0

How do I represent this statement using symbolic notation?

This is my attempt at it. $$ \forall n \in \Bbb{Z}, \forall m \in \Bbb{Z}, (n = 2q + 1) \wedge (m = 2k + 1) \Longrightarrow (m + n = 2l) $$

In English, I would read it as:

For all $n$ in the set of integers, and for all $m$ in the set of integers, if $n$ is odd and $m$ is odd, then $m + n$ is even.

Is this the right way to show it symbolically?

epimorphic
  • 3,219
user125535
  • 559
  • yup, that works, if you construct L = q + k + 1. You should also state the existence of such integers q and k (for all n, there exists a q such that n = 2q+1 etc.) – user130512 Feb 27 '14 at 20:00
  • 3
    The formula is : $\forall n \forall m [\exists q \exists k (n = 2q +1 \land m = 2k +1) \rightarrow \exists l (m+n = 2l) ]$, where all variables are "restricted" to the domain $\mathbb Z$, and your reading is right. – Mauro ALLEGRANZA Feb 27 '14 at 20:06

1 Answers1

2

Rather: $$ \forall n \in \Bbb{Z}, \forall m \in \Bbb{Z} \Big(\big(\exists q\in\mathbb Z (n = 2q + 1)\big) \wedge \big(\exists k\in\mathbb Z(m = 2k + 1)\big) \Longrightarrow \exists \ell\in\mathbb Z(m + n = 2\ell)\Big) $$ Note. Final version improved according to the suggestions of Henning Makholm.

Yiorgos S. Smyrlis
  • 83,933