num property, hard and official one

Hi there!

emilianoIT wrote: To come to my conclusion i did divide both side by the same amount t2, not by t1/t2...

Your question was the following:

p1*(t1/t2) > p2
Since t1/t2 > 1 --> then p1 > p2.

The hypothesis is in red, the thesis is in blue. If you had divided both sides (of the hypothesis) by t2, you would get p1*(t1/t2^2) > p2/t2 and this inequality is not the one you would like to obtain (as the thesis). In other words, you did not divide both sides by t2.

In our situation we are dividing by t2 that is a sales tax vale, and it must be a positive value since it is a sales tax. How can a sales tax be negative?

t2 cannot be negative, for sure.

You used "general" a, b, c and d and with them you made a statement. Nobody must assume you mean d=t2... My obligation (to all readers) is to be careful, therefore for your statement to be true (once you substitute the word "equation" by "inequality"), the fact that (a "general") d must be positive is IMHO sufficiently important to be mentioned.

Hi Fabio and thanks again for your quick answer!

Actually my hypothesis was the following:

(1) t1 > t2
(2) p1*t1 > p2*t2

that is the one stated by the problem and which equals to the following:

(1) t1/t2 > 1
(2) p1*(t1/t2) > p2

in which (1) is obtained by dividing each side of the inequality by t2, and (2) is obtained by dividing each side of the inequality by t2.

Now my point is the following:
Is it true that if we divide both sides of an inequality by the same positive value the result is still true?

Thanks man,
E.

emilianoIT wrote: Is it true that if we divide both sides of an inequality by the same positive value the result is still true?

Yes, it is true. Look below:

Proposition 01: if a>b and c>0 then ac > bc

Proof:

a>b implies a-b > 0

From the fact that c>0 then (a-b).c >0 (product of 2 positive numbers is positive)

Therefore ac - bc > 0, hence ac > bc.


Proposition 02: if a>b and c>0 then a/c > b/c

Proof: c>0 implies 1/c >0 therefore by Proposition 01 we know that a*(1/c) > b*(1/c)

...
(1) t1/t2 > 1
(2) p1*(t1/t2) > p2

..., and (2) is obtained by dividing each side of the inequality by t2.

Understood (and THIS is correct), but please read my FIRST post: you cannot conclude from (1) and (2) above that p1 > p2 , as I have shown through the counter-example: t1 = 110, t2 = 100 , p1 = 100 and p2 = 105.

Please note that a>b>0 and c>d>0 does NOT imply a/c > b/d ... Got it?

Cheers,
Fabio.

P.S.: now I guess I understood the misunderstanding...

When you´ve asked this:

emilianoIT wrote:p1*(t1/t2) > p2

Since t1/t2 > 1 --> then p1 > p2.

Am I wrong?

I thought you´ve asked if you were wrong to conclude the red part. (You were). You´ve probably asked if you were right to conclude the bold part. (You were.)

Hi Fabio!

I got your example:

Understood (and THIS is correct), but please read my FIRST post: you cannot conclude from (1) and (2) above that p1 > p2 , as I have shown through the counter-example: t1 = 110, t2 = 100 , p1 = 100 and p2 = 105.

Even if p1*t1 > p2*t2, doesn't mean that p1 > p2. Clear!

Thanks again man,

E.

That´s it, E.

(My pleasure!)

Regards,
Fabio.