- #1

Mr. Fest

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## Homework Statement

Prove that [itex]\sqrt{6}[/itex] is irrational.

## Homework Equations

## The Attempt at a Solution

[itex]\sqrt{6}[/itex] = [itex]\sqrt{2}[/itex]*[itex]\sqrt{3}[/itex]

We know that [itex]\sqrt{2}[/itex] is an irrational number (common knowledge) and also this was shown in the textbook.

So, let's assume [itex]\sqrt{6}[/itex] and [itex]\sqrt{3}[/itex] are both rational.

[itex]\Rightarrow[/itex] [itex]\sqrt{2}[/itex]*[itex]\sqrt{3}[/itex] = [itex]\frac{a}{b}[/itex]

Also since, [itex]\sqrt{3}[/itex] is assumed to be rational it can be written as [itex]\frac{c}{d}[/itex] [itex]\Rightarrow[/itex] [itex]\sqrt{6}[/itex] = [itex]\sqrt{2}[/itex]*[itex]\frac{c}{d}[/itex] = [itex]\frac{a}{b}[/itex] [itex]\Leftrightarrow[/itex] [itex]\sqrt{2}[/itex] = [itex]\frac{d}{c}[/itex]*[itex]\frac{a}{b}[/itex] which is a contradiction since a, b, c and d are integers and [itex]\sqrt{2}[/itex] is not a rational number but [itex]\frac{d}{c}[/itex]*[itex]\frac{a}{b}[/itex] would be a rational number. Therefore, we can conclude that [itex]\sqrt{6}[/itex] is an irrational number.

Would this solution be correct? Or rather is this solution enough?

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