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Mathematics 19 Online
OpenStudy (anonymous):

Why anything to the power zero is 1? for example, 55^0=1

OpenStudy (anonymous):

Okay first of all \[ x^a\times x^b = x^c \]What is \(c\)?

OpenStudy (anonymous):

In terms of \(a\) and \(b\), what is \(c\)?

OpenStudy (anonymous):

@tryder Do you know?

OpenStudy (anonymous):

not really.

OpenStudy (anonymous):

Experiment.

OpenStudy (anonymous):

\[ 2^2\times 2^3=2^c \]

OpenStudy (anonymous):

\[ 2^2\times 2^2 = (2\times 2)\times (2\times 2\times 2) = 2^c \]

OpenStudy (anonymous):

In this case, what is \(c\)?

OpenStudy (anonymous):

Hint: count.

OpenStudy (anonymous):

The empty product is 1. When you multiply no numbers together you get 1. If the empty product way not 1, multiplication would not make sense. Multiplying 5 * 2, is the same as multiplying no numbers, then multiplying the result by 5 and then by 2. So we have: 5 * 2 = empty product * 5 * 2 And the only way this is true, is if the empty product is 1. So when raising to the power of zero, we are multiplying a number by itself zero times, which is the same as the empty product, so is equal to 1.

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

Oh wow, lol.

OpenStudy (anonymous):

When you multiply, the exponents are added: \[ x^a\times x^b=x^{a+b} \]

OpenStudy (anonymous):

Suppose one of the exponents are \(0\):\[ x^0\times x^b=x^{0+b}=x^b \]

OpenStudy (anonymous):

\[ x^0\times x^b=x^b \]Divide both sides by \(x^b\):\[ x^0 = 1 \]

OpenStudy (anonymous):

thank you @wio for helping. sorry I didn't understand you in the beginning. I am not good at math

OpenStudy (anonymous):

In the beginning, there was no definition of the \(0\) exponent. The definition was chosen based on what would be consistent with current properties of exponents.

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