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The question I got on my exam was $$\text{If}\;a > 0 \;\text{then}\; (b > 0 \Leftrightarrow ab > 0) $$ The answer I put was "recall the mutiplication rule from the inequality axioms of real numbers". I will put it for reference.

$$a < b \Leftrightarrow ac < bc \;\text{if}\; c > 0 $$ $$a < b \Leftrightarrow ac > bc \;\text{if}\; c < 0 $$ Proof:

This is a direct consequence of the multiplication law.

  • If $b>0$ then if $0 < a \implies ab > 0 $ by the multiplication axiom given.

  • Similarly, if $ ab > 0$ and if $a > 0 \implies b > 0$ by the multiplication axiom again.

I am failing to see how is not a simple proof by using the definition of axiom for this problem. Apparently my professor gave me no points at all for this question and just simply said this is not a proof.

Also I will like to see why this solution is not a proof either:

$$\text{If}\; x \neq 0 \;\text{then}\; x^2 > 0$$

Proof:

If $x \neq 0 $ then x is a positive real number or x is a negative real number

Case 1 : x is positive
then by definition of posistive

$$0 < x \Leftrightarrow 0 < x * x = x^2 $$ if $x > 0$ (By multiplication axiom given)

Case 2: x is negative then by definition of negative $$x < 0 \Leftrightarrow x * x = x^2 > 0 $$ ( By the multiplication axiom again) if $x < 0$

So in both cases $0 < x^2$ and $x^2 > 0$. I am failing to see how these are not correct proofs to these problems when I am simply using the definitions of axioms given. All the variables are for real numbers btw.

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Recker12
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  • I would suggest asking your professor, through email or in person, about this. – 79037662 Oct 09 '19 at 03:58
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    Although your proofs seem basically OK, it is important to keep in mind that Math.SE is not an arbitrator for professor-student disputes. If you find out in more detail what your professor finds lacking in your proofs, then Math.SE may be able to better explain this. Under no circumstances would I recommend that you take something said here to challenge your professor; it is unlikely to provide satisfaction. – Brian Tung Oct 09 '19 at 03:59
  • Your first proof is essentially proving the axiom using the axiom itself. That's circular reasoning and not allowed. When one prove something as fundamentally as that, you need to be very clear 1) what exactly you are using 2) what is the difference between that and what you are going to prove. – achille hui Oct 09 '19 at 04:13

1 Answers1

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There is a missing part after "Similarly". You first need to observe that if $a > 0$, then $a^{-1} > 0$. Then you can use the multiplication rule, by multiplying $ab$ and $a^{-1}$ to get $b$. That is, if $ab > 0$ and $a^{-1} > 0$, then $b =(ab)a^{-1} > 0$.

J.-E. Pin
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