# Prologue - Problem 1

We have to prove the following assertions:

(i) If $ax = a$ for some number $a\ne0$, then $x = 1$. \begin{align*} ax &= a \\ aa^{-1}x &= aa^{-1} && \text{| P8} \\ x &= 1 && \text{| P7} \end{align*}

(ii) $x^2 - y^2 = (x - y)(x + y)$. \begin{align*} (x - y)(x + y) &= x^2 + xy - yx - y^2 && \text{| P9} \\ &= x^2 + xy - xy - y^2 && \text{| P8} \\ &= x^2 - y^2 && \text{| P3} \end{align*}

(iii) If $x^2 = y^2$, then $x = y$ or $x = -y$. \begin{align*} x^2 &= y^2 \\ x^2 - y^2 &= 0 && \text{| P3} \\ (x + y)(x - y) &= 0 && \text{| see (ii)} \end{align*} Now we have 2 options. First: \begin{align*} (x + y)(x - y)(x - y)^{-1} &= 0(x - y)^{-1} \\ (x + y) &= 0 && \text{| P7 and 0a = 0} \\ x &= -y && \text{| P3} \end{align*} Second: \begin{align*} (x + y)(x - y)(x + y)^{-1} &= 0(x + y)^{-1} \\ (x - y) &= 0 && \text{| P8, P7 and 0a = 0} \\ x &= y && \text{| P3} \end{align*}

(iv) $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$. (From here on I will only occasionally reference the central properties.) \begin{align*} (x - y)(x^2 + xy + y^2) &= x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 \\ &= x^3 - y^3 \end{align*}

(v) $x^n - y^n = (x - y)(x^{n - 1} + x^{n - 2}y + ... + xy^{n - 2} + y^{n - 1})$. This one is a bit tricky, because we have to be aware that the right-hand side of the equation is generated by the following sum: $(x - y)\sum_{j = 0}^{n - 1}x^jy^{n - 1 - j}$ Let's take $n = 3$ as an known example: $\sum_{j = 0}^{2}x^jy^{2 - j} = y^2 + xy + x^2 \text{ | } x^0 = 1$ This is equivalent to what we have in (iv). With that we can now rewrite the problem to: \begin{align*} (x^n - y^n) &= (x - y)\sum_{j = 0}^{n - 1}x^jy^{n - 1 - j} \\ &= \sum_{j = 0}^{n - 1}x^{j + 1}y^{n - 1 - j} - \sum_{k = 0}^{n - 1}x^ky^{n - k} \\ &= x^n + \sum_{j = 0}^{n - 2}x^{j + 1}y^{n - 1 - j} - y^n - \sum_{k = 1}^{n - 1}x^ky^{n - k} \\ &= x^n - y^n + (\sum_{j = 0}^{n - 2}x^{j + 1}y^{n - 1 - j} - \sum_{k = 1}^{n - 1}x^ky^{n - k}) \end{align*} We now have to show that $(\sum_{j = 0}^{n - 2}x^{j + 1}y^{n - 1 - j} - \sum_{k = 1}^{n - 1}x^ky^{n - k}) = 0$. \begin{align*} \sum_{j = 0}^{n - 2}x^{j + 1}y^{n - 1 - j} - \sum_{k = 1}^{n - 1}x^ky^{n - k} &= 0\\ \sum_{j = 0}^{n - 2}x^{j + 1}y^{n - 1 - j} &= \sum_{k = 1}^{n - 1}x^ky^{n - k} && \text{| } k - 1 = j \\ &= \sum_{j = 1 - 1}^{n - 1 - 1}x^{j + 1}y^{n - (j + 1)}\\ &= \sum_{j = 0}^{n - 2}x^{j + 1}y^{n - 1 - j} \end{align*}

(vi) $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$. Replacing $y$ with $-y$ produces: $x^3 + (-y)^3 = (x - y)(x^2 + xy + y^2)$ That is identical to (iv).