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An Exponent Theorem

Let us consider the statement \(b^n - b^{n-1} = b^{n-1} (b-1)\).

By quotient property of exponents,

\(b^n \div b^{n-1} = b\)

Subtracting 1 from both sides,

\((b^n \div b^{n-1}) - 1 = b - 1\)

\((b^n \div b^{n-1}) - (b^{n-1} \div b^{n-1}) = b - 1\)

\( (b^n - b^{n-1}) \div b^{n-1} = b - 1\)

Multiplying both sides by \( b^{n-1} \),

\(\boxed{b^n - b^{n-1} = b^{n-1} (b-1)}\)


Let \(b = \big\{x | x \in \mathbb{R}\big\}\).

If \(b = 0\), let \(n - 1 = \big\{x | x \in (1,+\infty)\big\}\).

Note by Adriel Padernal
1 year, 6 months ago

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You can use the property of the exponents. Provided that \(b \neq 0\), \[b^n-b^{n-1} = b^{n-1}(\frac{b^n}{b^{n-1}}-\frac{b^{n-1}}{b^{n-1}})\] \[b^n-b^{n-1} = b^{n-1}(b^{n-(n-1)}-b^{(n-1)-(n-1)})\] \[b^n-b^{n-1} = b^{n-1}(b^1-b^0) = b^{n-1}(b-1)\] \[Q.E.D.\] Kay Xspre · 1 year, 6 months ago

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@Kay Xspre Thank you very much, I have now provided more supportive evidence for my theorem. Adriel Padernal · 1 year, 5 months ago

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@Adriel Padernal Kay's comment isn't just "supportive evidence". It is a proof of the statement which you stated. Calvin Lin Staff · 1 year, 5 months ago

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@Calvin Lin Thank you, I understand. At first, I was doubtful in a way that others might correct me for calling it "proof" so I decided to call it "supportive evidence." Adriel Padernal · 1 year, 5 months ago

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@Adriel Padernal Thanks for updating the note!

It can take a while for you to become familiar with mathematical rigor. We all have to start somewhere, and I'm glad that you're learning! Calvin Lin Staff · 1 year, 5 months ago

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Note that you have not proven the theorem. You have only demonstrated that it is true in 5 specific cases. Calvin Lin Staff · 1 year, 6 months ago

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@Calvin Lin Thank you for commenting your thoughts on this. This is my first note on Brilliant. Thank you again. Adriel Padernal · 1 year, 6 months ago

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