# Fraction calculation, can't find mistake

1. Sep 30, 2016

### late347

1. The problem statement, all variables and given/known data

I did some homework and wolfram alpha afterwards gave different result than what I have. Apparently I must have done some error, but I couldn't find error after I doublechecked about two times. So I decided to see if you guys can find an error. And what is the correct answer also. I was pulling my hair out after doing this problem for some time

$$\frac{\frac{1}{s}-s}{\frac{1}{s}-\frac{1}{s-1}} ~~~~~~$$

2. Relevant equations

3. The attempt at a solution

here we expanded
$(\frac{1}{s}-\frac{s^2}{s})/(\frac{s-1}{s(s-1)}-\frac{s}{s(s-1)})$

here I tried to get the plusses and minuses right, but I'm not sure I succeeded
$[\frac{1}{s}+\frac{-s^2}{s}] / [\frac{s-1}{s(s-1)}+\frac{-s}{s(s-1)}]$

$(\frac{1-s^2}{s}) / [\frac{(s-1)-s}{s(s-1)}]$

$[ \frac{1-s^2}{s}] ~~*~~ [\frac{s(s-1)}{-1} ]$

$\frac{(s-1)(s+1)}{s} * \frac{s(s-1}{-1}$

$\frac{(s-1)^2(s+1)(s)}{-s}$

$(-1)(s-1)^2(s+1)$

2. Sep 30, 2016

### Staff: Mentor

What is $(1-s^2)$ factored?

3. Sep 30, 2016

### SammyS

Staff Emeritus
Looks OK.

What's the problem you're having?

4. Sep 30, 2016

### late347

basically $(1-s^2) = (1^2-s^2)$

so that is (1+s) (1-s)

= 1-s +s -s^2
= 1- s^2

5. Sep 30, 2016

### Staff: Mentor

Yes.

6. Sep 30, 2016

### late347

can you bold print my mistake in quote for example

7. Sep 30, 2016

### Staff: Mentor

You wrote
$[ \frac{1-s^2}{s}] ~~*~~ [\frac{s(s-1)}{-1} ] = \frac{(s-1)(s+1)}{s} * \frac{s(s-1}{-1}$
But $(s-1)(s+1) \neq (1-s)(1+s)$

8. Sep 30, 2016

### late347

I think I got it right this time.

(s-1)(s+1)

= s^2 +s -s -1

$=~~ (+s^2~~-1)$

the correct part said that the factors were as follows

$(1-s) (1+s) = 1 +s -s -s^2$
$= -s^2 +1$

the latter case is the "counter number" to the other number (multiplied by minus one)

9. Sep 30, 2016

### SammyS

Staff Emeritus
Oh!

not quite OK !

sign mistake

10. Sep 30, 2016

### Staff: Mentor

The other way around. Until $1-s^2$ everything was fine, and by the way, afterwards, too. You simply changed the sign by substituting $1-s^2 = (s-1)(s+1)$ which is wrong by $-1$. You should have used $1-s^2=(1-s)(1+s)$ or $1-s^2=(-1)(s-1)(s+1)$

11. Sep 30, 2016

### late347

looks like the tricky math prob got under your skin at first also !! xD

12. Sep 30, 2016

### Staff: Mentor

By the way. Me, too. I only found out as I calculated it on my own. At first sight I assumed a false input in Wolfram.

Last edited: Sep 30, 2016
13. Sep 30, 2016

### late347

ok i will try to post the new answer in full, but it will be tedious because of Latex code

14. Sep 30, 2016

### late347

$\frac{1/s-s}{1/s-[1/(s-1)]}$

$\frac{(1-s^2)/s}{(-1)/[s(s-1)]}$

$\frac{(1-s)(1+s)}{s}*\frac{s(s-1)}{-1}$

$\frac{(1-s)(1+s)(s)(s-1)}{(-1)(s)}$

$\frac{(-1)(s-1)(s)(s-1)(s+1)}{-1s}$

$\frac{(s-1)^2(s)(s+1)}{s}$

$(s-1)^2(s+1)$

15. Sep 30, 2016

### SammyS

Staff Emeritus
What does Wolfram give?

Did I somehow miss that in one of these posts?

16. Sep 30, 2016

### late347

I think I did a wrong factorization for the term $(1^2 -s^2)$

evidently wolfram alpha did not become fooled by that.

this skewed my result from early on. I tried to use the memory formula for the $(a^2-b^2)$ for finding factors

17. Sep 30, 2016

### Staff: Mentor

No. The OP hasn't said and I haven't done. Far too many parenthesis to type in.

18. Sep 30, 2016

### late347

code for wolfram alpha

&& inside these &&

&& (1/s-s)/(1/s-1/(s-1)) &&

19. Sep 30, 2016

### late347

looks like I did the problem correctly as post # 14 will show the same alternative form result as wolfram alpha.

$(1-s)(1-s^2)$

or

$(s-1)^2(s+1)$

20. Sep 30, 2016

### Staff: Mentor

I wouldn't have relied on this ambiguous inline coding. I prefer to use parenthesis to avoid false interpretations.
Concerning the sign error all of us blundered into: it happens. Nothing to worry about.