A much easier math question (with a lesson)

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Lemon law

Lifer
Nov 6, 2005
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To start out with, the reason all of us have problems with math, is that we fail to understand that 2=1. Which I will proceed to prove.

Lets start out with the assumption that

A=B

now multiply each side of the equation by A

So

A squared=AB

Now subtract B squared from each side of the equation

So A squared minus B squared =AB-B squared

Now factor each equation to get

(A+B)x(A-B)=B(A-B)

Now since we have a common term, namely (A-B), on each side of the equation, we can factor out (A-B), and we get:

(A+B)=B

Now since we originally defined that A=B, then we can substitute a B for an A to get.

(B+B)=B which can also be written 2B=B

Now divide by B

To get 2=1

And this therefore explains why we all have difficulties with math.

( Of course I cheated, but can you figure out where I cheated?)
 

Juncar

Member
Jul 5, 2009
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You factored out something from 0, which can be anything. Your equation has infinite solutions.
 

Cogman

Lifer
Sep 19, 2000
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Now since we have a common term, namely (A-B), on each side of the equation, we can factor out (A-B), and we get:

(A+B)=B

A-B = 0, you divided by zero. (this is the oldest trick in the math world btw.)
 

Lemon law

Lifer
Nov 6, 2005
20,984
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You factored out something from 0, which can be anything. Your equation has infinite solutions.
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IMHO, not very well stated, Juncar, but you stumbled on the solution. Even if you are wrong in saying the end equation of 2=1 has an infinite number of solutions.

If A=B, then the (A-B) I divided by must then equal zero, and if you just let me divide by zero, I can prove the most amazing and invalid things. And its why dividing by zero is a prohibited and undefined mathematical operation. If (A-B) does not equal zero, then (A-B)/ (A-B) equals one. IF (A-B) does equal zero, there is no way to tell what it equals.

Edit in, while I was writing this post, Cogman said it perfectly. And even if Cogman says, " (this is the oldest trick in the math world btw.) ", it always ensnares new victims.

Part of the lesson lies in computers, that approximate many complex mathematical operations, especially using non-integer numbers, and if the computer divides by a very small non zero number, a tiny error in approximation can become huge on the output.
 
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Lemon law

Lifer
Nov 6, 2005
20,984
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You got to zero earlier than that.

b=b

multiply both sides by b

subtract b^2 from both sides.
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No, DominionSeraph, as this quick example will show.

Lets say A=2, then B=2---any non zero number would do as well.

So 2=2

The multiply by A

2x2=2x2

then subtract B squared

2x2-4=2x2-4

Both sides of the equation may now equal 0, but its still a mathematically valid equation.

And so if the factoring step which results in (A+B) (A-B)= B(A-B)

And even if (2+2) times 0=2 times 0, its still a mathematically valid equation. It still comes down to 0=0.

It only becomes a mathematically invalid operation after I try to symbolically divide by A-B, aka, in this case 0 by initial definition.

And we end up with 4=2, the first equation in the series that yields an incorrect result and inequality.
 

iCyborg

Golden Member
Aug 8, 2008
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I have a more difficult one:

Euler's formula says that e^(i*Pi)=-1. From here it's easy to get e^(2*i*Pi)=1, and generally e^(2N*i*Pi)=1 for integer N.

Now we have
e = e^(1+2*i*Pi) =
[e^(1+2*i*Pi)]^(1+2*i*Pi) =
e^(1+4*i*pi-4*Pi^2) =
e^(1-4*Pi^2).

And e=e^(1-4*Pi^2) is clearly false. So, what's wrong?
 

C1

Platinum Member
Feb 21, 2008
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More problematic to the average person (because they are more common) are problems formulated such as :

2 + 18 ÷ 4 - 3 x 4 = ? (Ans: -5.5)
 
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