Intro to proofs in real analysis

Question

Hey I am m having trouble proving this OFC10:

If $x < y$ and $z < 0$, then $yz < xz$.

I started with $x+z < y+0$ by OFC8, but right now I'm stuck. Any suggestions?

Note: OFC10 means Ordered Field Consequence 10.

Answer 1

$z<0$ means $0<-z$ and $x<y$ means $0<y+(-x)$.

You may want to prove $(-a)b = -(ab), \,(-a)(-b)= ab$ beforehand, but you may also prove them directly in your proof:

The product of two positive numbers is positive, thus

\begin{align} & 0<(-z)(y+(-x)) \\ & = (-z)y + (-z)(-x) \\ \end{align} You can jump to the last line if you've already proved the aforementioned statements, otherwise: \begin{align} & = (-z)y + (-z)(-x) + -(zy) + zy \\ & = y(z+(-z)) + (-z)(-x) + -(zy)\\ & = y\cdot0 + (-z)(-x) + -(zy) \\ & = y\cdot0 + (-z)(-x) + -(zy)+ y\cdot0 + -(y\cdot0) + x\cdot0 + -(x\cdot0) \\ & = y\cdot(0+0) -(y\cdot0)+ (-z)(-x) + -(zy) + x\cdot(0+0) + -(x\cdot0) \\ & = y\cdot0 -(y\cdot0)+ (-z)(-x) + -(zy) + x\cdot0 + x\cdot0 -(x\cdot0) \\ & = (-z)(-x) + -(zy) + x\cdot0\\ & = (-z)(-x) + -(zy) + x(z+(-z))\\ & = (-z)(-x) + -(zy) + (-z)x+zx \\ & = (-z)(-x + x) + -(zy) + zx \\ &= (-z)\cdot0 + (-z)\cdot0 + -((-z)\cdot0) + -(zy) + zx \\ &= (-z)\cdot(0+0)+ -((-z)\cdot0) + -(zy) + zx \\ & = -(zy) + zx \end{align}

which by definition means, $zy<zx$.

(Note that this proof can be shortened a lot if we already knew $a\cdot0=0$.)

Here, I've used these axioms of $\mathbb R$.

Note that $x\in \mathbb {R}^+\iff x>0$. Using the definition of $x>y$ as $x+(-y)\in \mathbb R^+$, this shouldn't be hard to prove.

Answer 2

Hint: If $x-y < 0$ and $z<0$ then $(x-y)z>\ldots$?