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**nikad****Member**- Registered: 2006-01-21
- Posts: 3

I have the following:

a-b=X

I need one single formula (that uses only addition, substraction, multiplication or division) where if X is a positive number the result will be 1 and if X is a negative number the result will be -1 or 0

Is this possible?

Thanx in advance!

nik

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

How about |X| ÷ X?

That formula will give 1 if X is positive, -1 if X is negative and won't work if X is 0.

Edit: |X| ≡ abs (X). But you figured that out anyway.

Why did the vector cross the road?

It wanted to be normal.

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**nikad****Member**- Registered: 2006-01-21
- Posts: 3

mathsyperson wrote:

How about |X| ÷ X?

That formula will give 1 if X is positive, -1 if X is negative and won't work if X is 0.

Excuse my ignorance, but if i do X ÷ X I get 1 if X is positive, but if X is negative I still get 1, unless |X| means absolute value which I am not sure it will be recognized on a computer.

I appreciate your help

nik

*Last edited by nikad (2006-01-21 12:07:02)*

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**nikad****Member**- Registered: 2006-01-21
- Posts: 3

I will try to compute the absolute value somehow, you ruuuuuuuuuuuuuule!

nik

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**irspow****Member**- Registered: 2005-11-24
- Posts: 455

Here you go:

x³ / ((x²)^9)^(1/6)

That will work but I don't know how you would do it in computer language.

x³ / √(x^6)

That might be easier.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Wouldn't it be simpler to just have x/√x² if you're doing it that way?

Why did the vector cross the road?

It wanted to be normal.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 455

Some people......just kidding. Of course it would, but I am not the sharpest knife in the drawer you know?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Hehe, sorry. I'll try to go easy on you from now on.

Why did the vector cross the road?

It wanted to be normal.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 455

Pour it on. I am a glutton for punishment. That is why I keep posting answers when I know that I will be corrected eventually.

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

[x+abs(x)]/2x

positive = 1

negative = 0

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Assuming that you are programming this...

unless |X| means absolute value which I am not sure it will be recognized on a computer.

Absolute value is extremely easy to do on a computer. All it requires is to switch the very first bit of a signed number and then subtract that from the highest possible signed value. Most languages support an abs() function, but it also works if you just do: x = (unsigned int) x or something similar.

But if you're on a computer, the above isn't needed.

if (X < 0) return -1; return 1;

x³ / ((x²)^9)^(1/6)

That will work but I don't know how you would do it in computer language.

Pretty easy actually:

(x*x*x) / pow((pow(x, 18)), 1/6)

If your language doesn't support a power function, you can write it yourself, although doing non integral powers is a bit tricky. But, as I said, you don't need to do all this.

*Last edited by Ricky (2006-01-21 20:35:20)*

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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