Equality of two sets written differently

Question

$$A = \{2m+ 1:m \text{ exists in } \mathbb{Z}\}$$

$$B= \{2n + 3:n \text{ exists in } \mathbb{Z}\}$$

For this question, it seems that $A=B$, and we know it's equal because we can just plug in numbers from the universe of all integers, but my question is if it is possible to REWRITE $B$ so that $A=B$ for $m=n$ from universe of all integers. Right now, it's only equal if I plug in different $m$ and $n$. In other words, is it possible to rewrite and simplify the equation $2n+3$ into $2n+1$?

Note that $2n + 3 = 2(n-1) + 1$, so we can rewrite B with $m = n-1$ — Mathmo123, Jul 01 '14 at 21:00

evinda · Accepted Answer · 2014-07-01T21:35:19.450

0

Notice that $2(m-1)+3=2m-2+3=2m+1$.

edited Jul 01 '14 at 21:35

answered Jul 01 '14 at 21:01

evinda

7,823

I'm assuming you isolated n and moved everything to the side of m? – Belphegor Jul 01 '14 at 21:14

score 0 · Answer 2 · answered Jul 01 '14 at 21:30

0

$$A=\{2m+1:m \in \mathbb{Z}\}=\{2(m-1)+3:m-1 \in \mathbb{Z}\}=\{2k+3:k \in \mathbb{Z}\}=B$$

answered Jul 01 '14 at 21:30

Surb

55,662

Equality of two sets written differently

2 Answers2