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06 Jan 2011, 13:37
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Would someone explain how or perhaps where I could learn how
\(m^n - m^n^-^1 = m^n^-^1 (m-1)\)
Mostly I don't understand how \(m^n (1-m^-^1)\) becomes \(m^n^-^1 (m-1)\)
\(m^n - m^n^-^1 = m^n^-^1 (m-1)\)
Mostly I don't understand how \(m^n (1-m^-^1)\) becomes \(m^n^-^1 (m-1)\)
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06 Jan 2011, 14:14
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Cinna
Negative powers:
\(a^{-n}=\frac{1}{a^n}\)
Operations involving the same bases:
Keep the base, add or subtract the exponent (add for multiplication, subtract for division)
\(a^n*a^m=a^{n+m}\)
\(\frac{a^n}{a^m}=a^{n-m}\)
Now, as for your problem: you can directly factor out \(m^{n-1}\) from \(m^n-m^{n-1}\) and get \(m^{n-1}*(m-1)\) as \(m^{n-1}*(m-1)=m^{n-1}*m-m^{n-1}=m^n-m^{n-1}\).
Or the long way: \(m^n-m^{n-1}=m^n-m^n*m^{-1}\) -> factor out \(m^n\): \(m^n(1-m^{-1})=m^n(1-\frac{1}{m})=m^n(\frac{m-1}{m})=\frac{m^n}{m}*(m-1)=m^{n-1}*(m-1)\).
Check Number Theory chapter of Math Book for more on exponents: math-number-theory-88376.html
Hope it's clear.
06 Jan 2011, 15:13
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I certainly see the short method and understand that, I'm still stuck though on the long hand.
When I get to this step, \(m^n (\frac{m-1}{m})\) I don't recognize the rules that let us move the division to the other part of the forumla to finish it.
When I get to this step, \(m^n (\frac{m-1}{m})\) I don't recognize the rules that let us move the division to the other part of the forumla to finish it.
