An Exponent Theorem

Let us consider the statement bnbn1=bn1(b1)b^n - b^{n-1} = b^{n-1} (b-1).

By quotient property of exponents,

bn÷bn1=bb^n \div b^{n-1} = b

Subtracting 1 from both sides,

(bn÷bn1)1=b1(b^n \div b^{n-1}) - 1 = b - 1

(bn÷bn1)(bn1÷bn1)=b1(b^n \div b^{n-1}) - (b^{n-1} \div b^{n-1}) = b - 1

(bnbn1)÷bn1=b1 (b^n - b^{n-1}) \div b^{n-1} = b - 1

Multiplying both sides by bn1 b^{n-1} ,

bnbn1=bn1(b1)\boxed{b^n - b^{n-1} = b^{n-1} (b-1)}


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

If b=0b = 0, let n1={xx(1,+)}n - 1 = \big\{x | x \in (1,+\infty)\big\}.

Note by Adriel Padernal
3 years, 11 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 - 3 years, 11 months ago

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

Adriel Padernal - 3 years, 11 months ago

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You can use the property of the exponents. Provided that b0b \neq 0, bnbn1=bn1(bnbn1bn1bn1)b^n-b^{n-1} = b^{n-1}(\frac{b^n}{b^{n-1}}-\frac{b^{n-1}}{b^{n-1}}) bnbn1=bn1(bn(n1)b(n1)(n1))b^n-b^{n-1} = b^{n-1}(b^{n-(n-1)}-b^{(n-1)-(n-1)}) bnbn1=bn1(b1b0)=bn1(b1)b^n-b^{n-1} = b^{n-1}(b^1-b^0) = b^{n-1}(b-1) Q.E.D.Q.E.D.

Kay Xspre - 3 years, 11 months ago

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

Adriel Padernal - 3 years, 10 months ago

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

Calvin Lin Staff - 3 years, 10 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 - 3 years, 10 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 - 3 years, 10 months ago

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