2 Equals 1 - It's True

electriic ink

Dec 9 2006, 07:23 PM

In light of the topic, 2+2=5 methought I'd create my own puzzle:

Let x = y

Therefore x² = xy

Therefore x² - y² = xy - y²

If we factorise these equations we end up with (x + y)(x - y) = y(x - y)

We now have a common term on both sides "(x - y)," which we will divide off to make our problem easier to solve
x + y = y

But x = y so y + y = y thus 2y = y

Then, if we divide off the common value "y," 2 = 1

Simple! Where's the flaw?

husker

Dec 9 2006, 09:59 PM

I remember this from last year in my math class. It was in the step when you canceled (x-y) I believe, because (x-y) is 0, therefore you are multiplying by 0 which would be 0. You can't cancel (x-y). Good one!

Plenoptic

Dec 10 2006, 12:03 AM

Well like husker said (x-y) would equal zero and it's the fact that you can't divide by zero not multiply like husker said. So that would really mean that it ends up as 0 = 0 because if you plug in the 0 of (x - y) , that would multiply with the rest of the equation to make 0 because it is all multiplication, no addition or subtraction except between parentheses. There may be a way though so keep working at it.

bhavesh

Dec 10 2006, 05:37 AM

As I have said in similar topic 2+2=5, I would again say that this type of questions are raised by people with little knowledge of Mathematics.

As explained by Husker and Plenoptic as x and y are equal therefore x-y is zero and whenever zero comes it cant be cancelled from both sides, if this is possible then all numbers will be equal to each other, like this.

0=0
or, 0*a=0*b
i.e. a=b
where a and b are any number.
therefore zeros from both sides cannot be cancelled.

electriic ink

Dec 10 2006, 09:03 AM

Congrats! The answer was you can't divide by 0 or x - y. You can prove anything when you divide by 0.

Terciel Silvi

Dec 20 2006, 04:36 PM

Here is another way of proving that 1 = 2

QUOTE

Everyone knows this:

-1/1= 1/-1

Now we will square root both sides:

√-1/1 = √1/-1

Now we break up the roots:

√-1 √1
--- = ---
√1 √-1

The square root of a negative 1 is i and the square root of 1 is 1. In other words:

i/1 = 1/i

Now we multiply the entire thing by 1/2:

i/2 = 1/2i

Now let's add 3/(2i) to this to make the math easier.

i/2 + 3/2i = 1/2i + 3/2i

Now we can multiply the entire thing by i:

i(i/2 + 3/2i) = i(1/2i + 3/2i)

So now we expand this beast:

1^2/2 + 3i/2i = i/2i + 3i/2i

We know that the square root of -1 is i, so i^2 must be -1.:

-1/2 + 3i/2i = i/2i + 3i/2i

Now we simplify the i's

-1/2 + 3/2 = 1/2 + 3/2

Let's calculate this thing:

2/2 = 4/2

And so

1 = 2

That does work as far as I can see. And with this proof you can prove that dividing by zero is possible, because

anygivennumber^0 = 1

Source

I know the source isnt reliable, but the 1-=2 thing I posted does work as far as I can see.

rsf

Dec 20 2006, 08:09 PM

1 will never equal 2 by definition. So I'm going to bother with the second proof because there's definately a flaw somewhere. If it was correct then there would be a flaw in our addition, multiplication, etc.

Remember you can't divide by variables unless you can prove they aren't 0 like y = x^2 + 1 where x is a real number (no i's

)

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2 Equals 1 - It's True - 2=1

2 Equals 1 - It's True - 2=1