Imagine that 2 + 2 = 5. Starting with that, can you prove to me that 1 = 1 and that, at the same time, 1 = 42?

Vel.

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Imagine that 2 + 2 = 5. Starting with that, can you prove to me that 1 = 1 and that, at the same time, 1 = 42?

Vel.

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January 18, 2010 at 6:33 pm

everything equals itself so 1=1

on the other hand 0=2+2-4=5-4=1

so 0=0*41=1*41=41

so 1=1+0=1+41=42

basically if two different numbers (2+2=4, and 5) equal eachother then you can do something similar what i did to show that 1=0 and then x=x*1=x*0=0 so the whole number system collapses to one number

January 18, 2010 at 8:00 pm

Yep, I’d give you the points for the second part, and bonus marks for doing it algebraically, but I’m not sure just saying ‘everything equals itself’ can be counted as a proof ðŸ˜€

January 20, 2010 at 10:42 pm

If 2+2 = 5 and 1= 1 and 1 = 42 then

(2+2) * 1 = 5 * 1 and (2 + 2) * 42 = 5 * 42

All whole numbers must equal each other

All whole numbers must exist at the same point.

So all whole numbers must equal one

So 1 = 42

This is a circle with a circumference of 1. Each value is the distance traveled around the edge.