Question

Prove that,
√7 is a non-core number

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Answer 1

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Answer 2

Answer:

Let us assume that is rational √7

It can be written as √7 = a/b,

Where 'a' and 'b' are integers in the lowest terms and "b not equals to zero"

If 'a' and 'b' are integers in the lowest terms, it means that donot share any common factors.

So, our equation can be written as

⇒ 7 = a²/b²(by squaring both sides)

⇒ 7b² = a²

Since, both sides are equal, if 7 divides the L.H.S., then it must also divide R.H.S. So we can rewrite 'a' as '7k', where 'k' is some integer and "k is not equals to zero".

⇒ 7b² = (7k)²

⇒ 7b² = 49k²

⇒ b² = 7k²

Similarly, b is also divisible by 7.

Thus, 'a' and 'b' are divisible by 7.

But, 'a' and 'b' are in the lowest terms and donot share any common factors. Thus, we have a contradiction which means our assumption is wrong.

⇒ ${\boxed{\orange{\text{Therefore,√7 is an irrational number.}}}}$