%\addcontentsline{toc}{section}{Activity 3.2}
\section*{Activity 3.2}

\begin{enumerate} \item \begin{enumerate}[$\rhd$] \item To prove that $\sqrt{2}$ is irrational let us suppose that $\sqrt{2}$ is rational, then there exist two natural number $m$ and $n$ such that $\sqrt{2} = \frac{n}{m}$ where $m \neq 0$ and the greatest common divisor of $m$ and $n$ is 1. Now, $\sqrt{2} = \frac{n}{m}$ implies $2 = \frac{n^2}{m^2}\ \Rightarrow n^2 = 2m^2$. The number $n^2$ is even because is equal to 2 times something, also $n$ is even, since even number times even number is always an even number. This means that $n$ can be written as $2k$ for some $k\in \textit{N}$. Substituting $n = 2k$ in $2 = \frac{n^2}{m^2}$ yield $2 = \frac{4k^2}{m^2} \Rightarrow m^2 = 2k^2$, …show more content…
\end{enumerate} \item There are infinitely many prime numbers. To prove this suppose there are finitely many prime numbers, let us say there are $n$ of them, it means we can list them as $\{p_{1},p_{2},...,p_{n}\}$. Now, $p = p_{1}\times p_{2}\times ...\times p_{n} + 1$ is larger than any of the primes and $p$ is not a prime number. Thus $p$ must be divisible by at least one of our prime(s). But if we divide $p$ by one of $p_{i}$ for $i\in \{1;2;...n\}$ we get a reminder 1, therefore this contradict our assumption that there are finitely many prime. Hence, there are infinitely many prime numbers. \item Suppose that $\log_{2}5$ is rational, then there exist two positive integers $a$ and $b$ such that $\log_{2}5 = \frac{a}{b}$ where $b\neq 0$ and the greatest common divisor of $a$ and $b$ is 1. From $\log_{2}5 = \frac{a}{b}$ it follows that $2^{\frac{a}{b}} = 5 \Rightarrow 2^a = 5^b$, and this means that $a = b = 0$ but $b$ cannot be 0. Therefore, $\log_{2}5$ is irrational. \item Pythagoras triples are generated by formula $z^2 = x^2 + y^2$, where $x = 2mn;\ y = m^2 - n^2$ and $z = m^2 + n^2$.\\ LHS: $(m^2 + n^2)^2 = m^4 + 2m^2n^2 +