Question

(show the supposition, proof and conclusion)

Use proof by contradiction to show that If a and b are rational numbers with b ≠ 0 and x is an irrational number, then a + bx is irrational.

Answer 1

Explanation:

We are given that a and b are rational numbers where
`b\\eq0` and x is irrational number .

We have to prove a+bx is irrational number by contradiction.

Supposition:let a+bx is a rational number then it can be written in
`(p)/(q)` form

`a+bx=(p)/(q)` where
`q\\eq0` where p and q are integers.

Proof:
`a+bx=(p)/(q)`

After dividing p and q by common factor except 1 then we get

`a+bx=(r)/(s)`

r and s are coprime therefore, there is no common factor of r and s except 1.

`a+bx=(r)/(s)` where r and s are integers.

`bx=(r)/(s)-a`

`x=((r)/(s)-a)/(b)`

When we subtract one rational from other rational number then we get again a rational number and we divide one rational by other rational number then we get quotient number which is also rational.

Therefore, the number on the right hand of equal to is rational number but x is a irrational number .A rational number is not equal to an irrational number .Therefore, it is contradict by taking a+bx is a rational number .Hence, a+bx is an irrational number.

Conclusion: a+bx is an irrational number.