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Nataly_w [17]
3 years ago
12

(show the supposition, proof and conclusion)

Mathematics
1 answer:
larisa86 [58]3 years ago
5 0

Answer:

Step-by-step explanation:

We are given that a and b are rational numbers where b\neq0 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 \frac{p}{q} form

a+bx=\frac{p}{q} where q\neq0 where p and q are integers.

Proof:a+bx=\frac{p}{q}

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

a+bx=\frac{r}{s}

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

a+bx=\frac{r}{s} where r and s are integers.

bx=\frac{r}{s}-a

x=\frac{\frac{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.

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