6

I am not cheating my homework. This was already handed in, and handed back with a grade. I'm not sure if my teacher is a harsh grader, or if I really am doing as bad as she makes it out to be. I figured a few people here might be able to help me get this a little more solid since I'm at a loss as to what to do.

$A = \{n \in \mathbb{Z}|\exists k \in \mathbb{Z}|n=4k+1\}$

$B = \{n \in \mathbb{Z}|\exists j \in \mathbb{Z}|n=4j-7\}$

Prove $A=B$

Proof: We first show that $A \subseteq B$. Let $n \in A$ and show that $n \in B$. Since $n \in A$ and $n=4k+1$ for some fixed $k$ it implies that $n=4j-7$ for some unknown $j$. We set the two equations equal to eachother and solve for $j$.

$$4j-7=4k+1$$ $$4j=4k+8$$ $$j=k+2$$

Thus any $n \in A$ must be in $B$. We then show that $B \subseteq A$. Let $n \in B$ and show that $n \in A$. Since $n \in B$ and $n=4j-7$ for some fixed $j$ it implies that $n=4k+1$ for some unknown $k$. We set the two equations equal and solve for $k$.

$$4k+1=4j-7$$ $$4k=4j-8$$ $$k=j-2$$

Thus any $n \in B$ must be in $A$.

Thus $A \subseteq B$ and $B \subseteq A$ which proves that $A=B$ as required.

symplectomorphic
  • 18,604
Michael Rector
  • 127
  • 4
    The proof is almost perfect, but you made a writing mistake (twice). "Since $n \in A$ and $n=4k+1$ for some fixed $k$ it implies that $n=4j-7$ for some unknown $j$". At this point in the proof, what you want to do is prove that it does imply that $n=4j-7$, for some integer $j$, but you can't just say it implies. Better would be something like "we'd like to prove there exists an integer $j$ such that $n=4j-7$. By this and that, it's enough to consider $j=k+2$", otherwise very nice job. – Git Gud Oct 25 '16 at 22:40
  • 2
    For example, you should say that if $n\in A$ then there is a $k$ such that $n=4k+1$. And since $4k+1=4(k+2)-7$ then $n$ is also in $B$ because there is an element of $\mathbb{Z}$, namely $k+2$ such that $4(k+2)-7=n$. Similarly for the other direction. – John Wayland Bales Oct 25 '16 at 22:44
  • 1
    something along the lines of $n = 4k+1 = 4(k+2)-7$ more cleanly shows that $n\in A\implies n\in B$ – Doug M Oct 25 '16 at 22:48

3 Answers3

1

The problem is that your proof is kind of backwards. You assume $4j-7=4k+1$ and then solve for $j$. This is a valid way to find the $j$ you should choose, but in proofs, by assuming that equation, you are implicitly assuming that $A \subseteq B$ instead of proving it. Therefore, you need to start with $j=k+2$ and then show that such value of $j$ leads to $4j-7=4k+1$.

This might seem odd because you're basically working in the reverse order of how you solved the problem, but often, how you originally solve your problem and what your proof is go in reverse order because we solve problems by assuming and manipulating equations in order to get down to something that is obviously true while we solve proofs by starting with those obviously true statements and then showing those equations are true.

To make this more clear, here's a quick re-writing of your proof:

We first show that $A \subseteq B$. Let $n \in A$ so that $n=4k+1$ for some integer $k$. Then, choose $j=k+2$ so that $j$ is also an integer. By multiplying this equation by $4$ and subtracting by $7$, we see $4j-7=4k+1$. This means $n=4j-7$ for some integer $j$, implying that $n \in B$.

We then show that $B \subseteq A$. Let $n \in B$ so that $n=4j-7$ for some integer $j$. Then, choose $k=j-2$ so that $k$ is also an integer. By multiplying this equation by $4$ and adding by $1$, we see $4k+1=4j-7$. This means $n=4k+1$ for some integer $k$, implying that $n \in A$.

Thus $A \subseteq B$ and $B \subseteq A$ which proves that $A=B$ as required.

Noble Mushtak
  • 18,402
  • 28
  • 44
  • Thank you very much for this explanation and re-write. This helped make a more sense of the comment the teacher had given me, but she didn't quite explain it. – Michael Rector Oct 25 '16 at 23:51
1

$A=\{4k+1, \forall k \in \mathbb{Z}\} =\{4(k+2)-7, \forall k \in \mathbb{Z}\} =\{4j-7, \forall j \in \mathbb{Z}\} =B$.

Dean C Wills
  • 414
0

Well, we have that $1\equiv -7\mod4$. So, the equivalence classes (via the equivalence relation $a$ related to $b$ if $a-b$ is a multiple of $4$, which is precisely $a\equiv b \mod 4$) in which $1$ and $-7$ are , are the same.

Arthur
  • 750