Prove, from the given axioms for the real numbers, that for real numbers \(a\), \(b\), \(c\), \(d\)

  1. \(a(bc) = c(ba)\);
  2. \(-(a+b) = (-a) + (-b)\);
  3. if \(ab = ac\) and \(a \neq 0\), then \(b = c\);
  4. if \(a < b\) and \(c < d\) then \(a + c < b + d\);
  5. if \(a \leq b\) and \(c \leq d\) then \(a+c = b+d\) only if \(a = b\) and \(c = d\).
  1. \begin{align*}a(bc)&=a(cb)&\text{(· is commutative)}\\&=(cb)a&\text{(· is commutative)}\\&=c(ba)&\text{(· is associative)}\end{align*}
  2. First we prove that $0·a=0$. Indeed by distributive axiom we have $0·a=(0+0)·a=0·a+0·a$, then by adding the additive inverse of $0·a$ to both sides we find $0·a=0$. Now we have $(-1)·a=-a$, in fact: $0=(1+(-1))·a=a+(-1)·a$, so $(-1)·a=-a$. Finally $(-a)+(-b)=(-1)·a+(-1)·b=(-1)·(a+b)=-(a+b)$.
  3. $ab=ac\Rightarrow a(b-c)=0$, and $a\ne0$, so $b-c=0$, so $b=c$.
  4. $b-a>0$ and $d-c>0$, so $(b-a)+(d-c)>0$, so $a+c\lt b+d$.
  5. $a+c=b+d\Rightarrow a-b=d-c$ but $a-b\le0$ and $d-c\ge0$, we have $a-b=d-c=0$, so $a=b$ and $c=d$.


Prove the following assertions, for real numbers \(a\), \(b\), \(c\):

  1. if \(a < b\), then \(ac > bc\) if and only if \(c < 0\);
  2. \(a^2 + b^2 = 0\) if and only if \(a = b = 0\);
  3. \(a^3 < b^3\) if and only if \(a < b\).
  1. ⇒: If $a\lt b$ and $c\lt 0$, we have $-(a-b)\gt 0$ and $-c\gt 0$, so $-(a-b)(-c)\gt 0$, so $ac\gt bc$.
    ⇐: If $a\lt b$ and $ac\gt bc$, we have $(a-b)c\gt 0$ and $-(a-b)^{-1}\gt 0$, so $-(a-b)^{-1}(a-b)c\gt 0$, so $c\lt 0$.
  2. ⇒: If $a=b=0$, we have $a^2=b^2=0$, so $a^2+b^2=0$.
    ⇐: If $a^2+b^2=0$, we assume $a\ne 0$, then $a^2>0$, and $b^2=0$, so $a^2+b^2=0$, a contradiction, so $a=0$, similarly $b=0$.
  3. ⇒: If $a^3\lt b^3$, we have $(a-b)(a^2+ab+b^2)=a^3-b^3\lt0$, and $a^2+ab+b^2=\frac12(a^2+b^2+(a+b)^2)\ge0$, so $a-b\lt0$, so $a\lt b$.
    ⇐: If $a\lt b$, we have $a^2+ab+b^2\ge0$, so $(a-b)(a^2+ab+b^2)=a^3-b^3\lt0$, so $a^3\lt b^3$.


  1. Prove that \((a^m)^{-1} = (a^{-1})^m\), for all \(a \in \mathbb{R}\setminus \{0\}\) (for \(m = 1\), \(2\), ...).
  2. Prove that \(a^{k+1} = a^k a\) for \(a \neq 0\) and \(k = -1\), \(-2\), \(-3\), ....
  3. Derive the law of indices: \(a^m a^n = a^{m+n}\) for \(a \neq 0\) and \(m\), \(n \in \mathbb{Z}\).
  1. Use induction on $m$. For $m=1$, both sides are $a^{-1}$. Assume the proposition is true for $m-1$, we have
    $(a^{-1})^ma^m=a^{-1}(a^{-1})^{m-1}a^{m-1}a=a^{-1}(a^{m-1})^{-1}a^{m-1}a=a^{-1}a=1\Rightarrow(a^{-1})^m=(a^m)^{-1}$. By induction, the theorem is true for all $m\ge1$.
  2. $a^{k+1}=(a^{-k-1})^{-1}=(a^{-1})^{-k-1}=(a^{-1})^{-k}(a^{-1})^{-1}=(a^{-k})^{-1}a=a^ka$.
  3. Induct on $n$. For $n=0$, $a^ma^0=a^m=a^{m+0}$. Assume the proposition is true for $n-1$, then $a^ma^n=a^ma^{n-1}a=a^{m+n-1}a$, by definition and part b, $a^{m+n-1}a=a^{m+n}$, so $a^ma^n=a^{m+n}$. By induction, the proposition is true for $m\in\mathbb Z$ and $n\in\mathbb Z^{\ge0}$. For $n\lt0$, $a^{m+n}a^{-n}=a^m$, so $a^{m+n}=a^m(a^{-n})^{-1}=a^ma^n$.


For \(n = 1\), \(2\), \(3\), ..., let $a_n = \left(1 + \frac{1}{n}\right)^n$ and $b_n = \left(1 + \frac{1}{n}\right)^{n+1}$.

  1. Show that the inequality \(a_n \leq a_{n+1}\) can be rearranged as $\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1} \geq \frac{n}{n+1}$. By applying Bernoulli's inequality to the left-hand side, verify this inequality.
  2. Show that \(b_{n+1} \leq b_n\) for all \(n\).
  3. Show that \(a_n \leq a_{n+1} \leq b_{n+1} \leq b_n\) for all \(n\). Deduce that \(a_n < 3\) for all \(n\).
  1. $a_n\le a_{n+1}\Leftrightarrow \frac{(n+1)^n}{n^n}\le\frac{(n+2)^{n+1}}{(n+1)^{n+1}}\Leftrightarrow\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1} \geq \frac{n}{n+1}$. By Bernoulli's inequality, $\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}=\left(1-\frac1{(n+1)^2}\right)^{n+1}\geq1-\frac1{(n+1)^2}\cdot(n+1)=\frac{n}{n+1}$.
    Here is another proof that $\{a_n\}$ increases:\begin{align*} \left(1+\frac1n\right)^n &= 1+1 +\dots + \binom{n}{p} \frac {1}{n^p} + \dots + \frac{1}{n^n}\\ &=1+\dots +\left(1-\frac1n\right)\dots\left(1-\frac{p-1}n\right)\frac{1}{p!} + \dots + \frac{1}{n^n}\\ \left(1+\frac1{n+1}\right)^{n+1} &= 1+\dots +\left(1-\frac1{n+1}\right) \dots \left(1-\frac{p-1}{n+1}\right)\frac{1}{p!} + \dots + \frac{1}{(n+1)^{n+1}}.\end{align*}The last sum has one more (positive) term than the preceding term and the $p$th term of the last sum is larger than the preceding $p$th term since each factor is larger (subtract $k/(n+1)$ from 1 as opposed to subtracting $k/n$). Hence$$\left(1+\frac{1}{n+1}\right)^{n+1} > \left(1+\frac{1}{n}\right)^n.$$
  2. $b_{n+1}\le b_n⇔\frac{(n+2)^{n+2}}{(n+1)^{n+2}}\le\frac{(n+1)^{n+1}}{n^{n+1}}⇔\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+1}\ge\frac{n+2}{n+1}⇔\left(1+\frac1{n(n+2)}\right)^{n+1}\ge1+\frac1{n+1}$.
    By Bernoulli's inequality, it suffices to prove $\frac1{n(n+2)}\ge\frac1{(n+1)^2}$, which is clearly true.
    Here is another proof that $\{b_n\}$ decreases:
    It is equivalent to that $\left(1-\frac{1}{n}\right)^n < \left(1-\frac{1}{n+1}\right)^{n+1}$ holds for $n\in\mathbb N^+$
    $\left(1-\frac{1}{n}\right)^n<\left(1-\frac{1}{n+1}\right)^{n+1}$$\Leftrightarrow \left(\frac{n-1}{n}\right)^n < \left(\frac{n}{n+1}\right)^{n+1} $$\Leftrightarrow \left(\frac{n-1}{n}\right)^n \left(\frac{n+1}{n}\right)^{n+1} \lt 1$ $\Leftrightarrow \left(\frac{n^2 -1}{n^2}\right)^n \left(\frac{n+1}{n}\right)^1 \lt 1$ $\Leftrightarrow \left(\frac{n^2 -1}{n^2}\right)^{\frac{n}{n+1}} \left(\frac{n+1}{n}\right)^{\frac{1}{n+1}} \lt 1$
    By weighted AM-GM , $\left(\frac{n^2 -1}{n^2}\right)^{\frac{n}{n+1}} \left(\frac{n+1}{n}\right)^{\frac{1}{n+1}} \lt \frac{n^2 - 1}{n^2} \times \frac{n}{n+1} + \frac{n+1}{n} \times \frac{1}{n+1} = 1$, and we are done.
  3. $\frac{b_n}{a_n}=1+\frac1n>1\Rightarrow b_n>a_n$, combining with part a and b, we see $a_n\le a_{n+1}\le b_{n+1}\le b_n$ for all $n$.
    We know $b_i$ is an upper bound for $\{a_n\}$ for all $i\in\mathbb Z^{\ge1}$ but $b_5=\left(1+\frac15\right)^6\approx2.9860\lt3$⇒$a_n\lt$3
    Another proof:
    Using binomial theorem, $\displaystyle\left(1+\frac{1}{k}\right)^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$
    When $r\gt1$, we have $\displaystyle{k\choose r}\frac{1}{k^r}=\frac{k!}{r!(k-r)!}\frac{1}{k^r}=\frac{1\left(1-\frac{1}{k}\right)\left(1-\frac{2}{k}\right)\cdots\left(1-\frac{r-1}{k}\right)}{r!}\lt \frac{1}{r!}\le\frac{1}{r(r-1)}$
    Thus, $\displaystyle\left(1+\frac{1}{k}\right)^k\lt1+1+\frac1{1⋅2}+\frac1{2⋅3}+\frac1{3⋅4}+\cdots+\frac1{k(k+1)}=3-\frac1{k+1}\lt3$


Define \(\max(a,b)\), the maximum of two real numbers \(a\) and \(b\), saying which axiom(s) show that your specification is well defined. Using your definition, prove that \(\displaystyle \max(a,b) = \frac{1}{2}(a+b) + \frac{1}{2}|a-b|\), and write down an analogous formula for \(\min(a,b)\).

$\max(a,b):=\begin{cases}a&a\ge b\\b&a\lt b\end{cases}$
The law of trichotomy shows that $a\ge b$ and $a\lt b$ cannot both be true, so the specification is well-defined.
When $a\ge b$, we have $a-b\ge0$, so $|a-b|=a-b$, so $\max(a,b)=a=\frac12(a+b)+\frac12(a-b)$; when $a\lt b$, we have $|a-b|=-a+b$, so $\max(a,b)=b=\frac12(a+b)+\frac12(-a+b)$.
$\displaystyle \min(a,b) = \frac{1}{2}(a+b) - \frac{1}{2}|a-b|$