Supply a proof for the assertion 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.
 A: 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 $(-x>0)$
$$x^2=(-x)(-x)>-x>1$$ 
A: 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 $x^2>1$.
A: 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 $\;x\;$,
$$\calc
    x^2 > 1
\op\equiv\hints{take square root on both sides, using $\;\sqrt{x^2}=\left| x \right|\;$}
         \hints{-- allowed because both sides are non-negative,}
         \hint{and the simplest way I know to go from $\;x^2\;$ to $\;x\;$}
    \left| x \right| > 1
\op\equiv\hint{basic property of $\;\left| \cdot \right|\;$}
    x > 1 \;\lor\; -x > 1
\op\equiv\hint{arithmetic: rewrite RHS -- to match our goal}
    x > 1 \;\lor\; x < -1
\op\when\hint{logic: strengthen -- to match our goal}
    x < -1
\endcalc$$
A: $$x+1<0\implies(x+1)(x-1)=x^2-1>0$$
because $-$ by $-$ gives $+$.
