Anything multiplied by 0 is 0 right so why does x number raised to the power of 0 = 1? isnt it x^0 = x*0 (im turning grade 10 and i asked my teacher about this he told me its because its just what he was taught 💀)
In: Mathematics
Great question! There’s a good, simple answer to this one, but unfortunately textbooks and teachers don’t seem to know it.
You don’t know what 2^0 should be [1]. So let’s take some positive power of 2, say 2^3. We can rewrite 2^3 as 2^(3+0). Because 2^(3+0) should be equal to 2^3 * 2^0, we get this equation: 2^3 = 2^(3+0) = 2^3 * 2^0.
So 2^0 should be the number that gives 2^3 when multiplied by 2^3. That number is, of course, 1.
There’s nothing special about 2, the same argument can be applied to any nonzero number [3].
[1] The word “should” is doing a lot of heavy lifting here.
There are two interesting patterns that occur for exponents that are positive integers:
– If you have something like x^3 * x^4, you end up multiplying seven x’s together, so it’s just x^7. In general, x^a * x^b = x^(a+b).
– If you have something like (x^3)^4, it’s (x * x * x)^4, you end up multiplying four groups of three x’s each, so the total number of x’s you multiply is 3 * 4. In general, (x^a )^b = x^(a * b).
We define 2^0 = 1, because we want the pattern to continue for 0.
So when I say “2^0 should be equal to 1”, it’s a kind of short-hand to say “defining 2^0 to be 1 is what makes the pattern continue, so let’s define it that way.” [2]
Negative exponents are defined in a similar way: Once we know 2^0 = 1, we can write 1 = 2^0 = 2^(3 + -3) = 2^3 2^-3, so 2^-3 should be the number that gives 1 when multiplied by 2^3. Which is of course 1/(2^3 ). There’s nothing special about 2 and 3, in general x^-a = 1/(x^a).
You can even do the same for fractional exponents: 8 = 8^1 = 8^(1/3 + 1/3 + 1/3) = 8^(1/3) * 8^(1/3) * 8^(1/3). So 8^(1/3) should be the thing that, when multiplied by itself 3 times, gives you 8. This is of course the cube root of 8. In general x^(1/n) is the nth root of x.
Then to get fractional exponents with a numerator that’s something other than 1, you can figure out what x^(m/n) should be by raising x^(1/n) to the mth power.
[2] There’s also a bit of peer pressure at work: Virtually all mathematicians like these patterns to continue, so they’ve embraced the definitions for hundreds of years. So these definitions have worked their way into a lot of existing mathematical infrastructure (textbooks, research papers, programming languages). You’re free to decide to you think 2^0 should be something else if you want — undefined or 753 or whatever — but that decision would cause a lot of confusion and miscommunication if you ever work with anybody else.
[3] Of course 0 is a special case, 0^0 is undefined (but some people disagree with this definition).
Another way to see it is that for (non-negative) whole-numbered exponentiation defined as repeated multiplication, the definition is actually:
x • y^n := x multiplied by y, n times (regardless of what x is).
So x • y^0 is then x multiplied by y, 0 times, which is obviously just x.
Hence y^0 must be 1 for all y, notably even when y is 0 (if you define whole numbered exponentiation this way).
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