Problem 5

Prove the following:

(i) If $a < b$ and $c < d$, then $a + c < b + d$. With $b - a \in P$ and $d - c \in P$ we get: \[ \begin{align*} b - a + d - c &> 0 \\ b + d - a - c &> 0 \\ b + d &> a + c \end{align*} \]

(ii) If $a < b$, then $-b < -a$. \[ \begin{align*} a &< b \\ 0 &< b - a \\ -b &< -a \end{align*} \]

(iii) If $a < b$ and $c > d$, then $a - c < b - d$. \[ \begin{align*} b - a + c - d &> 0 \\ b - d + c - a &> 0 \\ b - d &> a - c \end{align*} \]

(iv) If $a < b$ and $c > 0$, then $ac < bc$. \[ \begin{align*} b - a &> 0 \\ bc - ac &> 0 \\ bc &> ac \end{align*} \]

(v) If $a < b$ and $c < 0$, then $ac > bc$. \[ \begin{align*} a(-c) &< b(-c) && \text{| based on (iv)} \\ bc &< ac \\ \end{align*} \]

(vi) If $a > 1$, then $a^2 > a$. \[ \begin{align*} a^2 &> a \\ a^2 - a &> 0 \\ a - 1 &> 0 \\ a &> 1 \\ \end{align*} \]

(vii) If $0 < a < 1$, then $a^2 < a$. \[ \begin{align*} a^2 &< a \\ a - a^2 &> 0 \\ 1 - a &> 0 \\ a &< 1 \\ \end{align*} \] When $a = 0$ then $a^2 = a$ which means we have to avoid that case: $0 < a$. What happens when $a < 0$? \[ \begin{align*} a &< 0 \\ a^2 &> 0 \\ a &< a^2 && \text{| for any } a \\ \end{align*} \]

(viii) If $0 \leq a < b$ and $0 \leq c < d$, then $ac < bd$. \[ \begin{align*} b - a &> 0 \\ bc - ac &> 0 \\ bc &> ac \\ \end{align*} \] Then: \[ \begin{align*} d - c &> 0 \\ bd - bc &> 0 \\ bd &> bc \\ \end{align*} \] Both can now be combined through $bc$: \[ \begin{align*} bd &> ac \\ ac &< bd \\ \end{align*} \]

(ix) If $0 \leq a < b$, then $a^2 < b^2$. (Use (viii))

From (viii) we have $ac < bd$ if $a < b$ and $c < d$. If $c = a$ and $d = b$ then we get $aa < bb = a^2 < b^2$.

(x) If $a, b \geq 0$ and $a^2 < b^2$, then $a < b$. (User (ix) backwards.)

$a^2 < b^2 = aa < bb$ and based on (viii) $a = c$ and $b = d$ and with that $c < d$ or $a < b$.

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