To prove that [tex]\( 3 \sqrt{5} \)[/tex] is irrational, let's proceed with a proof by contradiction.
Assume that [tex]\( 3 \sqrt{5} \)[/tex] is rational. If [tex]\( 3 \sqrt{5} \)[/tex] is rational, it can be written as a ratio of two integers in the form [tex]\( \frac{a}{b} \)[/tex], where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are integers with no common factors (other than 1) and [tex]\( b \neq 0 \)[/tex].
[tex]\[ 3 \sqrt{5} = \frac{a}{b} \][/tex]
Next, isolate [tex]\( \sqrt{5} \)[/tex] by dividing both sides of the equation by 3:
[tex]\[ \sqrt{5} = \frac{a}{3b} \][/tex]
Now, [tex]\( \frac{a}{3b} \)[/tex] is also a ratio of two integers, indicating that [tex]\( \sqrt{5} \)[/tex] is rational. This means that [tex]\( \sqrt{5} \)[/tex] can be written as the ratio of two integers, which implies [tex]\( \sqrt{5} \)[/tex] is a rational number.
However, it is a well-known mathematical fact that [tex]\( \sqrt{5} \)[/tex] is irrational. The irrationality of [tex]\( \sqrt{5} \)[/tex] has been proved via various methods, including proof by contradiction similar to the one we are using here.
Since our initial assumption that [tex]\( 3 \sqrt{5} \)[/tex] is rational led us to the false conclusion that [tex]\( \sqrt{5} \)[/tex] is rational, our initial assumption must be incorrect. Therefore, we conclude that [tex]\( 3 \sqrt{5} \)[/tex] is irrational.
So, [tex]\( 3 \sqrt{5} \)[/tex] is indeed irrational.
