Question

Math question
You can prove 1 = 2

Answer 1

let a variable a = b
multiply both sides by a
\[a^2=ab\]
square subtract by b^2 on both sides
\[a^2-b^2=ab - b^2\]
the left side of the eqn is a difference of squares
\[(a+b)(a-b) = b(a-b)\]
We can divide both sides by a-b
\[a+b = b\]
Remember a = b, so
\[2b=b\]
divide both sides by b
\[2=1\]!?!?!?!

Answer 2

Math question, I would like to hear if you think this is right, or if wrong how come.
¯\_(ツ)_/¯ @mxddi3 @justus @ramen

Answer 3

I can tell you were bored, but lemme see

Answer 4

This is a classic example, however, the only problem is that since we assume a = b, that means a-b = 0

We can't divide both sides by 0

Answer 5

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ
This is a classic example, however, the only problem is that since we assume a = b, that means a-b = 0

We can't divide both sides by 0
\(\color{#0cbb34}{\text{End of Quote}}\)
xD yea u kinda got it
@ljtheballer16 remember when we divided both sides by a-b?
so we know a = b
so basically we divided both sides by either a-a (if we substitute a for b) or b-b (by substituting b for a), both of these equal 0. What I did between step 3 and 4 was divide both sides by a-b, so this is actually not valid and is a trick, Nice job @AZ for figuring it out!

Answer 6

\(\color{#0cbb34}{\text{Originally Posted by}}\) @darkknight
\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ
This is a classic example, however, the only problem is that since we assume a = b, that means a-b = 0

We can't divide both sides by 0
\(\color{#0cbb34}{\text{End of Quote}}\)
xD yea u kinda got it
@ljtheballer16 remember when we divided both sides by a-b?
so we know a = b
so basically we divided both sides by either a-a (if we substitute a for b) or b-b (by substituting b for a), both of these equal 0. What I did between step 3 and 4 was divide both sides by a-b, so this is actually not valid and is a trick, Nice job @AZ for figuring it out!
\(\color{#0cbb34}{\text{End of Quote}}\)
ohhh i gotchu