solakis
- 19
- 0
How do we formalize the following:
For all natural Nos : n^2 is even\Rightarrow n is even
For all natural Nos : n^2 is even\Rightarrow n is even
Rasalhague said:Here's one way:
(\forall n \in \mathbb{N})[((\exists p \in \mathbb{N})[n^2=2p])\Rightarrow((\exists q\in \mathbb{N})[n=2q])].
Here's another:
\left \{ n \in \mathbb{N} \; | \; (\exists p \in \mathbb{N})[n^2 = 2p] \right \} \subseteq \left \{ n \in \mathbb{N} \; | \; (\exists q \in \mathbb{N})[n=2q] \right \}.
The symbol \mathbb{N}, like the name "the natural numbers", is ambiguous, so you might want to specify whether you mean the positive integers, \mathbb{N}_1=\mathbb{Z}_+ = \left \{ 1,2,3,... \right \}, or the non-negative integers, \mathbb{N}_0 =\mathbb{Z}_+ \cup \left \{ 0 \right \}=0,1,2,3,....
Rasalhague said:1. \enspace (\exists p \in \mathbb{N}_0)[n^2 = 2p]
\Rightarrow ((\exists a \in \mathbb{N}_0)[n=2a]) \vee ((\exists b \in \mathbb{N}_0)[n=2b+1]).
2. \enspace (\exists b \in \mathbb{N}_0)[n=2b+1])
\Rightarrow (\exists q \in \mathbb{N}_0)[n^2=(2b+1)^2=4b^2+4b+1=2q+1]
\Rightarrow \neg (\exists p \in \mathbb{N}_0)[n^2 = 2p].
3. \enspace\therefore ((\exists p \in \mathbb{N}_0)[n^2 = 2p]) \Rightarrow (\exists a \in \mathbb{N}_0)[n=2a] \enspace\enspace \blacksquare
Or, to see the argument clearer, let A = (\exists a \in \mathbb{N}_0)[n=2a] and let B = (\exists b \in \mathbb{N}_0)[n=2b+1], and suppose (\exists p \in \mathbb{N}_0)[n^2 = 2p]. Then A \vee B. But \neg B. So A.