Checking my proof for the following question

Question

I wanted to check my proof if it is correct. 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 $0<a<b$ and $0<c<d$ then $ac<bd$.

Since $a<b$ and $0<c$, then by (iv), $ac<bc$. Moreover, as $c<d$ and $0<b,$ $bc<bc,$ also by (iv). Therefore we can conclude, by (ii) $ac<bd$.

Answer 1

Remember that in this formal axiomatic approach, every little detail matters.

So, when iv) says that if $x<y$ and $z>0$, then $xz<yz$, then that is not the same as $zx<zy$

Accordingly, the result of the second step is $cb<db$, rather than $bc<bd$. You will need to use two applications of A5 to change that into $bc<bd$.

And frankly, I am surprised that there is no explicit axiom that says that $x<y$ if and only if $y < x$, because while you are given that $0<c$, what you really need to apply iv) is $c>0$ ... I think you should point this out in your answer: a good professor will give you extra credit!

Otherwise good.

Answer 2

Property (ii) tells you that $0<b$ (you should mention it).

From $a<b$ and $c>0$, property (iv) tells you that $ac<bc$.

From $c<d$ and $b>0$, property (iv) tells you that $cb<db$. Applying A5 twice, you conclude that $bc<bd$ (you have a typo and don't mention the usage of A5).

Finally, property (ii), together with $ac<bc$ and $bc<bd$, tells you that $ab<bd$.

You could prove as a lemma the property (iv)$'$: if $x<y$ and $z>0$, then $zx<zy$. This is not the same as (iv) and requires applying A5.

Comment. In a less formal proof, the applications of (ii) and A5 would not be mentioned, because they're easy to see. In your assignment, however, you should be fussy about what axioms you're using, because that's the purpose of the exercise.