• Globulart
    16·
    3 years ago

    This isn’t strictly speaking a proof, but it did help me to accept it as it demonstrates the function that makes it 1.

    2^3 = 2x2x2

    2^2 = 2x2

    (23)/(22) = (2x2x2)/(2x2) = 2

    = 2^(3-2)

    In general terms:

    (xa)/(xb) = x^(a-b)

    If a and b are the same number this is x^0 and obviously (xa)/(xa) is one because anything divided by itself is 1.

    Hope that helps

    • hemmesEnglish
      72·
      3 years ago

      Yes, of course, obviously…JFC, what??

      • Globulart
        3·
        3 years ago

        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).

        • hemmesEnglish
          2·
          3 years ago

          Rule = #^0 = # x 1

          Don’t ask why…got it.

              • Globulart
                2·
                3 years ago

                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 #.

      • Flumsy@feddit.de
        31·
        3 years ago

        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…