# Inequality implication?

1. Apr 28, 2008

### ice109

if

$$s^{'}_{n} = s_n + b$$

and

$$S^{'}_{n} = S_n + b$$

and

$$s_n < A < S_n$$

does that imply that

$$s^{'}_{n} < A+b < S^{'}_{n}$$

?

ahh why can't i delete, i think this is probably obvious

the actual point of contention is if A is the only number between $s_n$ and $S_n$
is A+b the only number between $s^{'}_{n}$ and $S^{'}_{n}$ ? Surely it is but does it obviously follow?

Last edited: Apr 28, 2008
2. Apr 28, 2008

### rock.freak667

If you add b to everything you will get the last inequality. So it doesn't really imply it but you can get it by addition of b.

3. Apr 28, 2008

### Hurkyl

Staff Emeritus
Could you be more precise in what you're asking?

4. Apr 28, 2008

### ice109

$s_n$ and $S_n$ are the inner and outer rectangles of an integral of f(x) and $s^{'}_{n}$ and $S^{'}_{n}$ are similarly for f(x)+b. I want to use the results of the proof of the first integral for the second one.

5. Apr 29, 2008

### HallsofIvy

Yes, as rockfreak667 said, if you start with $s_n< A< S_n$ and add b to each part, $s_n+ b< A+ b< S_n+ b$ so $s'_n< A+ b< S'_n$.

If you want to be more specific, separate $s_n< A< S_n$ into $s_n< A$ and $A< S_n$. Adding b to each side of those, $s_n+ b< A+ b$ and $A+ b< S_n+ b$ so $s'_n< A+ b$ and $A+ b< S'_n$ which combine to $s'_n< A+ b< S'_n$.

6. Apr 29, 2008

### Hurkyl

Staff Emeritus
How are you ordering rectangles? Or did you mean s_n and S_n are their height? So I presume A, s_n, and S_n are all supposed to be real numbers? Then (no matter what their values are!) it cannot possibly be true that A is the only number between s_n and S_n.

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