**Question 1. Prove that √5 is irrational.**

**Solution :**Let us assume, to the contrary, that √5 is rational.

That is, we can find integers a and b (≠ 0) such that :

a | ||

√5 | = | |

b |

So, b √5 = a⋅

Squaring on both sides, and rearranging, we get 5b

^{2}= a

^{2}

[Note: As we know that if p is a prime number and If p divides a2, then p also divides a, where a is a positive integer.]

Therefore, a

^{2}is divisible by 5

Then it a is also divisible by 5.

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

Substituting for a, we get 5b

^{2}= 25c

^{2}, that is, b

^{2}= 5c

^{2}

This means that b

^{2}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 √5 is rational.

So, we conclude that √5 is irrational.

**Question 2. Prove that 3 + 2 √5 is irrational.**

**Solution :**Let us assume, to the contrary, that 3 + 2 √5 is rational.

That is, we can find coprime a and b (b ≠ 0) such that :

a | ||

3 + 2 √5 | = | |

b |

a | ||||

2 √5 | = | − | 3 | |

b |

Or

a | 3 | |||

√5 | = | − | ||

2b | 2 |

Since a and b are integers, we get :

a | 3 | ||

− | is rational | ||

2b | 2 |

⋅ But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our incorrect assumption that 3 + 2 √5 is rational.

So, 3 + 2 √5 is an irrational number.

**Question 3. Prove that the following are irrationals :**

1 | |||||

(i) | (ii) | 7√ 5 | (iii) | 6 + √2 | |

√ 2 |

**Solution (i)**Let us assume, to the contrary, that

1 | ||

is rational | ||

√ 2 |

1 | a | |

= | ||

√ 2 | b |

1×√ 2 | a | |

= | ||

√ 2×√ 2 | b |

Or

√ 2 | a | |

= | ||

2 | b |

Rearranging, we get :

2a | ||

√ 2 | = | |

b |

2a | |

is rational, So √ 2 is rational | |

b |

So, we conclude that -

1 | ||

is irrational | ||

√ 2 |

**Solution (ii)**Let us assume, to the contrary, that 7√5 is rational

That is, we can find coprime p and q (q ≠ 0) such that :

p | ||

7√5 | = | |

q |

p | ||

√5 | = | |

7q |

p | |

is rational, So √5 is rational | |

7q |

So, we conclude that 7√5 is rational

**Solution (iii)**Let us assume, to the contrary, that 6 + √2 is rational.

That is, we can find coprime p and q (q ≠ 0) such that :

p | ||

6 + √2 | = | |

q |

p | ||||

√2 | = | − | 6 | |

q |

Since p and q are integers, we get :

p | ||

− 6 | is rational | |

q |

⋅ But this contradicts the fact that √2 is irrational.

Our assumption that 6 + √2 is rational, is incorrect.

So, 6 + √2 is an irrational number.

## Points to remember

- A number ‘s’ is called irrational if it cannot be written in the form p/q , where p and q are integers and q ≠ 0. Examples : √2, √3, √15, π etc.
- For p being a prime number , If p divides a2, then p also divides a, where a is a positive integer.
- The sum or difference of a rational and an irrational number is irrational and
- The product and quotient of a non-zero rational and irrational number is irrational.

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