Question

How do you prove very simple things by contradiction? For example:

Prove that \(2 + 2 \ne 5\).

We start by assuming that \(2 + 2 = 5\), but then we can't count sticks and prove by contradiction.

Can anyone guide me through this process?

Answer 1

i dont know what you are looking for
but i would say

2+2=4

4≠5

2+2≠5

Answer 2

But first we have to prove that \(2 + 2 = 4\). How'd we do so?

Answer 3

what can we assume?

Answer 4

Apoorv?

Answer 5

@badreferences ?

Answer 6

We can complete this proof by using Spivak's 12 axioms w/ binary operations, I think. Starting with simple ideas like:\[a=b\iff a+x=b+x\]

I'm a bit lazy, but give me some time and I might be willing to type out a rigorous response. In any case, I'd like to note that the concept of 5 is defined by performing the operations \(((((1+1)+1)+1)+1)\), which you can demonstrate rigorously is not four using the axioms\[a=b\iff a+x=b+x\\(a+b)+c=a+(b+c)\]

Answer 7

One can be defined and rigorously separated from zero using the distributive axiom\[a\cdot(b+c)=ab+ac\\a-0=a\]but they take a bit of work, so I'll have to digress until I feel less lazy.

Answer 8

Hmm. So \(2 + 2\) is defined by \(((1 + 1) + 1) + 1\), and \(5\) is defined by \((((1 + 1) + 1) + 1) + 1\), but then isn't that like counting sticks?

Answer 9

Is there a logical way to get a contradiction in this?

Answer 10

There is an advanced way to get a contradiction, but that relies on Godel's incompleteness theorems which essentially postulates that axioms aren't binary in truthfulness.

For all pragmatic (non-philosophical) purposes, all math is defined by axioms. Without any basic assertions of truth, logical operations cannot be performed with any meaningful consequence. So we use axioms like\[a=b\iff a+x=b+x\]and so on. What I like about these axioms, though, is that they don't really use any number.

Answer 11

I see. Can you prove it for me? :P

Answer 12

I will list Michael Spivak's 12 axioms, and see if you can prove it. How about that? Everything in analytical mathematics can be demonstrated from these twelve axioms.

Answer 13

I'd love to know.

Answer 14

1. \(a+(b+c)=(a+b)+c\)
2. \(a+0=a\)
3. \(a+(-a)=0\)
4. \(a+b=b+a\)
5. \((a\cdot b)\cdot c=a\cdot(b\cdot c)\)
6. \(a\cdot1=a,1\neq0\)
7. \(a\cdot b=b\cdot a\)
8. \(a\cdot a^{-1}=1,\forall a\neq0\)
9. \(a\cdot(b+c)=a\cdot b+a\cdot c\)

The last three are mostly irrelevant (they deal with inequalities), so I'll leave them out.

Answer 15

These are Spivak's axioms. Bertrand Russell also has his own axioms, which I suspect work at a much deeper level. But I haven't studied them in any non-cursory manner, so I don't think I'm qualified to talk about them.

Answer 16

I still can't find a way to prove this in a logical way :P

Answer 17

Hmm, wait! I found one!

Answer 18

\( \color{Black}{\Rightarrow 2 + 2 \Longrightarrow 2 (2)}\) because \(a \times b = a + a + a (\text{b times, b }\in \mathbb{N})\)

Answer 19

As an even number is \(2k\) where \(k \in \mathbb{Z}\), then \(2(2)\) is also an even number.

Answer 20

But we have assumed that \(2 + 2 = 5\), and \(5\) is odd.

Answer 21

I'd say there's an easier method to provide proofs. If we hypothesize \(2+2=5\), we can test this by noting that:\[2+2=5\iff(1+1)+(1+1)=5\]Which is, of course, not true. The only necessary steps are to break down all operations into binary form, because this is how the axioms are demonstrated.

If you actually axiomatically assume that \(2+2=5\), there's not much I can do, because logic works "above" axioms, so to speak.