# 4 = 5 ### Help Support The Rocketry Forum: #### rocketsonly

##### Well-Known Member
Some super genius at my school taught me this little trick to prove 4 = 5. He'll be attending college after his junior year at Greenhill School! But honestly, I don't see why someone would do that. Senior year is supposed to be the best!

a + b = c

Form two separate equations, and multiply one by 4, and the other by 5.

4a + 4b = 4c .............. 5a + 5b = 5c

Subtract 4c from both sides on the first equation, and subtract 5c from both sides on the second equation.

4a + 4b  4c = 0 ................. 5a + 5b  5c = 0

Because both now equal 0, they must equal the same.

4a + 4b  4c = 5a + 5b  5c

Factor out the 4 on one side, and the 5 on the other.

4(a + b  c) = 5(a + b  c)

Now divide each side by a + b  c

4 = 5

Cool eh?

Can someone spot the mistake?

EDIT: spacing not correct

#### illini

##### Well-Known Member
The mistake is that from your first line, a + b = c, we also know that a+b-c = 0. Therefore, in the step where you divide both sides by a+b-c, you are in fact dividing both sides by 0 (not to mention already multiplying both sides by 0). In reality, several steps before that you had already stated that 4 x 0 = 5 x 0...which is no great surprise to anyone.

#### MetMan

##### Well-Known Member
Is that the term I'm looking for? If you want to solve a system of linear equations, you need to have indepent equations. a+b=c or 4a+4b=4c or 5a+5b=5c are not independent--they're the same thing.

So to solve for a, b, and c you'd need three independent linear equations.

#### rocketsonly

##### Well-Known Member
Originally posted by illini
The mistake is that from your first line, a + b = c, we also know that a+b-c = 0. Therefore, in the step where you divide both sides by a+b-c, you are in fact dividing both sides by 0 (not to mention already multiplying both sides by 0). In reality, several steps before that you had already stated that 4 x 0 = 5 x 0...which is no great surprise to anyone.
CONGRATULATIONS!

#### rocketsonly

##### Well-Known Member
Originally posted by MetMan
Is that the term I'm looking for? If you want to solve a system of linear equations, you need to have indepent equations. a+b=c or 4a+4b=4c or 5a+5b=5c are not independent--they're the same thing.

So to solve for a, b, and c you'd need three independent linear equations.
Maybe, I don't know.

#### illini

##### Well-Known Member
Originally posted by rocketsonly
CONGRATULATIONS!
Once a geek, always a geek.

#### illini

##### Well-Known Member
Originally posted by MetMan
Is that the term I'm looking for? If you want to solve a system of linear equations, you need to have indepent equations. a+b=c or 4a+4b=4c or 5a+5b=5c are not independent--they're the same thing.

So to solve for a, b, and c you'd need three independent linear equations.
It is true that you need three independent equations to solve for three variables. However, his problem as posed was not to solve for a, b, and c, but to provide a proof (or, rather, critique a proof) of something else. The real error in the proof was when it started playing games with zero.

#### shrox

##### Well-Known Member
Yes, it's just a trick. You can substitute any number as the 4 and 5. You will end up with 7=9. or 10.7654=3.1416 or whatever number you multiply by.

#### DynaSoar

##### Well-Known Member
Originally posted by rocketsonly
Some super genius at my school taught me this little trick to prove 4 = 5.
Ask him to do it in Boolean algebra.
Numbers: 0, 1
Operators: and, or, not

#### rocketsonly

##### Well-Known Member
Originally posted by DynaSoar
Ask him to do it in Boolean algebra.
Numbers: 0, 1
Operators: and, or, not
Haha, sure, I'll see what he says.