# Simple math

1. Nov 6, 2006

### ergonomics

if i have two functions |g(x)|<e, |f(x)|<=1, and i perform the following operation |g(x)|*|f(x)| will the result be |f(x)||g(x)|<e or |f(x)||g(x)|<=e

how do i even go about proving such things?

2. Nov 6, 2006

### HallsofIvy

Staff Emeritus
If x< a, y< b, all numbers positive, then, multiplying on both sides of the first inequality by y, xy< ay. Multiplying on both sides of the second inequality by a, ay< ab. By the transitive law, then, ax< ab.

3. Nov 6, 2006

### ergonomics

Sorry, i forgot to put a question mark on the general question.
if i understand what you are saying correctly, the result of the multiplication of |f(x)|<=1 by |g(x)|<e will be |f(x)||g(x)|<e
not |f(x)|g(x)|<=e ?

4. Nov 6, 2006

### Office_Shredder

Staff Emeritus
Suppose |f(x)|=1, the maximum it can be. Then no matter what g(x) is, you can't quite get to |f(x)||g(x)| = e

So it's a strong inequality. But you can use the weak one if you want, since it certainly is true too

5. Nov 6, 2006

### HallsofIvy

Staff Emeritus
If |f(x)|<= 1 and |g(x)|< e, then |f(x)||g(x)|<= |g(x)|< e.
|f(x)||g(x)|< e.

6. Nov 6, 2006

### ergonomics

thank you
for your help, all of you
i really feel stupid for not seeing it before, i usually go off even if it takes me days to work out problems, but my mind has been awafully loaded lately
i am really sorry for bothering you with such pettiness
again thank you.