Anything to zero power equals one?

Question

I am in Adv. Algebra 2 and I have a question. Firstly, would like to say I haven't taken algebra in a year due to geometry (stupid order they do but oh well) and I have a question understanding this: $(x+5)^{0}$. That would be $x^{0} + 5^{0}$ which then, wouldn't that be $1 + 1$ since anything that has a power of $0 = 1$? Maybe I misunderstood but that's what I got.

Answer 1

First, exponents do not distribute over addition. To start with the simplest example, $(a+b)^2=(a+b)(a+b)$. Applying the distributive rule, we see that this is the same as $a(a+b)+b(a+b) = a^2 + ab + ab + b^2$, which is different from $a^2+b^2$. This can be seen geometrically, too: A square built on a side of length $a+b$ has greater area than the square with side length $a$, combined with the square with side length $b$.

You can also see your result if you consider order of operations, and substitute an actual number for $x$. Let's consider $x=3$. Then we have:

$$(x+5)^0=(3+5)^0=8^0=1,$$

because Parentheses come before Exponents in PEMDAS.

Answer 2

Exponents do not distribute for addition for any power other than 1.

Consider a numerical example to prove this:

$(a+b)^{\ n}$

Choose any two real numbers for $a$ and $b$. Let's choose $a = 4$ and $b=2$ arbitrarily. For $n$, let's bound it to $2$ for simplicity.

By your logic:

$$(4+2)^2 = 4^2 + 2^2 = 20$$

But as you should know in your classroom, PEMDAS mandates terms under parentheses are operated on first. Thus,

$$(6)^2 = 36$$

And this can be demonstrated by using binomial expansion. For $n$ = 2, the answer is in the form:

$$a^2 + b^2 + 2ab$$

$$16 + 4 + 2(4)(2) = 36$$

Answer 3

As others have pointed out, you do not distribute exponents through parentheses. Positive integer exponents are a short hand for repeated multiplication: $x^n = x\cdot x \cdot ...\cdot x$ n times.

We also have a rule that says that any negative exponentiation can be rewritten as positive exponentiation of the reciprocal: $x^{-n} = \frac{1}{x^n}$.

Finally, addition of exponents can be expressed as follows: $x^n * x^m = x^{n+m}$.

So what is $x^0$? Zero can be written as n + -n. So $x^0 = x^{n + -n}$. But that can be rewritten as $x^n * x^{-n}$. That's the same as $\frac{x^n}{x^n} = 1$

Answer 4

The Freshman's dream. There's a bit to take in here.

First, you must remember that exponentiation is multiplication. This is the same way we think $2^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 16$. Now, with binomials, the same deal holds:

$$ (x+y)^4 = (x+y)(x+y)(x+y)(x+y) $$

But with the power zero, it's something we call an empty product, which we take to be equal to one. That is,

$$ (x + y)^0 = 1 $$

You've done what many elementary algebra students do, and think that distributing the exponents through the binomial is allowed. This is only allowed in a more advanced sense, when $x$ and $y$ are members of a commutative ring of characteristic $p$ - a prime. Of course, this is done in abstract algebra, a course math majors take in their junior or senior year of college.

Point is, anything to the zero is one, and you cannot distribute exponents through a binomial (for now).