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- 1 decade agoFavorite Answer
You can prove it by de-unifying the number scale, that is, dividing by zero, since any number divided by zero is equal to any other number.

But in REAL math, it is impossible.

Here's the divide by zero thing:

0 * (2+2) = 0 * (5) (true)

(0 * (2+2)) / 0 = (0 * (5)) / 0 (divide both sides by zero, and the zeros cancel out)

(2+2) = (5) (so it looks like they equal each other)

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- 1 decade ago
It's possible for 2 + 2 = 5 if you change standard conventions.

If you assume a false proposition, you can deduce anything from it and make it true, for one, so it's logically admissible to say "all cows fly, therefore 2 + 2 = 5".

Also, you can change the meaning of the symbols 2, +, = and 5 so they represent things that will make the statement true. For example, let = be still =, let 2 still be 2, let + mean / and let 5 mean 1. I shouldn't need to say there are an infinity of ways to do this.

Finally, you can suppose it's admissible to divide by cero in some system, which would mean you're allowed to say 1 = 0, so it's easy to deduce that 2 + 2 = 5.

Had fun yet? Unless you can do something useful with 2 + 2 = 5, it's kinda silly to try and force it to be that way. Oh, and by the way, this is a maths question, I don't think physics can answer it.

Source(s): I'm studying to become a mathematician.- Login to reply the answers

- PatLv 51 decade ago
It could be, depending on how you define 2, + ,= , and 5.

What if that 5 became our current 4? So that counting up yields 1,2,3,5,4. There's really nothing stopping you from using this convention besides confusing other people. You can get even crazier and say that + is actually what is known as the subtraction operator, 5 is actually 0 and so 2 + 2 = 5 becomes equivalent to 2 - 2 = 0. It all depends on how you define it.

After all, they're just symbols to an underlying meaning.

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- ridderLv 43 years ago
it somewhat is a sort of programming code, yet incomplete. it potential whilst (2+2 = real) then 5, a variable gets the cost 5 if 2+2 = real, in Perl is any term which has the effect >=a million real and the effect 0=fake, so 2+2 = real (in Perl) then a variable which you probably did no longer point out gets the cost 5

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- 1 decade ago
Unless this is somekind of trick question,

No.

Source(s): MPhys- Login to reply the answers

- Anonymous1 decade ago
possibly, who knows?

anything could be possible.

anything.

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- 1 decade ago
Yes...

I am sure you can prove this with a very lengthy prove. I can't...

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- Anonymous1 decade ago
not really unless u are blind

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