# Thread: Discrete Math - Last Problem. Need Help!

1. ## 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 n² ≥ n.” What kind of proof did you use?

2. P(a) is the proposition “If n is a positive integer, then n² ≥ 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.

3. Originally Posted by clic-clac
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.

4. Mhh ok no problem
First I wanted to know :
where P(a) is the proposition “If n is a positive integer, then n² ≥ n.”
Is it P(a) or P(n)?

5. Originally Posted by clic-clac
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.

6. 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.

7. Originally Posted by clic-clac
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