Question

show that squareroot2 is irrational ​

Answer 1

Assume $\sqrt{2}$ is rational, i.e. it can be expressed as a rational fraction of the form $\frac{b}{a}$, where $a$ and $b$ are two relatively prime integers. Now, since $\sqrt{2}=\frac{b}{a}$, we have $2=\frac{b^2}{a^2}$, or $b^2=2a^2$. Since $2a^2$ is even, $b^2$ must be even, and since $b^2$ is even, so is $b$. Let $b=2c$. We have $4c^2=2a^2$ and thus $a^2=2c^2$. Since $2c^2$ is even, $a^2$ is even, and since $a^2$ is even, so is a. However, two even numbers cannot be relatively prime, so $\sqrt{2}$ cannot be expressed as a rational fraction; hence $\sqrt{2}$ is irrational. $\blacksquare$

Euclid's proof of the infinitude of primes

Assume there exists a finite number of primes $p_1, p_2,\ldots, p_n$. Let $N=p_1p_2p_3...p_n+1$. $N$ is not divisible by any of the known primes since it will leave a remainder of one upon division by any one of them. Thus, $N$ must be divisible by some other prime not in our list, which contradicts the assumption that there is a finite number of primes. $\blacksquare$

Answer 2

Refer to the attachment ^