Prove of Statement, Calculus, Basic Properties of numbers

[c.teixeira]

hi there!

If ab > 0, then (a > 0 and b > 0) or (a < 0 and b < 0). This statement I can prove, just with the basic properties of numbers!

Then, 1$/$b is defined as b$^{-1}$ right?

So, how does one prove that if $\frac{a}{b}$ > 0, then (a > 0 and b > 0) or (a < 0 and b < 0)?

Can you give me the complete proog of that? Thanks!

For example, how does one prove that if $\frac{x+1}{x-1}$ > 0, then

x > 1 or x < -1?

Regards,

 

[micromass]

So $\frac{a}{b}>0$ implies that $ab^{-1}>0$. Now apply the property you could prove.

 

[c.teixeira]

So $\frac{a}{b}>0$ implies that $ab^{-1}>0$. Now apply the property you could prove.

So (a > 0 and b$^{-1}$ >0) or (a < 0 and b$^{-1}$ < 0 ). That is easy of course, but how do I know what b$^{-1}$ is?

In the example I made, how do I relate the variavle x, with that fact (x-1)$^{-1}$ < 0, or (x-1)$^{-1}$ > 0?

 

[micromass]

So (a > 0 and b$^{-1}$ >0) or (a < 0 and b$^{-1}$ < 0 ). That is easy of course, but how do I know what b$^{-1}$ is?

In the example I made, how do I relate the variavle x, with that fact (x-1)$^{-1}$ < 0, or (x-1)$^{-1}$ > 0?

I don't understand your problem. Do you mean that you can't prove that b>0 if $b^{-1}>0$??

 

[c.teixeira]

I don't understand your problem. Do you mean that you can't prove that b>0 if $b^{-1}>0$??

No, that is not what I am saying. I clearly understand the ab$^{-1}$ > 0.

Well, I better just try to explain my self using the example.

So,we have, $\frac{x-1}{x+1}$ > 0 $\Leftrightarrow$ (x-1)(x+1)$^{-1}$ > 0. So [(x-1) > 0 $\vee$ (x+1)$^{-1}$ > 0 ] $\wedge$ [(x-1) < 0 $\vee$ (x+1)$^{-1}$ < 0 ]. Right?

My question is what good is it to know that for example (x+1)$^{-1}$ > 0 ?

How do I relate this with x < something or x > something?

If I was trying to solve this without this without all the ( proof. based on properties), I would just say that if $\frac{x-1}{x+1}$ > 0, then ( x-1) > 0 and ( x+1) > 0( or the other way around).

Is my doubt clear now?

regards,

cteixeira

 

[c.teixeira]

No, that is not what I am saying. I clearly understand the ab$^{-1}$ > 0.

Regarding my last post.

Actually I think that is it! How can I prove that b > 0 given b$^{-1}$ > 0 ?

 

[micromass]

If b<0, then from $b^{-1}>0$ would follow $bb^{-1}<0$ or 1<0. This is a contradiction. So b>0 (since b=0 is clearly not allowed).

 

[c.teixeira]

If b<0, then from $b^{-1}>0$ would follow $bb^{-1}<0$ or 1<0. This is a contradiction. So b>0 (since b=0 is clearly not allowed).

That was exactly my doubt! Thanks, my question is answered!

So, b and b$^{-1}$ have always to have the same sign (given any b), right?

Regards,

 

[micromass]

So, b and b$^{-1}$ have always to have the same sign (given any b), right?

Right!