Question
Show that $5-\sqrt{3}$ is irrational.
✓
Answer
Let us assume, to the contrary, that $5-\sqrt{3}$ is rational.
That is, we can find coprime numbers a and $b \left( b \neq 0\right.$ ) such that $5-\sqrt{3}=\frac{a}{b}$
Therefore, $5-\frac{a}{b}=\sqrt{3}$
Rearranging this equation, we get $\sqrt{3}=5-\frac{a}{b}=\frac{5 b-a}{b}$
Since a and b are integers, we get $5-\frac{a}{b}$ is rational, and so $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ 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.
That is, we can find coprime numbers a and $b \left( b \neq 0\right.$ ) such that $5-\sqrt{3}=\frac{a}{b}$
Therefore, $5-\frac{a}{b}=\sqrt{3}$
Rearranging this equation, we get $\sqrt{3}=5-\frac{a}{b}=\frac{5 b-a}{b}$
Since a and b are integers, we get $5-\frac{a}{b}$ is rational, and so $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ 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.
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