why X⁰ is equal to 1

422 viewsMathematicsOther

X² is X times X.
X¹ is X.
In X^Y Y is the factor that decides how many Xs are getting multiplied.

So why is X⁰ 1? It comes out of nowhere. Zero Xs are getting multiplied, so there’s no numbers being multiplied, meaning there’s nothing.

In: Mathematics

10 Answers

Anonymous 0 Comments

As you say, exponentiation is defined (at first) by “repeated multiplication.” So:

* X^3 = X * X* X
* X^2 = X * X
* X^1 = X

So that definition makes sense for what we call the *natural numbers* or the *counting numbers.* But when you try to extend this concept to other numbers, things get a bit unsatisfying. If you say that X^0 = 0, that means that the multiplication and division pattern in the previous sequence **stops working**. By that what I mean is:

* To get from X^1 to X^2, you multiply by an extra “X.”
* To get from X^2 to X^3, you multiply by an extra “X.”

If X^0 = 0, then you *can’t get* from X^0 to X^1 through multiplication. Uh oh! That property of exponents is one of the core properties that makes them useful, as it turns out! So, rather than break the pattern, we *define X^0* so that it maintains it:

* To get from X^0 to X^1, you multiply by an extra “X,” so X^0 must be equal to 1.

This also then allows us to define negative exponents (X^(-1) = 1/X^(1) because to get from X^(-1) to X^0 you need to multiply by an extra “X”) and fractional exponents (a bit more complicated to explain, but it follows another pattern of exponents).

——

tl;dr – “Repeated Multiplication” is just the starting point for exponents. The myriad of other properties that exponents have end up being more valuable than that starting point, so other non-standard exponential values get defined in order to ensure those properties stay accurate, no matter the exponent!

Anonymous 0 Comments

When you go from x¹ to x² to x³, you multiply by x each time.

If you do it in reverse, going from x³ to x² to x¹, you’re dividing by x each time. Do it one more time, to go from x¹ to x⁰: you get x/x, which is always 1.

Anonymous 0 Comments

Here’s another way to look at the multiplication rule

x^3 = 1 * x * x * x

x^2 = 1 * x * x

x^1 = 1 * x

x^0 = 1

If that seems arbitrary, here’s another explanation. When divide x^y by x, you get x^y-1 . So x^3 / x is x^2, x^2 / x is x. So when you divide x^1 / x, you get x^0 . But x^1 / x, is just 1, so x^0 = 1

Anonymous 0 Comments

So think of it this way:

X^4 / X^2 is just (XXXX)/(XX) = X^2

X^3 / X^2 is just (XXX)/(XX) = X

X^2 / X^2 is just (XX)/(XX) = 1

In each of these, X^a / X^b = X^(a-b)

If a = b, then we have X^0

This just means we have the same number of X’s in the numerator and denominator, so it’s just some number divided by itself

Anonymous 0 Comments

Yes, zero X’s are being multiplied. So we do nothing, correct. But nothing to what is the question, and what would do nothing. What is the baseline value? What is the baseline “identity” (the term used) for multiplication that does nothing?

Let start with addition. 2+3 = what? Well, 5 you say. Okay, but why? Why not 6? You can add 2 and then add 3 and end up at 6. It just means you started at one, the first natural number. Why is this wrong, or more importantly why is this not helpful way to define this operation? Well, because doing no additon would leave you with one. One does something to addition. Adding one to anything, well, changes the number. So what do we want to use as the starting point for addition? Zero, because adding zero to anything does nothing. So we start at zero, and doing no additon leaves us at zero. The identity value for addition is zero.

Now, you’re assuming zero should hold for multiplication, and hence exponentiation as well. Zero is nothing after all? Well, no, not for the multiplication operation. Zero does a lot, it turns everything to zero. If our base, our start, for multiplication was zero, every operation would be zero. The start point for multiplication, the value that does nothing, the identity, is one. X^0, doing no multiplication, is one. The identity value for multiplication is one. Not zero.

Anonymous 0 Comments

X⁰ = 1 because the 0 exponent means you multiply X 0 times, leaving it as X, not 0; dividing X¹ by X gives X⁰, and any number divided by itself is 1.

Anonymous 0 Comments

Terrence Howard, is that you? 🙂

Anonymous 0 Comments

Because 1 is the neutral element of multiplication.

Let’s look at how multiplication is defined by addition first. Multiplication says how many additions are performed:

* 2X = X+X
* 1X = X
* 0X =

Now that’s weird too, right? What does it mean that the right side is “empty”? To avoid that. we can expand it by adding 0 to the right side without changing the result. The fact that adding 0 doesn’t change the value is called 0 being the neutral element of addition.

* 2X = 0+X+X 
* 1X = 0+X
* 0X = 0

Better, right?

Now we can do the same thing with powers and multiplication. However, we have to think what is the neutral element of multiplication? Can’t be 0 because multiplying something with zero definitely changes its value. It has to be 1. 1 is the only number you can multiply some other value with, that doesn’t change the value.

So, let’s do the same thing with powers but expand the right side by the neutral element of multiplication:

* X^2 = 1XX
* X^1 = 1X
* X^0 = 1

There we have it!

Anonymous 0 Comments

The jump from multiplication to exponentiation is more complicated than it is first taught in school.

Some examples where the multiplication analogy falls apart include:

X^(½) is the same thing as the square root of X. If you choose X = 4, then X^(½) = 2.

X^(-1) is the same thing as 1/X. If you choose X = 4, then X^(-1) = ¼

——

However, you can make the multiplication analogy work for X^0 because 1 is the multiplicative identity (1 * X = X for any X). Therefore, this is a valid interpretation:

X^2 = 1 * X * X
X^1 = 1 * X
X^0 = 1

Anonymous 0 Comments

It’s just defined that way.

The *multiply X by itself* n *times* “definition” is a simplification of the definition of exponentiation to *natural* powers, not zero.

It is convenient to define it that way because if we multiply two numbers X^n and X^m for natural numbers *n* and *m*, then it is clear that it is equal to X^(n+m). With this, we can relate exponentiation to multiplication as we do multiplication to addition.

In choosing how to define exponentiation to negative integers, we use the property above. Because the natural numbers are closed under addition, for any natural number *k ≠ 1*, there exists *m* such that k = m + 1. Then, because we have defined X^k, we know that X^(k-1) = X^m. Because this is equal to a product of X’s, we know that X^m = X^k / X.

It follows from this that the inverse of a real number X under multiplication is X^(-1), i.e., X^(-1) is the number such that X * X^(-1) = 1 (they “cancel out”).

Applying the relationship above, X^n * X^m = X^(n+m), we have X * X^(-1) = X^1 * X^(-1) = X^(1-1) = X^0 = 1.

I am going to bed…