Question

proving that 1=2

Answer 1

Way 1:

\[a^2-a^2=a^2-a^2\]\[a(a-a)=(a-a)(a+a)\]\[a\cancel{(a-a)}=\cancel{(a-a)}(a+a)\]\[a=a+a\]\[a=2a\]\[1=2\]

Answer 2

Too obvious.

Answer 3

1=2
1+2=2+1
3=3

so 1=2

Answer 4

Of course. Simply define it to be so and it is! Nice. Very bold.

Answer 5

ty