Powers of 0 and 1

[mathdad]

If x is a number, then x^0 = 1 and x^1 = x. Does anyone know why that is true?

 

[Otis]

Here's one way to think about it. The exponent tells us how many base factors we've got. (In the expression x^4, the exponent is 4 and the base is x.)

Also, when we write a monomial, like x^4, we don't normally write the coefficient 1 that's in front, but it's understood to be there.

1*x^3 = 1*x*x*x (three base factors multiplied together)

1*x^2 = 1*x*x (two base factors)

1*x^1 = 1*x (one base factor)

1*x^0 = 1 (zero base factors)

?

 

[ksdhart2]

Here's another way to think about it, involving what exponents mean for integer powers. Let's consider a few examples of dividing various powers of \(x\):

\(\displaystyle \frac{x^5}{x^3} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot x \cdot x}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x}} = x \cdot x = x^2\)

\(\displaystyle \frac{x^6}{x^2} = \frac{\cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot x \cdot x}{\cancel{x} \cdot \cancel{x}} = x \cdot x \cdot x \cdot x = x^4\)

\(\displaystyle \frac{x^3}{x^4} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x}}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot x} = \frac{1}{x} = x^{-1}\)

These seem to suggest that we have the property that \(\displaystyle \frac{x^a}{x^b} = x^{a-b}\). And, indeed, this is one of the rules of exponents. But now the logical question becomes: What if the powers were the same?

\(\displaystyle \frac{x^2}{x^2} = \frac{\cancel{x} \cdot \cancel{x}}{\cancel{x} \cdot \cancel{x}} = \frac{1}{1} = 1\)

\(\displaystyle \frac{x^3}{x^3} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x}}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x}} = \frac{1}{1} = 1\)

The rule we just derived holds for any power, we know that \(\displaystyle \frac{x^n}{x^n} = x^{n-n} = x^0\) for all \(n\) and all non-zero \(x\). Then if we put the two together, that tells us that \(\displaystyle x^0 = 1\) for all non-zero \(x\).

 

[mathdad]

Here's one way to think about it. The exponent tells us how many base factors we've got. (In the expression x^4, the exponent is 4 and the base is x.)

Also, when we write a monomial, like x^4, we don't normally write the coefficient 1 that's in front, but it's understood to be there.

1*x^3 = 1*x*x*x (three base factors multiplied together)

1*x^2 = 1*x*x (two base factors)

1*x^1 = 1*x (one base factor)

1*x^0 = 1 (zero base factors)

?

Nicely explained.

 

[mathdad]

Here's another way to think about it, involving what exponents mean for integer powers. Let's consider a few examples of dividing various powers of \(x\):

\(\displaystyle \frac{x^5}{x^3} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot x \cdot x}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x}} = x \cdot x = x^2\)

\(\displaystyle \frac{x^6}{x^2} = \frac{\cancel{x} \cdot \cancel{x} \cdot x \cdot x \cdot x \cdot x}{\cancel{x} \cdot \cancel{x}} = x \cdot x \cdot x \cdot x = x^4\)

\(\displaystyle \frac{x^3}{x^4} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x}}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x} \cdot x} = \frac{1}{x} = x^{-1}\)

These seem to suggest that we have the property that \(\displaystyle \frac{x^a}{x^b} = x^{a-b}\). And, indeed, this is one of the rules of exponents. But now the logical question becomes: What if the powers were the same?

\(\displaystyle \frac{x^2}{x^2} = \frac{\cancel{x} \cdot \cancel{x}}{\cancel{x} \cdot \cancel{x}} = \frac{1}{1} = 1\)

\(\displaystyle \frac{x^3}{x^3} = \frac{\cancel{x} \cdot \cancel{x} \cdot \cancel{x}}{\cancel{x} \cdot \cancel{x} \cdot \cancel{x}} = \frac{1}{1} = 1\)

The rule we just derived holds for any power, we know that \(\displaystyle \frac{x^n}{x^n} = x^{n-n} = x^0\) for all \(n\) and all non-zero \(x\). Then if we put the two together, that tells us that \(\displaystyle x^0 = 1\) for all non-zero \(x\).

More good notes for my files.