this post was submitted on 11 Oct 2023
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submitted 1 year ago* (last edited 1 year ago) by hypertown to c/[email protected]
 
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[–] Globulart 3 points 1 year ago* (last edited 1 year ago) (1 children)

2^(a-b) = (2^a)/(2^b)

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

(2^5)/(2^3)=(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. (2^5)/(2^3)=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).

[–] hemmes 2 points 1 year ago (1 children)

Rule = #^0 = # x 1

Don't ask why...got it.

[–] Globulart 3 points 1 year ago (1 children)
[–] hemmes 2 points 1 year ago (1 children)
[–] Globulart 2 points 1 year ago* (last edited 1 year ago)

Yup

5^0 can be rewritten as 5^(2-2)

5^(2-2) = (5^2)/(5^2)

This is a number divided by itself so cancels to 1 every time, regardless of #.