Prove $\left(\begin{matrix}n\\1\end{matrix}\right)=n$ - Help Needed

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Discussion Overview

The discussion revolves around proving the combinatorial identity $\left(\begin{matrix}n\\1\end{matrix}\right)=n$. Participants explore different approaches to establish this identity, including factorial definitions and induction methods.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents the identity and attempts to prove it using the factorial definition, stating $\left(\begin{matrix}n\\1\end{matrix}\right)=\frac{n!}{1!(n-1)!}=\frac{n!}{(n-1)!}$.
  • Another participant reiterates the factorial approach and suggests considering the recursive definition of factorial to simplify the proof.
  • A later reply seeks clarification on the term "recursive definition of factorial," prompting an explanation of how factorial is defined recursively as $n! = (n - 1)! n$.
  • Participants discuss the implications of different definitions of factorial and suggest that induction could also be a valid method for proving the identity.

Areas of Agreement / Disagreement

Participants generally agree on the validity of the identity and the approaches to prove it, but there is no consensus on the best method to use, as different participants suggest various techniques and definitions.

Contextual Notes

Some assumptions about the definitions of factorial and the methods of proof are not explicitly stated, which may affect the clarity of the discussion.

bergausstein
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please help me with this

prove that

$\left(\begin{matrix}n\\1\end{matrix}\right)=n$

this is what I get when I try to prove it,

$\left(\begin{matrix}n\\1\end{matrix}\right)=\frac{n!}{1!(n-1)!}=\frac{n!}{(n-1)!}$

thanks!
 
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Re: Proving a corrolary

bergausstein said:
please help me with this

prove that

$\left(\begin{matrix}n\\1\end{matrix}\right)=n$

this is what I get when I try to prove it,

$\left(\begin{matrix}n\\1\end{matrix}\right)=\frac{n!}{1!(n-1)!}=\frac{n!}{(n-1)!}$

thanks!

$$n! = 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n - 1) \cdot n$$

$$(n - 1)! = 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n - 1)$$

What's left when you divide $n!$ by $(n - 1)!$? In other words, what's $n!$ in terms of $(n - 1)!$? (hint: recursive definition of factorial). If this is still unsatisfactory, consider proving by induction...
 
Re: Proving a corrolary

Bacterius said:
$$n! = 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n - 1) \cdot n$$

$$(n - 1)! = 2 \cdot 3 \cdot 4 \cdot \cdots \cdot (n - 1)$$

What's left when you divide $n!$ by $(n - 1)!$? In other words, what's $n!$ in terms of $(n - 1)!$? (hint: recursive definition of factorial). If this is still unsatisfactory, consider proving by induction...

I get your point here, but what is "recurisive definition of factorial"?
 
Re: Proving a corrolary

bergausstein said:
I get your point here, but what is "recurisive definition of factorial"?

The factorial is typically defined recursively as $n! = (n - 1)! n$ and $1! = 1$, for instance:
$$5! = (5 - 1)! 5 = 4! 5 = ((4 - 1)! 4) 5 = (3! 4) 5 = ((3 - 1)! 3)4)5 = ((2! 3)4)5 = (((2 - 1)! 2)3)4)5 = (((1! 2)3)4)5 = (((2)3)4)5 = 2 \times 3 \times 4 \times 5 = 120$$
If you follow this definition, your result is immediate. If not, you can start with whatever your definition of the factorial is to arrive at the same result. For instance, if your definition is instead "product of all integers up to $n$" then you could use the product argument I gave in my previous post, or use induction (slightly tautologically, I have to say). Does that make sense?
 

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