Let us consider the statement \(b^n - b^{n-1} = b^{n-1} (b-1)\).
By quotient property of exponents,
bn÷bn−1=b
Subtracting 1 from both sides,
(bn÷bn−1)−1=b−1
(bn÷bn−1)−(bn−1÷bn−1)=b−1
(bn−bn−1)÷bn−1=b−1
Multiplying both sides by bn−1,
bn−bn−1=bn−1(b−1)
Let b={x∣x∈R}.
If b=0, let n−1={x∣x∈(1,+∞)}.
#NumberTheory
#Exponents
#Theorem
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Note that you have not proven the theorem. You have only demonstrated that it is true in 5 specific cases.
Log in to reply
Thank you for commenting your thoughts on this. This is my first note on Brilliant. Thank you again.
You can use the property of the exponents. Provided that b=0, bn−bn−1=bn−1(bn−1bn−bn−1bn−1) bn−bn−1=bn−1(bn−(n−1)−b(n−1)−(n−1)) bn−bn−1=bn−1(b1−b0)=bn−1(b−1) Q.E.D.
Log in to reply
Thank you very much, I have now provided more supportive evidence for my theorem.
Log in to reply
Kay's comment isn't just "supportive evidence". It is a proof of the statement which you stated.
Log in to reply
Log in to reply
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!