Question

Prove √3 is irrational. Hence, show 15+ 17√3 is an irrational number.​

Answer 1

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Let assume √3 is a rational number.

∴ √3= $\frac{p}{q}$                      [p,q are integers, q≠0 and also

Squaring both sides, we get

3=$\frac{p^{2}}{q^{2}}$

⇒p²= 3q²                             ...(i)

⇒2 is a factor of p²

⇒2 is a factor of p also

⇒p is 2m where m is an integer

⇒p²=4m²         [squaring both sides]

⇒2q²=4m²      [from(i)]

⇒ 2 is a factor of q²

⇒2 is a factor of q also.

So, both p and q are even integers and they have their common factor 2, which contradicts the fact that p and q have no common factors.

So, our assumption was wrong.

Hence √2 is irrational.

we know that when a irrational number is divided by a rational, the product is irrational so,

17×√3= 17√3, which is a irrational.

we also know that when a rational is added with irrational, the answer is irrational. so,

15+17√3 is a irrational.

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