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**Yr7RockGod****Guest**

plz explain a°= 1

a to the power of zero

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,578

a° is a multiplied by a zero times.....

a¹ is a

a² is a x a

a³ is a x a x a and so on.

It is a convention that any number raised to the power zero is 1,

since, the law of indices states that

When m and n are equal,

We know

Therefore, by convention,

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Zmurf****Member**- Registered: 2005-07-31
- Posts: 49

Wow, I never knew exactly why that was until, i just accepted it, just like the idea that the moon is made of cheese and if you can yell loud enough ,the man in the moon will talk back.

*"When subtracted from 180, the sum of the square-root of the two equal angles of an isocoles triangle squared will give the square-root of the remaining angle squared."*

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,578

There's another interesting explanation....

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Note that powers that are positive real numbers work, but powers that are negative must be integers I think.

Or maybe there are certain negative fractional powers that work...

Let's see 16^(-1/2) = 1/4, so that works with a negative fractional power, hmm.

I thought my calculator didn't like something.

Oh yeah, it was a fractional power of a negative number.

But this is sometimes okay. It all depends I guess.

Like (-27)^(1/3) = -3, so that works.

But (-27)^2.5 would probably fail.

Let's see. (-27)^(10/4), yeah that works, but (-27)^(5/2) doesn't work.

Anyone see the flaw?

**igloo** **myrtilles** **fourmis**

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**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,578

John,

In (-27)^(10/4) and (-27)^(5/2), I guess the first caculation done is the numerator of the exponent, then the root is extracted.

In the former, since after the number is raised to the numerator, a positive number is got, there's no problem for the calculator to perform the next calculation, that is extracting the 4th root. But in the latter, the 5th power of -27 is a negative number and the operation of square-root of a negative number cannot be performed, unless the calculator works with complex numbers/imaginary numbers.

The calculator in the scientific mode in the system I am now using says 'Invalid input for function' when I input -27^(5/2) or -27^(10/4). I am not aware whether the fraction mode, i.e a/b is available, for input.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,673

Also see my pages: Exponents and, if you want, Fractional Exponents (has a nice graph at the bottom of the page, where you can set the exponent to zero)

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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

Ganesh is right. Calculators don't usually like fractional powers of negative numbers because the answers would have an imaginary part.

The exception to this is when the power can be written as a fraction that has an odd denominator.

e.g -8^(1/3) = -2.

If you think about it, that applies to integer powers as well, it's just that the denominator in that case is 1!

Or 1. Same thing. =P

Why did the vector cross the road?

It wanted to be normal.

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