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TopNewestI read your response.though my English language is not very well, but I could get the answer of the question that was in my mind for a year. Thanks

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Welcome, Juliet. :)

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They weren't wrong in teaching you that, calculus is a stepping stone into understanding mathematics that are/were developmental. It's a gradual process. 0^0 was indeterminate until other rules from different countries and mathematicians created new formulas, so you have to change from one school of thought to another to arrive at a solution. Math despite our best attempts isn't always uniform, it depends on who's method you are studying and why. Just my two cents.

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Yes i agree

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I would just like to state something that my teachers have taught me over the years. They always tell all of their students to look at exponents as a pattern, for example: 2^5 = 32 2^4 = 16 2^3 = 8 2^2 = 4 2^1 = 2 2^0 = 1 2^-1 = 1/2

... and so on and so forth. The way to get each descending term there is to divide by two to get your next term, which should make sense because you multiply the previous term each time to get the next ascending term. With this pattern in mind, 2^0 has to equal 1, and it works the same with any other number raised to the power of 0. Because of this there is no doubt in my mind that x^0 = 1.

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how the hell can 0^0 be indeterminate anyway? You take nothing and multiply it by nothing and should give you nothing. I always thought that any number ^0 = 0...

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Multiplication is just repetitive addition of the same number, which is why we evaluate it first before we do addition.

Let x be any real number.

3(x) = x + x + x

2(x) = x + x

1(x) = x

0 is the additive identity number of the set of real numbers, which means that it is always there in the world of addition, though we just take it for granted.

3(x) = 0 + x + x + x

2(x) = 0 + x + x

1(x) = 0 + x

You can add as many zeros you want, and it would be like you've just pumped in some air for the x's to breathe.

3(x) = 0 + 0 + x + x + x

2(x) = 0 + 0 + x + x

1(x) = 0 + 0 + x

And what happens when we don't add x to a world of zeros?

0(x) = 0 + 0

Exponentiation, on the other hand, is just repetitive multiplication of the same number, which is why we evaluate it first before we do multiplication.

x^3 = (x)(x)(x)

x^2 = (x)(x)

x^1 = x

1 is the multiplicative identity number of the set of real numbers, which means that it is always there in the world of multiplication, though we just take it for granted. Ever wondered why prime number is defined to be greater than 1?

x^3 = 1(x)(x)(x)

x^2 = 1(x)(x)

x^1 = 1(x)

You can multiply by as many ones you want, and it would be like you've just pumped in some air for the x's to breathe.

x^3 = (1)(1)(x)(x)(x)

x^2 = (1)(1)(x)(x)

x^1 = (1)(1)(x)

And what happens when we don't multiply a world of ones by x?

x^0 = (1)(1)

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@Ricky Tang , normally, any number to the 0th power is 1. Type 0^0 in both the Google calculator and Desmos.com/calculator (Hint: also graph x^x on desmos) and let me know what both say.

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I know what everyone is saying is true but it just doesn't make any sense. Anything times nothing should be nothing, considering any number multiplied by 0 is 0 so why isnt it the same for a number compound to 0? I love math but this rule simply doesn't make any sense.

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Nothing multiplied by nothing is nothing, which is a something. :)

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\( 0^0 \) is indeterminate.

Depending on the scenario, sometimes people make the simplifying assumption that \( 0 ^ 0 = 0 \) or 1, to help make their calculations work in special cases.

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Incorrect. The zero power axiom trumps the zero multiplicity rule.

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Could you please give a special case in which the simplifying assumption that 0^0 = 0 helped make a calculation work?

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Graph it for a second and see what it says.

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Graphing it, however, still begs the question of whether 0^0 is indeterminate or not. If it's indeterminate, then there is no point at x=0. The graph of ((x^2)-1)/(x-1), for example, has a hole at x=1, even though its limit as x approaches 1 equals 2.

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