Question

a=b
1=2
where is the fallacy?

Answer 1

\[a^{2}=b^{2}\]

Answer 2

<shrugs>
What are we to do with this given equation?

Answer 3

Prove it?

Answer 4

let me change the equation

Answer 5

so many solutions

Answer 6

second line is false

Answer 7

LOL...
I think you're holding out on us XD

Answer 8

hmm I'll make it longer

Answer 9

get a real question,

Answer 10

a=b.
a^2=ab
a^2+a^2=a^2+ab
2a^2

wait I'm being called by my mom I'll continue this later
http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html

Answer 11

more simple:
1º: a=b
2º: ab=b^2 (multiply bouth sides by b)
3º: ab-a^2=b^2-a^2 (rest a^2 from bouth sides)
4º: a(b-a)=(b-a)(b+a)
5º: a=b+a
6º: a= a+a (since a=b)
7º: a=2a or 1=2

the falacy is in step 4º to 5º, because (b-a)=0. Remmember a=b. So we actualy devided by 0, which is not alowed

Answer 12

it is, if a\(\neq =0\)

Answer 13

what is your concern?

Answer 14

i am curious, :)

Answer 15

@Loser66 ?

Answer 16

a=2a
a/a=2a/a
1=2

Answer 17

the problem is already with a=2a. But we got here because of step 4,5

Answer 18

it is imposible to happen that a=2a for \(a \neq 0\)