Results 1 to 7 of 7

Thread: Discrete Math - Last Problem. Need Help!

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    93

    Discrete Math - Last Problem. Need Help!

    Q: Prove the proposition P(1) , where P(a) is the proposition If n is a positive integer, then n ≥ n. What kind of proof did you use?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    P(a) is the proposition If n is a positive integer, then n ≥ n.
    If it's $\displaystyle a$ and not $\displaystyle n$, for all $\displaystyle a$ , $\displaystyle P(a)$ is a tautology and then $\displaystyle P(1)$ is true. (Let $\displaystyle n$ be a positive integer, $\displaystyle ( n=n \wedge n\geq 1 )\Rightarrow n^2=n\times n\geq 1\times n=n$

    If it's $\displaystyle n$ and not $\displaystyle a$ , $\displaystyle 1$ is a positive integer and $\displaystyle 1^2=1\geq 1$ , therefore $\displaystyle P(1)$ is true.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    Posts
    93
    Quote Originally Posted by clic-clac View Post
    If it's $\displaystyle a$ and not $\displaystyle n$, for all $\displaystyle a$ , $\displaystyle P(a)$ is a tautology and then $\displaystyle P(1)$ is true. (Let $\displaystyle n$ be a positive integer, $\displaystyle ( n=n \wedge n\geq 1 )\Rightarrow n^2=n\times n\geq 1\times n=n$

    If it's $\displaystyle n$ and not $\displaystyle a$ , $\displaystyle 1$ is a positive integer and $\displaystyle 1^2=1\geq 1$ , therefore $\displaystyle P(1)$ is true.
    thanks click-clac for helping me with this. As I go over what you wrote, I am still having difficulty understanding this.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Mhh ok no problem
    First I wanted to know :
    where P(a) is the proposition If n is a positive integer, then n ≥ n.
    Is it P(a) or P(n)?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    Posts
    93
    Quote Originally Posted by clic-clac View Post
    Mhh ok no problem
    First I wanted to know :
    Is it P(a) or P(n)?
    it is P(n). My bad. I accidentily looked at the question below it also. That is where P(a) came from.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Nov 2008
    From
    Paris
    Posts
    354
    Ok! I'll write $\displaystyle \mathbb{N}^*$ instead of $\displaystyle \mathbb{N}-\{0\}.$

    We have $\displaystyle P(n):\ n\in\mathbb{N}^*\Rightarrow n^2\geq n$
    Thus $\displaystyle P(1):\ 1\in \mathbb{N}^*\Rightarrow 1^2\geq 1$

    Consider the formula $\displaystyle F:\ A\Rightarrow B,$ if $\displaystyle A$ is false or if $\displaystyle B$ is true, then $\displaystyle F$ is true.
    $\displaystyle 1^2\geq 1$ is true, therefore the implication $\displaystyle 1\in \mathbb{N}^*\Rightarrow 1^2\geq 1$ is true: we've proved $\displaystyle P(1).$

    ***********

    Another way to prove $\displaystyle P(1)$ (and even more) is to say that the proposition $\displaystyle P(x)$ is a tautology (i.e. is always true). Indeed:
    $\displaystyle \circ$ If $\displaystyle x$ isn't a positive integer, then $\displaystyle x\in\mathbb{N}^*$ is false and the implication $\displaystyle x\in\mathbb{N}^*\Rightarrow x^2\geq x$ is true: $\displaystyle P(x)$ is true.
    $\displaystyle \circ$ If $\displaystyle x$ is a positive integer, $\displaystyle (x\geq x\ \text{and}\ x\geq 1)\Rightarrow x\times x\geq x\times 1\Rightarrow x^2\geq x$ because the multiplication on $\displaystyle \mathbb{N}$ is compatible with the order. So the second part of the implication is true, and again $\displaystyle P(x)$ is true.

    In conclusion, whatever is $\displaystyle x$ (a number, a matrix, a topological space...), $\displaystyle P(x)$ is true. In particular, $\displaystyle P(1)$ is true.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jan 2009
    Posts
    93
    Quote Originally Posted by clic-clac View Post
    Ok! I'll write $\displaystyle \mathbb{N}^*$ instead of $\displaystyle \mathbb{N}-\{0\}.$

    We have $\displaystyle P(n):\ n\in\mathbb{N}^*\Rightarrow n^2\geq n$
    Thus $\displaystyle P(1):\ 1\in \mathbb{N}^*\Rightarrow 1^2\geq 1$

    Consider the formula $\displaystyle F:\ A\Rightarrow B,$ if $\displaystyle A$ is false or if $\displaystyle B$ is true, then $\displaystyle F$ is true.
    $\displaystyle 1^2\geq 1$ is true, therefore the implication $\displaystyle 1\in \mathbb{N}^*\Rightarrow 1^2\geq 1$ is true: we've proved $\displaystyle P(1).$

    ***********

    Another way to prove $\displaystyle P(1)$ (and even more) is to say that the proposition $\displaystyle P(x)$ is a tautology (i.e. is always true). Indeed:
    $\displaystyle \circ$ If $\displaystyle x$ isn't a positive integer, then $\displaystyle x\in\mathbb{N}^*$ is false and the implication $\displaystyle x\in\mathbb{N}^*\Rightarrow x^2\geq x$ is true: $\displaystyle P(x)$ is true.
    $\displaystyle \circ$ If $\displaystyle x$ is a positive integer, $\displaystyle (x\geq x\ \text{and}\ x\geq 1)\Rightarrow x\times x\geq x\times 1\Rightarrow x^2\geq x$ because the multiplication on $\displaystyle \mathbb{N}$ is compatible with the order. So the second part of the implication is true, and again $\displaystyle P(x)$ is true.

    In conclusion, whatever is $\displaystyle x$ (a number, a matrix, a topological space...), $\displaystyle P(x)$ is true. In particular, $\displaystyle P(1)$ is true.
    I see now. Thanks clic-clac
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Need Help on a hard problem on discrete math.
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Jul 19th 2011, 03:17 PM
  2. Discrete Math Problem
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Nov 17th 2010, 11:29 PM
  3. Discrete Math Problem! HELP!!
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: Mar 30th 2008, 10:48 AM
  4. Discrete Math Problem - Relation
    Posted in the Discrete Math Forum
    Replies: 5
    Last Post: Feb 27th 2007, 08:08 PM
  5. Discrete Math problem
    Posted in the Discrete Math Forum
    Replies: 8
    Last Post: Feb 15th 2007, 01:57 PM

Search Tags


/mathhelpforum @mathhelpforum