Since the functions have non-overlapping domains, they are not simultaneous equations, and x can mean different things for each of them.

Interesting

Ok so 4>3. What is your point?

4>3.

What you're describing there is a composite function with the y-value 4 from x-values 1 to 2, and y-value 3 from x-values 3 to 4. In that case it's true that the range of F(x) (which only exists in the domain from 1 to 2) is always greater than the range of G(x) (which only exists in the domain from 3 to 4).

In the first pair of functions, you're describing a composite function with a linear curve of slope 3 from x-values 1 to 30, an undefined curve between 30 and 91, and a linear curve of slope 1 (starting at point 91,91) from x-values 91 and onward. In that case, the range of F(x) is always greater than the range of G(x). But there is no case where a given x is mapped both to F(x) and to G(x) at the same time, so you can't ever conclude that a given value is greater than 3 times that value.

And in any case, you can't compare functions like that, by their domains, or even by their ranges, which only coincidentally (or more likely intentionally by your choice of values) don't overlap in either case here. It's entirely possible that F(x) and G(x) could have non-overlapping domains and overlapping ranges (for example if your first G(x) = 4x instead), in which case it would not always be the case that F(x) > G(x). And they could have overlapping domains, and not intersect, in which case everywhere in the overlapping area would have two y-values mapped to each x-value.

Clearly F(x)>G(x) — Wittgenstein

The expression "F(x)>G(x)" is undefined, because there isn't one x for which both F(x) and G(x) are simultaneously defined.


The expression "F(x)>G(x)" does not evaluate to true or false if either F(x) or G(x) is undefined. Therefore, your premise "F(x)>G(x)" is always undefined. Hence, it is not possible to draw any legitimate conclusion from it.

I can't argue against what you have just written since it is right.

What if ?

Take a look into this

Clearly F(x)>G(x) , hence x>3x. — Wittgenstein

I think this doesn't follow.

f(x) = 7x and g(x) = 2x
if x = 1 for f(x) and x = 2 for g(x) then we have
f(1) =7 and g(2) = 4
f(1) > g(2) but 1 < 2

So, we have a situation where f(x) > g(x) but only when the input values are identical. If not then we can't conclude that f(x) > g(x) always implies the inputs have the exact relationship as the functions themselves.

I restricted

I restricted the domain of F(x)=x and G(x)=3x to [91, inf) and [1,30] respectively. Ofcourse it is a wrong conclusion and it we won't get any contradictions as long as the two functions we are comparing have domains that overlap. — Wittgenstein

Yes, the restriction precludes the conclusion 1 > 3. The inputs aren't identical

I think this doesn't follow. — TheMadFool

It is the starting point itself, F(x)>G(x), that is always undefined. The expression G(x)-F(x) is never defined because there is no overlap in the domains for F(x) and G(x). Absolutely nothing could ever follow from that.

Since the OP writer has deleted/changed the OP and other substantial posts, closed.