What is the contradiction in the proof for M/I\subseteqJ/I and M\subseteqJ?

[EV33]

Homework Statement

I am curious,
if I,J, and M are ideals of the commutative ring R, and M/I$\subseteq$J/I, then M$\subseteq$J

Homework Equations

M/I = { m+I : m is in M}

J/I = { j+I : j is in J}

I$\subseteq$R is an ideal if
1.) if a and b are in I then a+b is in I
2.) if r is in R and a is in I then a*r is in I

The Attempt at a Solution

Proof by contradiction:

Assume M/I$\subseteq$J/I,
and M is not contained in J.

Since M is not contained in J then there exists a m in M that is not in J.
Then m+I is in M/I but not in J/I.

This is a contradiction to the hypothesis that M/I$\subseteq$J/I. Thus M is contained in J. (QED)

Any flaws?

Thank you for your time.

 

[jbunniii]

Since M is not contained in J then there exists a m in M that is not in J.
Then m+I is in M/I but not in J/I.

This step isn't necessarily obvious. The coset m+I can go by many different names. How do you know it isn't the same set as j+I for some j in J?

I suggest assuming that m+I = j+I for some j in J, and deriving a contradiction.