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Wiley50 · Aug 19, 2010, 06:35 AM

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

Unknown008 · Aug 19, 2010, 09:14 AM

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

galactus · Aug 19, 2010, 10:15 AM

These are called 'mathematical fallacies'.

Google that and you will find many.

ebaines · Aug 24, 2010, 11:32 AM

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?

Unknown008 · Aug 24, 2010, 11:46 AM

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

Is that it? :)

ebaines · Aug 24, 2010, 11:51 AM

Originally Posted by Unknown008

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."

Unknown008 · Aug 24, 2010, 11:54 AM

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'.

ebaines · Aug 24, 2010, 12:22 PM

Originally Posted by Unknown008

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...

Unknown008 · Aug 24, 2010, 12:53 PM

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)

ebaines · Aug 24, 2010, 01:00 PM

Originally Posted by Unknown008

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.

Unknown008 · Aug 24, 2010, 01:06 PM

Finally! Thanks! :)