91Ó°ÊÓ

We are given the number \(\frac{3 \sqrt{2}}{5}\), and we want to show that it is irrational. We will assume, for the sake of contradiction, that this number is a rational number. In other words, let \(\frac{3 \sqrt{2}}{5} = \frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\). Now, let's manipulate this equation to isolate the irrational part, \(\sqrt{2}\).

To isolate \(\sqrt{2}\), we can multiply both sides of the equation by \(\frac{5}{3}\): \[ \frac{3 \sqrt{2}}{5} \cdot \frac{5}{3} = \frac{p}{q} \cdot \frac{5}{3} \] This simplifies to: \[ \sqrt{2} = \frac{5p}{3q} \] Now, let's focus on this equation and try to find a contradiction.

For a quick reminder, \(\sqrt{2}\) is a well-known irrational number. As for a proof of this topic, it has been shown using contradiction that it cannot be expressed as a fraction \(\frac{p}{q}\) with \(p\) and \(q\) having no common factors apart from 1.

From our previous step, we have: \[ \sqrt{2} = \frac{5p}{3q} \] Let's now square both sides of the equation: \[ 2 = \frac{25p^2}{9q^2} \] Next, we multiply both sides by \(9q^2\): \[ 18q^2 = 25p^2 \] We observe that the left side of this equation, \(18q^2\), is an even number because it is divisible by 2, which implies that \(25p^2\) must also be even. Since 25 is an odd number, this means that \(p^2\) must be even. Therefore, \(p\) must also be even. However, if \(p\) is even, that means it can be written as \(p = 2k\) for some integer \(k\). Substituting this back into our equation: \[ 18q^2 = 25(2k)^2 \] Which simplifies to: \[ 18q^2 = 100k^2 \] This means that \(q^2\) must be even (as it is divisible by 2) which in turn implies that \(q\) is also even.

In step 4, we determined that both \(p\) and \(q\) must be even. However, this contradicts our initial assumption that \(\frac{p}{q}\) is in its simplest form, meaning that \(p\) and \(q\) have no common factors apart from 1. Since both \(p\) and \(q\) are even, they share a common factor of 2. This contradiction shows that our initial assumption, that the given number \(\frac{3 \sqrt{2}}{5}\) is rational, must be false. Therefore, we conclude that the given number is indeed irrational.