1 Proof \( 1+\sqrt{2} \quad \) is irrational

To prove that \( 1 + \sqrt{2} \) is irrational, we can use a proof by contradiction. Assume, for the sake of contradiction, that \( 1 + \sqrt{2} \) is rational. This means that we can express it as a fraction of two integers: \[ 1 + \sqrt{2} = \frac{p}{q} \] where \( p \) and \( q \) are integers, and \( q \neq 0 \). Rearranging this equation gives: \[ \sqrt{2} = \frac{p}{q} - 1 \] This can be rewritten as: \[ \sqrt{2} = \frac{p - q}{q} \] Since \( p \) and \( q \) are integers, \( p - q \) is also an integer. Therefore, we can denote \( r = p - q \), where \( r \) is an integer. Thus, we have: \[ \sqrt{2} = \frac{r}{q} \] Squaring both sides results in: \[ 2 = \frac{r^2}{q^2} \] Multiplying both sides by \( q^2 \) gives: \[ 2q^2 = r^2 \] This equation implies that \( r^2 \) is even (since it is equal to \( 2q^2 \)). If \( r^2 \) is even, then \( r \) must also be even (since the square of an odd number is odd). Therefore, we can write \( r \) as: \[ r = 2k \] for some integer \( k \). Substituting this back into the equation \( 2q^2 = r^2 \) gives: \[ 2q^2 = (2k)^2 = 4k^2 \] Dividing both sides by 2 results in: \[ q^2 = 2k^2 \] This shows that \( q^2 \) is also even, which means that \( q \) must be even as well. Now we have established that both \( p \) and \( q \) are even. This contradicts the assumption that \( \frac{p}{q} \) is in simplest form (i.e., \( p \) and \( q \) have no common factors other than 1). Since our assumption that \( 1 + \sqrt{2} \) is rational leads to a contradiction, we conclude that: \[ 1 + \sqrt{2} \text{ is irrational.} \]