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If $x<-1$, then $x^2>1$.

Ok so I know this should be an easy proof but I have tried to work with algebraically and I am running into problems. Obviously when you square a number less than -1 you will get a number larger than 1 but I am having trouble proving it.

Algebraically I have $$x<-1$$ $$x^2<1$$ squaring both sides but you can see that raises some concern with the initial proof.

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    On the right side, when we are squaring ,we are multiplying by a negative number $-1$, so the inequality should flip. – 2017-01-25
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    On a side note, if $f:\mathbb{R}\to\mathbb{R}$ is a function, then $$ \forall x,y\in\mathbb{R}:x$f$ is strictly increasing. The function $x^2$ is not strictly increasing on $\mathbb{R}$, so you should be careful when implying $x^2<(-1)^2$ from $x<-1$. In this case this is a false inference. However, if $0\leq x$x\mapsto x^2$ is a strictly increasing function on $\mathbb{R}_{\geq0}$. Finally, you might want to take a look at [this](https://en.wikipedia.org/wiki/Ordered_field#Properties_of_ordered_fields) – 2017-01-25
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    You tagged this foundations. Asking for a formal proof from a foundations point of view is a _very_ different question than just asking for some mediocre handwaving proof. Which one is it? – 2017-01-26

4 Answers 4

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Remember that if a negative number multiplies an inequality this changes of sense. i.e:

$$x<-1\Rightarrow -x>1$$

Also you know that if $y>1$ then $y^2>y$. Whit this $(-x>0)$

$$x^2=(-x)(-x)>-x>1$$

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We may proceed as follows.

Assume that $x<-1$. Then $x-1<-2<0$ and $x+1<0$. This means that both $x-1$ and $x+1$ are negative. Thus, $$x^2-1=(x-1)(x+1)>0.$$ This implies that $x^2>1$.

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    Excellent choice to use the product of differences. – 2017-01-25
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    @Drew Christensen Thanks--:) – 2017-01-25
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The earlier answers are both more or less based on a 'trick', or some creative insight. Here is another perspective, in an attempt to give a proof with the minimum amount of 'magic' and surprise, and instead trying to be as 'constructive' as possible.$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\Ref}[1]{\text{(#1)}} \newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} $

To prove that $$ x < -1 \;\then\; x^2 > 1 $$ for all $\;x\;$, we will start at the most complex side, so the right hand side, and try towards simplify to the other side.

In other words, we calculate as follows: for all $\;x\;$, $$\calc x^2 > 1 \op\equiv\hints{take square root on both sides, using $\;\sqrt{x^2}=\left| x \right|\;$} \hints{-- allowed because both sides are non-negative,} \hint{and the simplest way I know to go from $\;x^2\;$ to $\;x\;$} \left| x \right| > 1 \op\equiv\hint{basic property of $\;\left| \cdot \right|\;$} x > 1 \;\lor\; -x > 1 \op\equiv\hint{arithmetic: rewrite RHS -- to match our goal} x > 1 \;\lor\; x < -1 \op\when\hint{logic: strengthen -- to match our goal} x < -1 \endcalc$$

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$$x+1<0\implies(x+1)(x-1)=x^2-1>0$$

because $-$ by $-$ gives $+$.