1. ## Set proof

My teacher gave me this to prove, could someone help me prove it? Thanks.

Proposition: {4m + 1 : m e Z} c {2n - 1 : n e Z}

e = element
c = subset
Z = set of all integers

Sorry if its a little hard to read, don't know how to make the element symbol, the subset symbol, or the set of all integers symbol.

2. Originally Posted by k1ll3rdr4g0n
My teacher gave me this to prove, could someone help me prove it? Thanks.

Proposition: {4m + 1 : m e Z} c {2n - 1 : n e Z}

e = element
c = subset
Z = set of all integers

Sorry if its a little hard to read, don't know how to make the element symbol, the subset symbol, or the set of all integers symbol.
What you need to show is that for every number of the form 4m + 1, that this number is also of the form 2n - 1, such that m, n are integers. This is simple enough, just set:
2n - 1 = 4m + 1

Given an m we should be able to find an n, and we can:
n = 2m + 1
(Since m is an integer, so is 2m. Since 2m is an integer so is 2m + 1. Thus n is also an integer.)

So every element of the set $\displaystyle \{ 4m+1|m \in \mathbb{Z} \}$ is an element of $\displaystyle \{ 2n-1|n \in \mathbb{Z} \}$.

In the event you need this to be a proper inclusion, note that not every element of $\displaystyle \{ 2n-1|n \in \mathbb{Z} \}$ is in $\displaystyle \{ 4m+1|m \in \mathbb{Z} \}$ because:
4m + 1 = 2n - 1
gives
m = (1/2)(n - 1)

Which means there are values of n such that m is not an integer. So we have:
$\displaystyle \{ 4m+1|m \in \mathbb{Z} \} \subset \{ 2n-1|n \in \mathbb{Z} \}$
but not the other way around. Hence this is a proper inclusion.

-Dan