Verifying proof for the following question

Question

I wanted to check my proof if it is correct or not. Using these field axioms:

(i)Trichotomy Property: Exactly one of $x<y$, $y<x$, or $x=y$ hold.

(ii)Transitivity: if $x<y$ and $y<z$ (which we could write in shorthand as $x<y<z$), then $x<z$.

(iii)If $x<y$ then $x+z < y+z$.

(iv) If $x<y$ and $z>0$ then $xz<yz$.

• (A1) Addition is commutative
• (A2) Addition is associative
• (A3) Addition has a neutral element $0$
• (A4) Any element has an additive inverse
• (A5) Multiplication is commutative
• (A6) Multiplication is associative
• (A7) Multiplication has a neutral element $1$
• (A8) Any non-zero element has a multiplicative inverse
• (A9) Multiplication distributes over addition

1. Prove that for all $a,\, b,\, c\in\mathbb{R}$, if $a<b$ and $c<0$ then $bc < ac$.

Since $a<b$, by (A4), we have that $$ 0 = a + (-a) < b + (-a) = b - a.$$ Since $b - a > 0$ and $c < 0$, then $c(b-a) < 0$. By (A9), this is equivalent to $cb-ca<0$. By (A3), $$ cb=cb-ca+ca<0+ca=ca.$$ Thus $cb<ca$.

Answer 1

Your proof has a gap. How do you know that $b-a \gt 0 \land c \lt 0 \Rightarrow c(b-a) \lt 0$? I'm not saying you're wrong -- I'm just saying you haven't yet proved it.

Edit to add:

Lemma: If $c \lt 0 \land x \lt y, \text{ then } xc \gt yc$.

Proof: $c+(-c) = 0 \land c \lt 0 \Rightarrow -c \gt 0$. (Use property (iii), adding $-c$ to both sides of the inequality.)

Thus, $-xc \lt -yc$. Add $xc + yc$ to both sides: $yc \lt xc$ and the Lemma is proved.

Answer 2

"Since b−a>0 and c<0, then c(b−a)<0"

Why do you state this. Either this is what you wish to prove, or (and I suspect this) you are using iv) but not saying so.

So you should say so.

Since $(b-a)> 0$ and $c < 0$ then by iv) $c(b-a) < 0\cdot (b-a)$

But notice. You don't have an axiom that $0\cdot x = 0$(!!!!)

You have to prove that.

Hint 1: $0d = (0+0)d$

Hint 2: $-0d$ must exist.

....

Once you prove that you have

$c(b-a) <0$

So by A9 $c(b-a) = c(b+(-a)) = cb +c(-a)$

So $cb + c(-a) < 0$ and

....

Now we need to prove $c(-a) = -ca$.

Hint: $c(-a) + ca = c(-a + a)$.

...

Now we have $cb - ca <0$ and ... you can finish your proof.