Question

Show that 5-√3 is irrational.​

Answer 1

$\bf\green{Question:-}$

$\bf{✪ \:Show \:that \:5-√3 \:is \:irrational.}$

$\bf\green{Solution:-}$

Let us assume, to the contrary, that $\bf{5-√3}$ is rational.

That is, we can find co-prime a and b (b is not equal to 0) such that $\bf{5-√3=a/b.}$

Therefore, $\bf{5-a/b=√3.}$

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

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

But this contradicts the fact that $\bf{√3}$ is irrational.

This contradiction has arisen because of our incorrect assumption that $\bf{5-√3}$ is rational.

So, we conclude that $\bf{5-√3}$ is irrational.

$\bf\green{Proved}$

Answer 2

Answer:

Let us assume, to the contrary, that $\sf{5-√3}5−√3$is rational.

That is, we can find co-prime a and b (b is not equal to 0) such that $\sf{5-√3=a/b.}5−√3=a/b.$

Therefore, $\sf{5-a/b=√3.}5−a/b=√3.$

Rearranging this equation, we get $\sf{√3=5-a/b=5b-a/b.}√3=5−a/b=5b−a/b.$

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

But this contradicts the fact that $\sf{√3}√3$ is irrational.

This contradiction has arisen because of our incorrect assumption that $\sf{5-√3}5−√3$is rational.

So, we conclude that $\sf{5-√3}5−√3$ is irrational.