I am looking for a trick math equation to where 3 = 1 like the trick math equation where 2 equals 1.

If you're taking it as the thread that Nestorian made some last year, well, you can have any sort of thing...

I'll take a chemistry example, regarding molecules.

2SO_2 + 2H_2O + O_2 \rightarrow 2H_2SO_4

There you end up with:

2 + 2 + 1 = 2

:p

These are called 'mathematical fallacies'.

Google that and you will find many.

Here's one: it actually "proves" that 1 = - 1, but if you then add 2 to both sides you get 3 = 1:


-1 = -1 \\
\frac 1 {-1} = \frac {-1} 1


Take the square root of both sides (yes - it is perfectly OK to take the square root of a negative number):



\sqrt {\frac 1 {-1}} = \sqrt {\frac {-1} 1}


Since \sqrt{a/b} = \sqrt a/ \sqrt b:


\frac {\sqrt 1} {\sqrt {-1}} = \frac {\sqrt {-1 }} {\sqrt 1}


Now cross multiply, using the rule for fractions that if a/b = c/d, then ad = cb:


\sqrt 1 \ \times \ sqrt 1 = sqrt{-1} \ \times \ \sqrt{-1}


Since \sqrt a \ \times \ \sqrt a = a :
1 = -1

And add 2 to both sides:

1+2 = -1+2
3 = 1

QED

Do you see where the fallacy is?

I would say... \sqrt{1} because this can take either of \pm1

Is that it? :)

I would say... \sqrt{1} because this can take either of \pm1

Is that it? :)

You're saying that \sqrt a \times \sqrt a = \pm a ? So if we simply chose \sqrt 1 \times \sqrt 1 = -1 that would fix it. True enough. But there's actually an error that occurs earlier in the "proof."

Well, I had suspicion about the

\frac{1}{-1} = \frac{-1}{1}

But I wasn't sure. This is the only thing that I don't think might be 'legal'.

Well, I had suspicion about the

\frac{1}{-1} = \frac{-1}{1}

But I wasn't sure. This is the only thing that I don't think might be 'legal'.

Nothing wrong with that - both sides equal -1, right?

Keep looking...

Or is it because you got

\frac{\sqrt1}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}

You have in fact, \frac{1}{i} = \frac{i}{1}

I don't know much about complex numbers, but cross multiplication might not be something allowed.. (taking only the positive root, for the time being)

Or is it because you got

\frac{\sqrt1}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}

You have in fact, \frac{1}{i} = \frac{i}{1}

I don't know much about complex numbers, but cross multiplication might not be something allowed... ? (taking only the positive root, for the time being)

Tha's it! Using the notation \sqrt {-1} = i, the left hand side is:


\frac{\sqrt1}{\sqrt{-1}} = \frac 1 i = -i


and the right hand side is


\frac{\sqrt{-1}}{\sqrt{1}} = i


So the error is:

\sqrt{\frac a b} \ \ne \ \frac {\sqrt a} {\sqrt b}

for negative numbers.

Finally! Thanks! :)