Question

prove that a-c√b is irrational where a and c are natural number and b is not a perfect square.​

Answer 1

Answer:

let, a and c are natural number and b is not a perfect square.

now, b is not a perfect square .

so, √b can not be represented as (p/q) ,where q doesn't equal to zero. then √b is a irrational number.

Multiplication of natural number and irrational number is always irrational number.

so, c√b is also irrational number.

Substation of natural number and irrational number is always irrational number.

so,a-c√b is irrational .

hence, a-c√b is irrational where a and c are natural number and b is not a perfect square.

[proved]

Answer 2

Answer:

Step-by-step explanation:

Given number is a-c√b

To prove,

a-c√b is a rational number

Solution:

Since b is not a perfect square, √b is not a rational number. That is √b is irrational

Let us assume the contrary,  a-c√b is rational

Then there exits two numbers 'p' and 'q'  such that

a-c√b = $\frac{p}{q}$, where p and q are coprime and q≠0

a-c√b = $\frac{p}{q}$ ⇒-c√b = $\frac{p}{q}$-a =

⇒-c√b = $\frac{p}{q}$-a = $\frac{p-aq}{q}$

⇒√b = $\frac{p-aq}{-cq}$

Since a and c are natural numbers and p and q are integers

$\frac{p-aq}{-cq}$ is a rational number

Thus √b is a rational number, which contradicts the fact that √b is an irrational number.

Hence our assumption that 'a-c√b is rational' is wrong. That is a-c√b is an irrational number

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