Why is anything the power of 0 always 1?

Now I know this is one known fact that students just accept. Why is it that anything raised to the power of 0 is 1? We know that 2^{3}=2 \times 2 \times 2 = 8, meaning we take the 2 and multiply it three times. So wouldn’t 2^0 mean we multiply 2 zeroth times, and get back 2? So thats baffling.

This can be solved intuitively, like how we looked at 0! previously. We can look at b^a as the number of ways to arrange a set of a number with numbers from 1 to b numbers.

Consider 2^3. This means we arrange the set of 3 numbers where the numbers are just 1 and 2. So we have \{1, 1, 1 \}, \{1, 1, 2 \}, \{1, 2, 2 \}, \{2, 2, 2 \}, \{1, 2, 1 \}, \{2, 2, 1 \}, \{2, 1, 1 \}, \{2, 1, 2 \}

And we find the number of ways to be eight!

Showing 2 comments
  • Aravindan Sai Narayanan
    Reply

    0^0 is undefined.It doesn’t make sense to say that 0^0 is always one.To overcome this problem when a function of the form f(x)^g(x) has f(x) approaching zero as x tends to a and thus,f(x) approaches zero and when x tends to a given point a ,g(x) approaches zero.Thus,as x tends to a ,we have f(x)^g(x) as 0^0.0^0 can be any value and to overcome this indeterminate form, we need to first transform it to 0*infinity form; followed by changing it to infinity/infinity or 0/0 form;afterwhich,we apply L Hospital’s Rule to evaluate the limit as x tends to a.Other indeterminate forms include infinity^0, 0^infinity,1^infinity ,0*infinity,infinity-infinity, infinity/infinity and 0/0.See the attached link for more and detailed information about indeterminate forms /Users/akshayleninsingapore/Downloads/Indeterminate_form.pdf

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