idk if this is a troll post or not but anyways........... Let us assume that $\sqrt{2}$ is rational. Let $\sqrt{2} = \frac{p}{q}$ where $p,q$ are co-prime numbers and $q \neq 0$ $\sqrt{2} = \frac{p}{q}$ On squaring both sides we get : $2 = (\frac{p}{q})^2$ \begin{align} \implies 2q^2 = p^2 \end{align}$\frac{p^2}{2} = q^2$ So $2 | p$ and $p$ is a multiple of $2$. $\implies p = 2m$ \begin{align} p^2 = 4m^2 \end{align}From equations $(1)$ and $(2)$ we get $2q^2 = 4m^2$ $\implies q^2 = 2m^2$ $\implies $q^2$ is a multiple of $2$. $\implies $q$ is a multiple of $2$ Hence $p,q$ have a common factor $2$. Hence $\sqrt{2}$ is an irrational number. $\blacksquare$

The problem can be proved with descente-infinite too.

Com10atorics wrote: Prove that $\sqrt 2$ is irrational. Assume that $\sqrt{2}$ is rational. Consider root of the equation $x^2-2=0$. From rational root theorem, the rational root of this equation can only be $1,2,-1$ or $2$ which is impossible because $1^2-2=-1$ $2^2-2=2$ $(-1)^2-2=-1$ $(-2)^2-2=2$ Therefore, there is no rational root of $x^2=2$ but $\sqrt{2}$ is the root of this equation and it is rational. Contradiction $\sqrt{2}\not\in\mathbb{Q}$ $\Box$

Com10atorics wrote: Prove that $\sqrt 2$ is irrational. Assume (p/q)²=2 where p/q is a rational no. => p²=2q² => p is even => p=2r for some positive integer r => 4r²= 2q² => 2r²=q² => (q/r)² = 2 => q/r = √2 = p/q And we can clearly see that p>q>r such that √2=p/q=q/r And now i can claim that there exist a positive integer s such p>q>r>s such that √2 = p/q =q/r=r/s but here we have p,q,r,s € N so we can't do this construction bcoz this decreasing sequence has to stop somewhere and thus we arrive to an contradiction which means √2 isn't rational

Amlan6 wrote: Com10atorics wrote: Prove that $\sqrt 2$ is irrational. Assume (p/q)²=2 where p/q is a rational no. => p²=2q² => p is even => p=2r for some positive integer r => 4r²= 2q² => 2r²=q² => (q/r)² = 2 => q/r = √2 = p/q And we can clearly see that p>q>r such that √2=p/q=q/r And now i can claim that there exist a positive integer s such p>q>r>s such that √2 = p/q =q/r=r/s but here we have p,q,r,s € N so we can't do this construction bcoz this decreasing sequence has to stop somewhere and thus we arrive to an contradiction which means √2 isn't rational Wasn't the exact same proof posted by @Bratin_Dasgupta?

Suppose that $a$ and $b$ are positive integers such that $\sqrt 2= \dfrac ab \rightarrow 2b^2=a^2$ The prime factorization of $b^2$ is in the form $b^2= 2^{2k_0}\cdot 3^{2k_1}\cdot 5^{2k_2}\cdots$ with $k_i$ non-negative integers. That is because any positive integer $n$ is a square if and only if all the exponents in his prime factorization are even. The prime factorization of $2b^2$ is $2b^2=2(2^{2k_0}\cdot 3^{2k_1}\cdot 5^{2k_2}\cdots)=2^{2k_0+1}\cdot 3^{2k_1}\cdot 5^{2k_2}\cdots$ This can't be a square because the exponent of 2 is odd, so we reach a contraddiction! In general, suppose we have $\sqrt x= \dfrac ab \rightarrow xb^2=a^2$ If $x$ is a perfect square, the exponents in its prime factorization will all be even, and so will be also the exponents in the prime factorization of $xb^2$ (since we are summing only even exponents when multiplying the two). Otherwise, if $x$ is not a square, at least one of the exponents in its prime factorization will be odd, meaning at least one of the exponents in the prime factorization of $xb^2$ will also be odd. So $\sqrt x$ is rational if and only if $x$ is a perfect square.

starchan wrote: Amlan6 wrote: Com10atorics wrote: Prove that $\sqrt 2$ is irrational. Assume (p/q)²=2 where p/q is a rational no. => p²=2q² => p is even => p=2r for some positive integer r => 4r²= 2q² => 2r²=q² => (q/r)² = 2 => q/r = √2 = p/q And we can clearly see that p>q>r such that √2=p/q=q/r And now i can claim that there exist a positive integer s such p>q>r>s such that √2 = p/q =q/r=r/s but here we have p,q,r,s € N so we can't do this construction bcoz this decreasing sequence has to stop somewhere and thus we arrive to an contradiction which means √2 isn't rational Wasn't the exact same proof posted by @Bratin_Dasgupta? ..

Suppose not. We know that $\sqrt{2}>0$ so suppose $\sqrt{2}=\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. So $p^2=2q^2$. Case 1: $q$ is odd. Then $p^2\equiv2\pmod4$, a contradiction. Case 2: $q$ is even. Then $p$ is also even, which implies $p$ and $q$ are not relatively prime.

what ? I mean Kosovo people solve these kind of olympiad problems ?? this used to be given in our school tests