The Proof Of Basic Inequalities I am really curious about the proofs of the following inequalities. I know they are really easy problems but I couldn't prove them by myself.


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*$a,b,c,d \in \mathbb R^+$, with $a<b$ and $c<d$ implies $a \times c < b \times d$

*$0 < a < b$  implies  $0 < a^n < b^n$

*$a < b < 0$  implies  $a^{2n-1} < b^{2n-1} < 0$

*$a < b < 0$  implies  $a^{2n} > b^{2n} > 0$

*$0 < x < 1$  implies  $0 < x^n < x < 1$ ($n\in \mathbb Z^+$)


Thank you for your time and effort.
 A: You're getting into into the logical foundations to define the reals.  One way of defining the real is that it is an ordered Dedekin-complete field.  Take note especially of properties


*

*[A1] if $x ≥ y$ then $x + z ≥ y + z$

*[A2] if $x ≥ 0$ and $y ≥ 0$ then $x\cdot y ≥ 0$.


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*Let's see first that if $0 < a < b$ and $c \in \mathbb R^+$, then $a \cdot c < b \cdot c$.  Note that by [A1], $ b > a $ if only if $ b - a > 0$.  Since $c>0$, this means that by [A2], $(b-a)\cdot c > 0$.  By distributivity it means that $b \cdot c - a \cdot c > 0$, hence $b \cdot c > a \cdot c$

*Using this, your proposition is easy:
$$
a \cdot c < b \cdot c < b \cdot d
$$


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*

*Just use recursion.  If $0 < x < y$ and we assume $x^{n-1} < y^{n-1}$, then using the previous proposition 1., you obtain that $x^n < y^n$ (by having $a = x^{n-1}$, $b = y^{n-1}$, $c = x$ and $d = y$ in that proposition).


*
*

*Observe that, $(-1)\cdot a = -a$ ($a$ times the opposite of the multiplication neutral $1$ equals the opposite of $a$) since
$$a + (-1)\cdot a = a\cdot(1 + (-1)) = 0.$$
But $-a$ is uniquely defined, which implies that $-a = (-1)\cdot a$.

*Similarly observe that $(-1)\cdot(-1) = 1$ since by distributivity,
$$(-1) + (-1)\cdot(-1) = (-1) \cdot (1 + (-1)) = (-1) \cdot 0 = 0.$$
By the unicity of the opposite, this implies that $(-1)^2 = 1$.

*Third observation, $x < y$ implies $-x > -y$.  Indeed $x < y$ if and only if $y - x > 0$.  But then, using the first observation that $-a = (-1)\cdot a$ and properties of fields,
\begin{gather}
(y-x)+(-(y-x)) > -(y-x),\\
0 > -(y-x)\\
0 > (-1)(y-x)\\
0 > (-1)y - (-1)x\\
(-1)x > (-1)y\\
-x > -y
\end{gather}

*Final observation, by [A1], if $a<0$, then $-a>0$.  By the third observation this also means that $a < b < 0$ implies $0 > -b > -a$.

*So by that last observation, we have $0 < -b < -a$.
By proposition 2 this means $(-b)^{2n-1} < (-a)^{2n-1}$.  But by our first observation, $(-b)^{2n-1} = \big((-1)\cdot(b)\big)^{2n-1}$ and by commutativity, $(-b)^{2n-1} = (-1)^{2n-1} \cdot b^{2n-1}$.  By recursively applying our second observation, we conclude that $(-1)^{2b-1} = -1$ so that $(-b)^{2n-1} = - (b^{2n-1})$.

*Putting it all together, we have $0 < -(b^{2n-1}) < -(a^{2n-1})$, which means that $b^{2n-1} > a^{2n-1} > 0$


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*It's simalar to 3.


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*I'll leave the details to you.  Try to show that $0 < x < 1$ iff $ \frac{1}{x} > 1$, that $0 < x < y$ iff $0 < \frac{1}{y} < \frac{1}{x}$, and using your first proposition that $x > 1$ implies $x^2 > x$.



