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This question will probably be subjective and somewhat disliked because of this, but I can't get it out of my head. In mathematics something that is often taken for granted and even used as insults to other people is making it look like someone else claims something as ridiculous as "$2+2=5$".


Does there exist any algebraic (easy-to-understand) as well as useful construction which allows us to assign $2+2=5$?


Own work : The thing I have come up with so far is counting steps on an escalator. I take 2 steps two times and end up having traveled a total distance of 5 steps with respect to inertial frame (because escalator is moving). $4$ steps is due to my own motion and $1$ step would be the distance the escalator "boosted" me with. What would such an algebra look like? What would it be called? Would a simple group suffice?

verret
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mathreadler
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4 Answers4

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By the axiomatic definition of the natural numbers, $2$ is the successor of the successor of $0$, and $5$ is the successor of the successor of the successor of the successor of the successor of $0$.

Then using the axioms, $2+2=5$ can be proven to be false, full stop.

Any attempt that you would make will involve other axiomatisations or other definitions of $2$ or $5$ or other definitions of addition, resulting in a formula $2+2=5$ not belonging to standard arithmetic and of no practical interest other than fooling the naives. You could as well write $$\aleph \bigoplus\oint\stackrel *=5.$$


Anyway, it is well known that connecting in series two alternative tension generators with a phase shift of $104.5°$, the amplitude are such that

$$2+2=3.$$

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    You might as well use clock arithmetics instead of mr Peano. There it is easy to prove that 11+2 is 1. No fooling of naiives necessary. – mathreadler Jul 18 '19 at 14:12
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    @mathreadler: which does not prove that $11+2=1$, of course, but that $11+2=1\mod12$. In other words, $\aleph \bigoplus\oint\stackrel *=5$. –  Jul 18 '19 at 14:13
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    @mathreadler That's not quite how clock arithmetic (in actual math) works. 11+2 = 13 in clock arithmetic too, it just so happens that 13 = 1. – Wojowu Jul 18 '19 at 14:14
  • @Wojowu: if you look at it algebraically there exist 12 elements in the set of our group. You can map $\mathbb Z$ to these 12 elements. – mathreadler Jul 18 '19 at 14:15
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$$ 2 + 2 = 5 \quad\text{(mod $1$)} $$

Ruben
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In most areas where numbers are used, $2$ is defined as $1+1$ and $4$ is defined as $1+1+1+1$. One of these, or both, might well be equal to zero (e.g. in the binary field) or we might have $2 = 5$.

To have $4=5$ while still obeying basic addition rules, we would need $1 = 0$:

$$1 + 1 + 1 + 1 = 4 = 5 = 1 + 1 + 1 + 1 + 1,$$

subtract four ones on both sides and you are left with $1=0$.

This is of course quite boring, because having $1 = 0$ implies that everything is equal to zero.

One way to avoid this is getting rid of some basic rules.
For example, who is telling us that $2 + 2 = (1 + 1) + ( 1 + 1)$ should be equal to $4 = 1 + 1 + 1 + 1$. This is given by the associative property, a very basic property that holds in many areas of mathematics. But, on the other hand, this tells us that to get an example of $2 + 2 = 5$, we should find a system that does not follow this rule, meaning that the order in which additions are evaluated might change things.

Your steps on the escalator might be a good example: For every addition we do, we get boosted one step. So doing $2 + 2$ will boost by one step, but doing $((1 + 1) + 1) + 1)$ would boost three times.

Just as a small warning, from an algebraic point of view: The associative property is one of the last to get rid off when studying general structures. You will loose a lot of properties that hold for our usual number systems. In fact, even the definition of $4 = 1 + 1 + 1 + 1$ becomes problematic in this case, because in which order do you add these ones?

Dirk
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Your example asserts that $$(2+2)+1 =5$$ which is correct.

You have taken two steps twice, which is the $(2+2)$ part and the escalator lifted you one step, so the total is $$(2+2)+1 =5$$

As you notice $$2+2=4$$ is still valid.