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 to be left with 2x2, or 2^2
The quicker way to calculate this is doing 2^(5-3) which when you resolve the bracket is just 2^2 or 2x2.
If both numbers in the bracket are the same it will always resolve to 0, and if both numbers are the same the top and bottom number in the fraction is the same, 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 to be left with 2x2, or 2^2
The quicker way to calculate this is doing 2^(5-3) which when you resolve the bracket is just 2^2 or 2x2.
If both numbers in the bracket are the same it will always resolve to 0, and if both numbers are the same the top and bottom number in the fraction is the same, 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…
It’s just 1+1-1 with more steps, I don’t see the problem.