Question

If 5 is rational and √3 is irrational, then what is 5 + √3 ?

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Answer 1

If 5 is rational and $\sqrt{3}$

is irrational, then 5 + $\sqrt{3}$ is an irrational.

Step-by-step explanation:

Let us assume to the contrary that 5 + √3 is rational.

That is we can find coprime a and b where b ≠ 0 .

such that 5 + √3 = $\frac{a}{b}$.

Therefore 5-a/b= -√3.

Rearranging this equation we get √3 = $\frac{a}{b}$ - 5 = $\frac{a - 5b}{b}$.

Since a and b are integers we get $\frac{a}{b}$ - 5[/tex] is rational and so √3 is rational.

But this contradicts the fact that √3 is irrational.This contradiction has arisen because of our incorrect assumption that 5 + √3 is rational.

So we conclude that 5 + √3 is irrational.

Answer 2

If 5 is rational and $\sqrt{3}$

is irrational, then 5 + $\sqrt{3}$ is an irrational.

Step-by-step explanation:

Let us assume to the contrary that $5 + \sqrt{3}$ is rational.

That is we can find coprime a and b where b ≠ 0 .

such that $5 + \sqrt{3}$ = $\frac{a}{b}$.

Therefore 5 - $\frac{a}{b}$= $-\sqrt{3}$.

Rearranging this equation we get $\sqrt{3}$ = $\frac{a}{b} - 5$ = $\frac{a - 5b}{b}$.

Since a and b are integers we get $\frac{a}{b} - 5$ is rational and so √3 is rational.

But this contradicts the fact that √3 is irrational.This contradiction has arisen because of our incorrect assumption that $5 + \sqrt{3}$ is rational.

So we conclude that $5 + \sqrt{3}$ is irrational.