Why does it equal 1?

[aca]

Why is any number that is raised to the power of zero equal to one? Anybody who can explain this to me please do, it is keeping me up at nights. I do not understand its logic.

[Paul C. Anagnostopoulos]

Because 1 is the identity for multiplication. If you're going to multiply a number by itself zero times, it's handy to end up with the identity. Similarly, if you add a number to itself zero times, you end up with 0, the identity for addition.

What about the pesky zeroth root of a number?

http://mathforum.org/library/drmath/view/55764.html

~~ Paul

[Kullervo]

An intuitive explanation: http://www.geocities.com/thesciencefiles/powers/powerszero.html

also: http://ccrma-www.stanford.edu/~jos/mdft/Exponent_Zero.html

Say you have 5 cubed divided by 5 squared. You can solve this by subtracting the exponent 2 from 3, getting 5 to the first, or 5.

125/25=5

So if the exponents are the same, subtraction will leave you with a zero exponent, so any real number to the zeroth power must be one.

Except zero to the zeroth, which is undefined.

[hgc]

Originally posted by Kullervo
An intuitive explanation: http://www.geocities.com/thesciencefiles/powers/powerszero.html

also: http://ccrma-www.stanford.edu/~jos/mdft/Exponent_Zero.html

Say you have 5 cubed divided by 5 squared. You can solve this by subtracting the exponent 2 from 3, getting 5 to the first, or 5.

125/25=5

So if the exponents are the same, subtraction will leave you with a zero exponent, so any real number to the zeroth power must be one.

Except zero to the zeroth, which is undefined.
Great explaination. That reminds me of the proof we did for this in 10th grade.

[Kullervo]

An unrelated, but perhaps more vexing question, is why 0! (zero factorial) is equal to one.

Jim Loy's explanation is not quite satisfying to me:

http://www.jimloy.com/algebra/zero-f.htm

[aca]

Originally posted by Kullervo
An unrelated, but perhaps more vexing question, is why 0! (zero factorial) is equal to one.

yes I would appreciate having this spelled out for me in laymen terms. Zero is the number I have the most problems understanding. I am racking my brains over what is essentially nothing.

[Kullervo]

Back to the redoubtable Jim Loy:
n! is also the number of permutations (ways of arranging) exactly n things. It makes sense to say that there is one way to arrange zero things. So again, 0!=1.

also this: http://www.zero-factorial.com/whatis.html

I don't see these as formal proofs but more as arguments by analogy to justify a definition, namely that 0! = 1

another nice link dealing with zero: http://mathworld.wolfram.com/Zero.html

[roger]

another way to arrive at 0! = 1:

given n! - (n-1)! = (n-1)! (n-1)

solve for n = 1

1! - 0! = 0!*0

let 0! = x

1! - x = 0x
1! - x = 0
1 = x
1 = 0!

[Tez]

0!=1 is simply a convenient convention...

as to why anything raised to the power zero equals 1, first you have to abandon the notion that raising to a power means multiplying the thing by itself that many times.

For example 2^Pi cannot mean "take two and multiply it by itself Pi times"

Once youve contemplated your navel in the mountains of tibet for a while about this, you should be comfortable with what we mean by 2^x for any real number x.

After that, it becomes easy to accept 2^0=1, since this is simply the limit as epsilon goes to zero of 2^epsilon. But as I said, first become comfortable with 2 raised to a non-zero real number!

[Stimpson J. Cat]

An unrelated, but perhaps more vexing question, is why 0! (zero factorial) is equal to one.

Jim Loy's explanation is not quite satisfying to me:

http://www.jimloy.com/algebra/zero-f.htm

One explanation is in terms of the Gamma Function (http://mathworld.wolfram.com/GammaFunction.html) .

Gamma(x) is a peicewise continuous function, defined for all x other than non-positive integers. For positive integers, Gamma(n) = (n-1)!.

Dr. Stupid

[Jon_in_london]

Originally posted by roger
1 = 0!

That might look bad in someones sig............

[Ziggurat]

Here's another aproach to x^0 = 1. The relationship

x^y * x^z = x^(y+z)

is easy to understand for positive integers y and z. Now just generalize: ie, make this hold true for any y and z, including negative numers. Then you can do things like show that

x^(-1) = 1/x

which then leads to

x^1 * x^(-1) = x^0
= x * 1/x
= 1

so x^0 = 1

[HarryKeogh]

y'know, i like to think of myself as a fairly bright person but then i'll stumble upon a thread like this and realize how incredibly stupid i am.

you guys are so smart it makes me sick!!! okay, i'm just jealous (and impressed)

[Brown]

Here's a simpler question, but one that can baffle some grade school students:

When you add zero to a number, you aren't adding anything at all to that number, so the number remains the same. Similarly, when you subtract zero from a number, you aren't taking anything at all away from that number, so the number remains the same.

So if you multiply a number by zero, you aren't multiplying the number by anything at all, so the number should remain the same. Shouldn't it?

[Crossbow]

Originally posted by Brown
Here's a simpler question, but one that can baffle some grade school students:

When you add zero to a number, you aren't adding anything at all to that number, so the number remains the same. Similarly, when you subtract zero from a number, you aren't taking anything at all away from that number, so the number remains the same.

So if you multiply a number by zero, you aren't multiplying the number by anything at all, so the number should remain the same. Shouldn't it?

Well, Brown, by the same token, IF one were to divide a number by zero, THEN the number should be the same, yes?

However, there is an intuitive proof why these cases are not so.

Take any number, say 15, and multiply it by 1, it equals 15.
Now take 15 and divide it by 1, it equals 15.

Now use 0.1 instead of 1, and the product is 1.5 and the quotient is 150.
Now use 0.01, and the product is 0.15 and the quotient is 1500.
Now use 0.001, and the product is 0.015 and the quotient is 15,000.
And so on.

As the variable gets closer to zero,
then the product gets closer zero, and
the quotient gets closer to infinity.

I hope this helps!

[Brown]

Originally posted by Crossbow
I hope this helps! All correct, of course!

But the real riddle is: how would you explain this to a grade schooler? Sure, the kid could test test it out on a calculator and verify it, but how can the subject be made clear to someone who is just learning multiplication? (Do kids still learn times tables, by the way?)

It can be difficult to explain that "multiplying the number by zero" is different from "not multiplying the number by anything at all."

I might try something like this. "You have zero apples. If you had three times as many apples, how many would you have?"

(True story: when I was first learning multiplication, I thought it was nonsense and a waste of time. Basically, I thought addition and subtraction were all you needed. After all, most of the problems that the teacher wanted us to solve using multiplication could be solved with addition and subtraction as well. When the teacher tried to teach us multiplication, I wondered, "Am I EVER going to USE this??" The turning point for me was a problem in which we had to determine the number of desks in a classroom. There were three rows of desks and four desks per row. I found that by multiplying, I could get the answer a lot faster than by counting!)

[swellman]

Originally posted by Brown
(Do kids still learn times tables, by the way?)



Yes, at least in our local schools.

I don't let my 3rd grader use a calculator yet.

[Ziggurat]

Originally posted by Brown
All correct, of course!

But the real riddle is: how would you explain this to a grade schooler?


That's a harder problem, and I don't think I remember how I learned math. But here's one though for how to approach it: if you can get them to accept that a number times 1 is that original number, then it might be easier to get them to accept a number times zero is still zero. The concept of rows might work pretty well too - just as it got you to realize the usefulness of multiplication (rather than just repeated addition), if you don't have any rows (zero) of, say, apples, then you don't have any apples.

[Paul C. Anagnostopoulos]

I said:If you're going to multiply a number by itself zero times, it's handy to end up with the identity.
Then Tez said:as to why anything raised to the power zero equals 1, first you have to abandon the notion that raising to a power means multiplying the thing by itself that many times.
Bada bing, bada boom!

~~Paul

[The Central Scrutinizer]

Originally posted by aca
Why is any number that is raised to the power of zero equal to one? Anybody who can explain this to me please do, it is keeping me up at nights. I do not understand its logic.

Because that is the way god wants it.

[Vorticity]

Originally posted by aca
Why is any number that is raised to the power of zero equal to one? Anybody who can explain this to me please do, it is keeping me up at nights. I do not understand its logic.
On a related note, consider the quantity 0^0. Usually we simply refer to this as "undefined". However, consider the function

f(x) = x^x (for x greater than or equal to zero)

Consider the limit of f(x) as x->0. A little fooling around will reveal that this limit is 1. Also, as x->Infinity, f(x) clearly also ->Infinity.
Which brings us to our [semi] interesting question:

What value of x (greater than or equal to zero) minimizes f(x)?

And for all you incurable math geeks out there, a couple of interesting-looking functions to play with are the real and imaginary parts of:

g(x) = (-x)^(-x) (for x>= 0)

[ImpyTimpy]

Sorry roger, but that proof seems incorrect to me...

You're saying 1! = 0 and then proving that 1! = 0 using what you just stated. In other words, no proof is ever offered.

Originally posted by roger
another way to arrive at 0! = 1:

given n! - (n-1)! = (n-1)! (n-1)

solve for n = 1

1! - 0! = 0!*0

let 0! = x

1! - x = 0x
1! - x = 0
1 = x
1 = 0!

[Vorticity]

Originally posted by ImpyTimpy
Sorry roger, but that proof seems incorrect to me...

You're saying 1! = 0 and then proving that 1! = 0 using what you just stated. In other words, no proof is ever offered.
I assume you meant to type 0!=1.
Where does he assume that? The proof seems valid to me. The only assumption at the beginning is:
n!-(n-1)!=(n-1)! (n-1)
which follows from the properties of factorials...

[Stimpson J. Cat]

Originally posted by ImpyTimpy
Sorry roger, but that proof seems incorrect to me...

You're saying 1! = 0 and then proving that 1! = 0 using what you just stated. In other words, no proof is ever offered.
--------------------------------------------------------------------------------


I assume you meant to type 0!=1.
Where does he assume that? The proof seems valid to me. The only assumption at the beginning is:
n!-(n-1)!=(n-1)! (n-1)
which follows from the properties of factorials...

The real issue is that his proof includes the statement that 1! = 1. If you take the recursive relation n! = n(n-1)!, then all you need is one value of n and n! to completely define the relation.

In other words, Roger's proof could simply have been:

1! = 1 = 0! * 1
0! = 1! = 1

Likewise if you start with 0! = 1, you can derive that 1! = 1. Or you could just as easily start with 5! = 120, and derive everything from that.

This is a standard thing in mathematics, not unlike needing a set of boundary conditions to completely define the solution to a differential equation.

To put it another way, it is rather ill-conceived to talk about "proving" that 0! = 1. The fact is that this is a part of the definition of factorials. 0! equals one because it is defined to.

Dr. Stupid

[Soapy Sam]

Tez said- "..first you have to abandon the notion that raising to a power means multiplying the thing by itself that many times."

(Grinds teeth). Darn ! I wish mathematicians would stop changing things! Miss Young TOLD me that's what it meant in 1967. I did not argue. NOBODY argued with Miss young. Not unless they wanted to be doing homework for the rest of their life. When did they change it? Who got to tell Miss Young? Is he out of hospital yet?
So dies another cozy belief. One day, my head will explode and YOU LOT will be to blame.

Off to play with my new (but deeply ugly) fact.

[BillyJoe]

Originally posted by Brown
It can be difficult to explain that "multiplying the number by zero" is different from "not multiplying the number by anything at all."

I might try something like this. "You have zero apples. If you had three times as many apples, how many would you have?"Ah, but the question was....

"You have three apples. If you had zero times as many apples, how many would you have?

BillyJoe

[Jimmy]

This is the way I learned multiplication:

Take the problem 2 x 3. You could say that you have 2 threes or 3 twos, which is 6 of course.

Consider 3 x 0. You could say that you have 3 zeros or 0 threes, which is 0.

Maybe that's too simplistic. <shrug>. I think a third grader who was good at addition would understand that, however.

[BillyJoe]

Jimmy,

I'm trying to do this without the maths.

When you say...
"You have zero apples. If you had three times as many apples, how many would you have?"
.....the answer is intuitively simple

But when you say.....
"You have three apples. If you had zero times as many apples, how many would you have?
.....the answer is much less intuitive

Sure you can simply do it by saying that the second question is equivalent to the first question

Billy.

[Jimmy]

My apologies BillyJoe.

That is as you say, less intuitive. That zero is a bugger, huh?

We can have either zero groups of three apples or, we can have three groups of zero apples. Of course this is the same thing I said before.

Is there an error in the question itself?

You have three apples. If you had zero times as many apples, how many would you have?

Mathematically, we know that any number multiplied by zero is zero, but for someone who doesn't realize this, figuring out how much 'zero times as many is' could be quite difficult. If I didn't now that any number multiplied by zero is zero, I might guess the answer to be 3 apples. I'll have to think about it some more.

[T'ai Chi]

0! = 1 comes from the gamma function.

The gamma function is:

gamma(n) = Integral t^(n-1) * e^(-t) dt, from t=0 to t=oo

A known relation is that gamma(n) = (n-1)!

So evaluate gamma(1) and you get (1-1)! = 0!

And gamma(1) = Integral * e^(-t) dt, from t=0 to t=oo =

e^-a - e^-b, where a=0 and b tends towards oo, =

1 - 0 = 1.

So 0! = 1.

[T'ai Chi]

Originally posted by Vorticity

f(x) = x^x (for x greater than or equal to zero)

What value of x (greater than or equal to zero) minimizes f(x)?



f(x) = x^x

f(x)' = (ln(x)+1)*x^x

Setting f(x)' = 0 and solving for x, reveals that x = e^-1.

I'll spare the details of proving it really is a minimum.

[Peter Morris]

Originally posted by aca
Why is any number that is raised to the power of zero equal to one? Anybody who can explain this to me please do, it is keeping me up at nights. I do not understand its logic.
Look at it this way:

10^4 = 10000
10^3 = 1000 (one tenth of 10^4)
10^2 = 100 (one tenth of 10^3)
10^1 = 10 (one tenth of 10^2)
10^0 = 1 (one tenth of 10^1)


2^4 =16
2^3 = 8 (half of 2^4)
2^2 = 4 (half of 2^3)
2^1 = 2 (half of 2^2)
2^0 = 1 (half of 2^1)

and so on


===============================

Here's another question that intrigued me.

Zero divided by any number is zero.
Any number divided by itself is 1.
Any number divided by zero is undefined.

What is zero divided by zero?

[Kullervo]

zero divided by zero is an indeterminate form. It may have a value if the two zeros are limit values of functions, but just plain 0/0 doesn't tell you enough to determine anything.

For instance, the limit of (x^2 - 1)/(x - 1) as x approaches 1 can be determined even though it is 0/0 when x = 1.