Relationship between inequalities in proofs

[Seacow1988]

Hi,

Could you clarify the relationship between proofs that use ≤ and those that use <?

For example, if it's already proven that "abs(b) ≤ a if and only if -a≤ b≤a" can we say this implies that "abs(b) < a if and only if -a< b<a"? It seems that since the first statement holds for all abs(b) "less than or equal to" a, its application could narrowed to the case where abs(b) was simply "less than" a.

Thanks!

 

[LCKurtz]

Hi,

Could you clarify the relationship between proofs that use ≤ and those that use <?

For example, if it's already proven that "abs(b) ≤ a if and only if -a≤ b≤a" can we say this implies that "abs(b) < a if and only if -a< b<a"? It seems that since the first statement holds for all abs(b) "less than or equal to" a, its application could narrowed to the case where abs(b) was simply "less than" a.

Thanks!

It happens to be so for that example. But you can't just change them willy-nilly.

For example consider this statement:

If a function f which is continuous on 0 ≤ x ≤ 1 satisfies the inequality f(x) < 1 for all x with 0 < x < 1, then f(1) ≤ 1.

You can't change the conclusion to f(1) < 1 without making the statement false.

 

[Unit]

Consider the fact that for any real number a in R, -|a| ≤ a ≤ |a| is true. It would be false it we replaced "≤" with "<". The symbol "≤" between two numbers x and y is logically equivalent to the statement "x < y OR x = y".