question 1 and folder structure

q1/q1.md

@@ -0,0 +1,18 @@
+## Question 1
+
+> Prove that $\sqrt[3]{361}$ is irrational.
+
+To determine whether the cube root of 361 is irrational, we need to examine whether it can be expressed as a rational number, which is a number that can be written as the quotient of two integers (where the denominator is not zero).
+Let us assume that $\sqrt[3]{361}$ is a rational number. As such, $\sqrt[3]{361}=\frac{p}{q}$, where p,q are co-prime integers, and $q\neq0$ (definition of a rational number).
+
+Taking the cube of both sides of this equation gives
+
+$$361&=\left(\frac{p}{q}\right)^{3} \\ &=\frac{p^{3}}{q^{3}} \\ p^3 &=361q^3 &(1)$$
+
+Since p and q are integers, $p^3$ and $q^3$ are also integers.
+
+Consider the prime factorisation $361=19\times19$.  19 is prime, appears twice in the factorisation, and also is the only number in the factorisation. It is not in the form $\frac{p^{3}}{q^{3}}$. Therefore, there is no way to express 361 as the cube of a rational number. 
+
+This contradicts our original assumption that $\sqrt[3]{361}$ is rational. 
+
+Hence, by contradiction, we can conclude that $\sqrt[3]{361}$ is irrational.