How can we prove that 3 plus the square root of 5 is an irrational number?

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To prove that the number \(3 + \sqrt{5}\) is irrational, we can start by examining the definition of rational and irrational numbers.

A rational number is any number that can be expressed as the fraction \(\frac{a}{b}\) where \(a\) and \(b\) are integers and \(b \neq 0\). An irrational number, on the other hand, cannot be expressed in such a way; it has a non-repeating, non-terminating decimal expansion.

We know that \(\sqrt{5}\) is an irrational number. This can be shown by contradiction:

  1. Assume \(\sqrt{5}\) is rational. Then it can be expressed as \(\sqrt{5} = \frac{m}{n}\), where \(m\) and \(n\) are integers with no common factors.
  2. Squaring both sides gives \(5 = \frac{m^2}{n^2}\), which simplifies to \(m^2 = 5n^2\).
  3. This means \(m^2
    dquo; is divisible by 5, and therefore \(m
    dquo; must also be divisible by 5 (since the square of a number is divisible by a prime if and only if the number itself is divisible by that prime).
  4. Let \(m = 5k\) for some integer \(k\). Then, substituting back, we get: \( (5k)^2 = 5n^2 \Rightarrow 25k^2 = 5n^2 \Rightarrow n^2 = 5k^2\).
  5. So, \(n^2
    dquo; is also divisible by 5, which means \(n
    dquo; must be divisible by 5 as well.
  6. Since both \(m
    dquo; and \(n
    dquo; are divisible by 5, this contradicts our assumption that they have no common factors. Thus, our initial assumption that \(\sqrt{5}\) is rational must be false.

Now that we have established that \(\sqrt{5}\) is irrational, we will show that adding a rational number (in this case, 3) to an irrational number (\(\sqrt{5}\)) results in an irrational number.

Let’s assume, for the sake of contradiction, that \(3 + \sqrt{5}\) is a rational number. If this were true, we could express it as:

 3 + \sqrt{5} = \frac{p}{q}

where \(p\) and \(q\) are integers. Rearranging this equation gives us:

 \sqrt{5} = \frac{p}{q} - 3

By putting both terms on the right side over a common denominator, we get:

 \sqrt{5} = \frac{p - 3q}{q}

This implies that \(\sqrt{5}\) can be expressed as a fraction, which contradicts our earlier proof of its irrationality.

Therefore, our assumption that \(3 + \sqrt{5}\) is rational must be incorrect. Thus, we conclude that \(3 + \sqrt{5}\) is indeed an irrational number.

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