(i) 1/√2
Let 1/√2 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
1/√2 = a/b
√2 = b/a
b/a is rational as a and b are integers.
Therefore, √2 is rational which contradicts to the fact that √2 is irrational.
Hence, our assumption is false and 1/√2 is irrational.
```
```(ii) 7√5
Let 7√5 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
7√5 = a/b
√5 = a/7b
a/7b is rational as a and b are integers. Therefore, √5 should be rational.
This contradicts the fact that √5 is irrational. Therefore, our assumption that 7√5 is rational is false.
Hence, 7√5 is irrational.
```
```(iii) 6+√2
Let 6+√2 is rational.
Therefore, we can find two integers a, b (b ≠ 0) such that
6+√2 = a/b
√2 = (a/b)-6
Since a and b are integers, (a/b)-6 is also rational and hence, √2 should be rational.
This contradicts the fact that √2 is irrational.
Therefore, our assumption is false and hence, 6+√2 is irrational.
```
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