# Fake Proof for 2-3=0 [closed]

I get that this doesn't make any sense but I'm not sure what exactly is wrong with it: Let's just let # = infinity |1+2+3...| = |-1-2-3...| so # = #/2 and 2# = # Since 2# = #, 3# = # and 2# = 3# Then, 2/# = 3/# and 2/# - 3/# = 0 2 - 3 = 0# and 2 - 3 = 0

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## closed as unclear what you're asking by Boris Novikov, Stefan Hamcke, hardmath, Pedro Tamaroff♦, EuYuNov 13 '13 at 16:49

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Regular arithmetic breaks down when including $\infty$ in your number system. There's nothing more to it. $\infty$ does not play nicely with the rules of addition and multiplication. I had a recent answer detailing this. See here for my answer. –  Cameron Williams Nov 13 '13 at 16:12
But is there a specific step that is wrong or just the idea in general –  Sam Nov 13 '13 at 16:37
The idea is wrong in general. The regular rules of arithmetic break down when dealing with infinity. –  Cameron Williams Nov 13 '13 at 16:39

They can be added, multiplied but rules are different. $R\cup \{+\infty,-\infty\}$ has monoid structure both with addition and multiplication. –  user52045 Nov 13 '13 at 16:24