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

Click For Summary
SUMMARY

The discussion centers on the mathematical implications of adding a constant \( b \) to two real numbers \( s_n \) and \( S_n \) while preserving their relative inequality. It establishes that if \( s_n < A < S_n \), then adding \( b \) results in \( s_n + b < A + b < S_n + b \), thus confirming that \( s'_n < A + b < S'_n \). The conversation also emphasizes the need for precision in mathematical statements and the importance of separating inequalities for clarity. The context involves the inner and outer rectangles of an integral of a function \( f(x) \).

PREREQUISITES
  • Understanding of real numbers and inequalities
  • Familiarity with integral calculus and the concept of rectangles in integration
  • Knowledge of mathematical proofs and logical reasoning
  • Basic understanding of functions and their transformations
NEXT STEPS
  • Study the properties of inequalities in real analysis
  • Learn about the application of integrals in calculus, specifically the use of upper and lower sums
  • Explore the concept of transformations in functions and their effects on inequalities
  • Investigate the role of constants in mathematical proofs and their implications
USEFUL FOR

Mathematicians, students of calculus, and anyone interested in the properties of inequalities and integrals in real analysis.

ice109
Messages
1,707
Reaction score
6
if

s^{&#039;}_{n} = s_n + b

and

S^{&#039;}_{n} = S_n + b

and

s_n &lt; A &lt; S_n

does that imply that

s^{&#039;}_{n} &lt; A+b &lt; S^{&#039;}_{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^{&#039;}_{n} and S^{&#039;}_{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^{&#039;}_{n} and S^{&#039;}_{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^{&#039;}_{n} and S^{&#039;}_{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&lt; A&lt; S_n and add b to each part, s_n+ b&lt; A+ b&lt; S_n+ b so s&#039;_n&lt; A+ b&lt; S&#039;_n.

If you want to be more specific, separate s_n&lt; A&lt; S_n into s_n&lt; A and A&lt; S_n. Adding b to each side of those, s_n+ b&lt; A+ b and A+ b&lt; S_n+ b so s&#039;_n&lt; A+ b and A+ b&lt; S&#039;_n which combine to s&#039;_n&lt; A+ b&lt; S&#039;_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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 24 ·
Replies
24
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K