Does Adding a Constant to Two Real Numbers Preserve Their Relative Inequality?

ice109
Messages
1,707
Reaction score
6
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:
Physics news on Phys.org
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.
 
ice109 said:
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?
Could you be more precise in what you're asking?
 
Hurkyl said:
Could you be more precise in what you're asking?

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.
 
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.
 
ice109 said:
s_n and S_n are the inner and outer rectangles
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.
 
Back
Top