Here is an elementary proof of the fact that \( \sqrt{2} \) is irrational. But what does that mean? An irrational number is, well, a number that is not rational! So let's begin with a definition of the set of rational numbers \( \mathbb{Q} = \{ \frac{p}{q} | p, q \in \mathbb{Z}, q \neq 0 \}\). As we saw earlier, this is to be read: \( \mathbb{Q} \) is the set of all fractions \( \frac{p}{q} \) such that both \( p \) and \( q \) are integers, \( q \) not being \( 0 \), and with \( p \) and \( q \) in lowest terms. That simply means that they have no prime factor in common. (Let us recall that we can always write every natural number uniquely as a product of a number of prime factors.) What this definition means, at least for us right now, is that we're not going to be looking at numbers like \( \frac{12}{10} \); we're going to assume that all numbers have been reduced to their lowest form, like \( \frac{6}{5} \).
Another minor note: by our definition, a number like \( \frac{1.5}{3} \) would not count as a rational number. This is not true, of course, because we know we can simply rewrite \( 1.5 \) as \( \frac{3}{2} \), and then our number becomes \( \frac{\frac{3}{2}}{3} \), which is the same as \( \frac{1}{2} \), which is clearly a rational number.
So when we say a number is irrational, we're saying it's crazy in some really fundamental way: there really is no way to write it as the ratio of integers. (If you're wondering why we write \( \mathbb{Q} \) for the rational numbers, it's from the word "quotient".)
The proof of \( \sqrt{2} \) being irrational is a proof by contradiction, that is to say that we shall suppose that it is in fact rational, and show that this necessarily leads to a logical impossibility.
So let us indeed suppose, for the sake of contradiction, that \( \sqrt{2} \) is rational. In other words, suppose we can write \( \sqrt{2} = \frac{p}{q} \), where \( p, q \) are non-zero integers that have no prime factors in common. Through elementary algebra, we can rearrange this to \( p^2 = 2q^2 \). Evidently \( p^2 \) is an even number.
Now if \( p^2 \) is an even number, it must be the case that \( p \) is also even. This is because if \( p \) were odd, then \( p^2 \) would also clearly be odd. (This may seem like a strange reason, but it’s actually a basic law of logic: the statement “A implies B” is the same as “not-B implies not-A”. Note that this is completely different from saying “the statement ‘A implies B’ is the same as ‘not-A implies not-B’.” This is not true in general!) Since \( p \) is even, we can write it as \( p = 2n \) for some integer \( n \). In that case, \( p^2 = 4 n^2 \).
But we already have \( p^2 = 2q^2 \). This must mean that \( 4 n^2 = 2q^2 \), or more simply, \( q^2 = 2n^2 \). But this is impossible! If \( q^2 \) were even, then \( q \) would also be even, which would mean that you could write \( q = 2m \) for some integer \( m \). But then \( p, q \) would have a prime factor in common, which violates our initial supposition! It must therefore be the case that our initial supposition was itself false, which means that \( \sqrt{2} \) has to be irrational.
Now you may not be entirely convinced by this sort of proof by contradiction. Suspicion is a good thing when trying to work through math proofs because it is so easy to inadvertently overlook something critical. In particular, it may not be entirely clear why it is so problematic that \( p \) and \( q \) both end up even. (Everything else flows quite naturally and simply, or so I hope!) The answer to that is simple. Suppose \( p, q \) are in fact even. Then, as we saw above, we can write \( p = 2n, q = 2m \). This means that \( \sqrt{2} = \frac{p}{q} = \frac{2n}{2m} = \frac{n}{m} \). But then we are back where we started! We can square both sides and rearrange to discover that \( n^2 = 2m^2 \), but we know by now that this will end up showing that both \( n \) and \( m \) are even. In other words, we would end up looping through this set of equations until kingdom come. This shows us, then, why it simply has to be the case that \( \sqrt{2} \) irrational.
Another minor note: by our definition, a number like \( \frac{1.5}{3} \) would not count as a rational number. This is not true, of course, because we know we can simply rewrite \( 1.5 \) as \( \frac{3}{2} \), and then our number becomes \( \frac{\frac{3}{2}}{3} \), which is the same as \( \frac{1}{2} \), which is clearly a rational number.
So when we say a number is irrational, we're saying it's crazy in some really fundamental way: there really is no way to write it as the ratio of integers. (If you're wondering why we write \( \mathbb{Q} \) for the rational numbers, it's from the word "quotient".)
The proof of \( \sqrt{2} \) being irrational is a proof by contradiction, that is to say that we shall suppose that it is in fact rational, and show that this necessarily leads to a logical impossibility.
So let us indeed suppose, for the sake of contradiction, that \( \sqrt{2} \) is rational. In other words, suppose we can write \( \sqrt{2} = \frac{p}{q} \), where \( p, q \) are non-zero integers that have no prime factors in common. Through elementary algebra, we can rearrange this to \( p^2 = 2q^2 \). Evidently \( p^2 \) is an even number.
Now if \( p^2 \) is an even number, it must be the case that \( p \) is also even. This is because if \( p \) were odd, then \( p^2 \) would also clearly be odd. (This may seem like a strange reason, but it’s actually a basic law of logic: the statement “A implies B” is the same as “not-B implies not-A”. Note that this is completely different from saying “the statement ‘A implies B’ is the same as ‘not-A implies not-B’.” This is not true in general!) Since \( p \) is even, we can write it as \( p = 2n \) for some integer \( n \). In that case, \( p^2 = 4 n^2 \).
But we already have \( p^2 = 2q^2 \). This must mean that \( 4 n^2 = 2q^2 \), or more simply, \( q^2 = 2n^2 \). But this is impossible! If \( q^2 \) were even, then \( q \) would also be even, which would mean that you could write \( q = 2m \) for some integer \( m \). But then \( p, q \) would have a prime factor in common, which violates our initial supposition! It must therefore be the case that our initial supposition was itself false, which means that \( \sqrt{2} \) has to be irrational.
Now you may not be entirely convinced by this sort of proof by contradiction. Suspicion is a good thing when trying to work through math proofs because it is so easy to inadvertently overlook something critical. In particular, it may not be entirely clear why it is so problematic that \( p \) and \( q \) both end up even. (Everything else flows quite naturally and simply, or so I hope!) The answer to that is simple. Suppose \( p, q \) are in fact even. Then, as we saw above, we can write \( p = 2n, q = 2m \). This means that \( \sqrt{2} = \frac{p}{q} = \frac{2n}{2m} = \frac{n}{m} \). But then we are back where we started! We can square both sides and rearrange to discover that \( n^2 = 2m^2 \), but we know by now that this will end up showing that both \( n \) and \( m \) are even. In other words, we would end up looping through this set of equations until kingdom come. This shows us, then, why it simply has to be the case that \( \sqrt{2} \) irrational.
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