If a+b≥x is known to be true does that mean a+b≥x−1 contradicts it?

[Sage_Rat]

I’m not a mathematician, so this is a question not an answer. But if you have proved the first statement, I would expect it to allow you to say that the second statement is so true, you can even swap the >= sign for a >?

[davidm]

I think what you’re trying to say is (assuming I’m getting the set notation right)

{a+b,…} = {x,…} is contradicted by {a+b,…} = {x-1,…}

If so, then you’re correct but using incorrect notation.

[Your_Great_Darsh_Face]

@Sage_Rat

I’m not a mathematician, so this is a question not an answer. But if you have proved the first statement, I would expect it to allow you to say that the second statement is so true, you can even swap the >= sign for a >?

Indeed you could. But a + b >= x - 1 is still perfectly true even if we have established that the strict equality can never hold.

[md2000]

@md2000

a+b≥x - true

a+b≥x−1 rewritten as:
a+b+1≥x

I don’t see a contradiction, except that the second set allows for a+b to be 1 less than the solution to the first.

So, for x=0 the first solution is any a+b≥0 but the second is any a+b≥-1
The first solution is a subset of the second?

The solution for a+b≥x is a subset of the solution of a+b≥x−1

Therefore any solution that is true for the first is true for the second, but not everything that is true for the second is true for the first.

All circles are ellipses, but not all ellipses are circles.
All ellipses are conic sections, but not all conic sections are ellipses.