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Suppose, I have two inequalities as follows:

$$x^2>1\tag{1}$$ $$y<-1\tag{2}$$

Is it possible to form a new inequality by combining $(1)$ and $(2)$ where the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers?

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  • $\begingroup$ A common axiom of an ordered filed is "if $0 ≤ a $ and $0 ≤ b$ then $0 ≤ a \times b$" which can easily be extended to "if $0 \lt a \lt c$ and $0 \lt b \lt d$ then $0 \lt a \times b \lt c \times d$" so get your inequalities into that form $\endgroup$ – Henry Jul 15 '18 at 20:04
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You can multiply $$x^2>1\tag{1}$$ by a positive number without changing the direction of inequality.

In this case since $$ y<-1\tag{2}$$ we have $$-y>1$$ so you may multiply $$x^2>1$$ by $-y$ to get $$-x^2y>-y>1\implies-x^2y>1 $$ which is the same as multiplying the two inequalities $$-y>1$$ and $$x^2>1$$

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  • $\begingroup$ You could then easily get to the question's the left-hand side would contain $x^2y$ and the right-hand side would contain only numbers with $x^2y \lt -1$ $\endgroup$ – Henry Jul 15 '18 at 20:07
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We have $y<-1\iff -y>1>0$. So $-y>1$ and $x^2>1$. Notice that both sides of these two inequalities are non-negative, thus we can multiply them. So we have $-y \cdot x^2>1 \cdot 1=1$.

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The two basic axioms are:

If $a < b$ then $a + c < b + c$ for all $c$

If $a < b$ and $c > 0$ then $ac < bc$.

From there there are some basic propositions that can be proven with those two axioms:

If $a > 0$ then $-a < 0$.

If $a < b$ and $c < 0$ then $ac > bc$.

And $1 > 0$. and for $c\ne 0$ then $c^2 > 0$.

....

So if $x^2 > 1$ and $y < - 1<0$ then $x^2\cdot y < 1\cdot y = y < -1$.

That's it.

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