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 $n - 1 = \big\{x | x \in (1,+\infty)\big\}$.

5 years, 10 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.

Staff - 5 years, 10 months ago

Thank you for commenting your thoughts on this. This is my first note on Brilliant. Thank you again.

- 5 years, 10 months ago

You can use the property of the exponents. Provided that $b \neq 0$, $b^n-b^{n-1} = b^{n-1}(\frac{b^n}{b^{n-1}}-\frac{b^{n-1}}{b^{n-1}})$ $b^n-b^{n-1} = b^{n-1}(b^{n-(n-1)}-b^{(n-1)-(n-1)})$ $b^n-b^{n-1} = b^{n-1}(b^1-b^0) = b^{n-1}(b-1)$ $Q.E.D.$

- 5 years, 10 months ago

Thank you very much, I have now provided more supportive evidence for my theorem.

- 5 years, 9 months ago

Kay's comment isn't just "supportive evidence". It is a proof of the statement which you stated.

Staff - 5 years, 9 months ago

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."

- 5 years, 9 months ago

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!

Staff - 5 years, 9 months ago