Uniqueness of Real z such that x Choose n+1 Equals y Choose n+1 Plus z Choose n
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Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $x, y \in \R$ be real numbers which satisfy:
- $n \le y \le x \le y + 1$
Then there exists a unique real number $z$ such that:
- $\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$
where $n - 1 \le z \le y$.
Proof
We have:
\(\ds \dbinom y {n + 1}\) | \(\le\) | \(\ds \dbinom x {n + 1}\) | Ordering of Binomial Coefficients | |||||||||||
\(\ds \) | \(\le\) | \(\ds \dbinom {y + 1} {n + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dbinom y {n + 1} + \dbinom y n\) | Pascal's Rule |
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Sources
- 1979: László Lovász: Combinatorial Problems and Exercises: Problem $13.31 \ \text {(a)}$
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: Exercise $66$