You can see this in the example above but perhaps it's better to use different powers to make things a bit clearer.
2^5=2x2x2x2x2
2^3=2x2x2
(25)/(23)=(2x2x2x2x2)/(2x2x2)
You can cancel 3 of the 2s from the top and bottom of the fraction to be left with 2x2, or 2^2.
I.e. (25)/(23)=2^2
The quicker way to calculate this is doing 2^(5-3) which when you resolve the bracket is obviously just 2^2 or 2x2.
If both numbers in the bracket are the same the bracket will always resolve to 0, which is the same as saying a number divided by itself, any number divided by itself is one so it follows that any number to the power 0 is also 1 (because it's essentially exactly the same calculation).
Yes, of course, obviously…JFC, what??
2^(a-b) = (2a)/(2b)
You can see this in the example above but perhaps it's better to use different powers to make things a bit clearer.
2^5=2x2x2x2x2
2^3=2x2x2
(25)/(23)=(2x2x2x2x2)/(2x2x2)
You can cancel 3 of the 2s from the top and bottom of the fraction to be left with 2x2, or 2^2.
I.e. (25)/(23)=2^2
The quicker way to calculate this is doing 2^(5-3) which when you resolve the bracket is obviously just 2^2 or 2x2.
If both numbers in the bracket are the same the bracket will always resolve to 0, which is the same as saying a number divided by itself, any number divided by itself is one so it follows that any number to the power 0 is also 1 (because it's essentially exactly the same calculation).
Rule = #^0 = # x 1
Don't ask why…got it.
No not quite, #^0 = 1
Wait, so 5^0 = 1??
Yup
5^0 can be rewritten as 5^(2-2)
5^(2-2) = (52)/(52)
This is a number divided by itself so cancels to 1 every time, regardless of #.
That was pretty complicated, here is a simpler answer I hsve come up with:
1=(2x2x2)/(2x2x2)=2³/2³=2³⁻³＝2⁰
If that makes sense to you…
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