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

2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 3 months ago

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

Staff - 2 years, 3 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."

- 2 years, 2 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 - 2 years, 2 months ago

Note that you have not proven the theorem. You have only demonstrated that it is true in 5 specific cases.

Staff - 2 years, 4 months ago

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

- 2 years, 4 months ago