Question

Prove the following as irrational numbers
1.√7
2.3+2√5
3.7√5
4.√3/5​

Answer 1

Answer:

Step-by-step explanation:

1) Let us assume that  

√7  is rational. Then, there exist co-prime positive integers a and b such that

√7 = a/b

⟹a=b  √7

​Squaring on both sides, we get

a  ^2 =7b ^2

 Therefore, a  ^2  is divisible by 7 and hence, a is also divisible by7

So, we can write a=7p, for some integer p.

Substituting for a, we get 49p^2  =7b  ^2

⟹b  ^2  =7p  ^2

. This means, b ^2  is also divisible by 7 and so, b is also divisible by 7.

Therefore, a and b have at least one common factor, i.e., 7.

But, this contradicts the fact that a and b are co-prime.

Thus, our supposition is wrong.

Hence,  

√7  is irrational.

2) Let take that 3+2  

√5  is rational number

→ so, we can write this answer as

⇒3+2√5 = a/b

​Here a & b use two co prime number and b ≠0.

⇒2 √5  = a/ b  −3

⇒2√5  =  a−3b /b

​∴   √5  =  a−3b /2b

Here a and b are integer so a−3b /2b  is a rational number so √5 should be rational number but √5 is a irrational number so it is contradict

Hence 3+2√5 is irrational

3) Let us assume that is 7√5 rational number

Hence 7√5 can be written in the form of a/b where a,b(b≠ 0) are co-prime

⟹7√5=a/b

⟹√5=a/7b

But here √5 is rational and a/7b is irrational

as Rational≠ Irrational

This is a contradiction  

so 7  √5  is a irrational number

4)

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