# If we assume m is the largest, finite number, what happens to mathematics?

So I've been thinking a lot about the philosophy of mathematics and I'm afraid my lack of formal training in the subject has led me to a lot of questions, so sorry in advance if this is a bad one!

Basically, I understand that mathematics is made up of axioms that we kind of just assume to be true because they work, and that different kinds of maths are made up of different kinds of axioms (like Euclidean and non-Euclidean geometry).

So my question is related to axioms of number theory. What happens to number theory or mathematics generally if we assume that there is a largest number, let us call it m, for example.

Basically, I mean that if you add 1 to m it still equals m, but m << ∞, i.e. it is finite. I know this doesn't work, but I'm curious about how it doesn't work and why it doesn't work? Does it lead to mathematical paradoxes, for example?

Thank you so much, I really appreciate any responses!

• There's a wikipedia page for that. en.wikipedia.org/wiki/Finitism – MCCCS Oct 19 '17 at 17:33
• Since $m$ is finite, it is invertible. But $m=m+1$, so you get $1=0$. – E. Joseph Oct 19 '17 at 17:34
• If m+1=m and m is a number you can do arithmetic with, then you can subtract m from both sides to get 1=0. Then since false implies anything, "If 1=0 then [any statement]" is true. And since we have 1=0, we can use that to prove literally any valid statement. – Mark S. Oct 19 '17 at 17:35
• As another way of putting E. Joseph's and Mark S.'s comments, you would have to abandon the cancellation property of addition, $a + c = b + c \Rightarrow a = b$. – Daniel Schepler Oct 19 '17 at 17:48
• Why would it start a chain of reaction? You will potentially build a different arithmetic, potentially not reflecting the reality. Then you will have to prove your arithmetic's axioms are consistent (not leading to statements which are true and false at the same time) – rtybase Oct 19 '17 at 19:22