I'm stuck proving (or disproving) the equality of two sets $M_1=M_2$.

$M_1=\{k\in\mathbb{N}\ |\ \exists x(x\in\mathbb{N}\ \wedge\ k^2+4k+4=2x+4)\}$

$M_2=\{l\in\mathbb{N}\ |\ \forall y(y\in\mathbb{N}\ \rightarrow y\leq l)\}$

Here are my steps so far:

Solving for $k$ I get

$M_1=\{k\in\mathbb{N}\ |\ \exists x(x\in\mathbb{N}\ \land\ (k=-2+\sqrt{4+2x}\ \lor\ k=-2-\sqrt{4+2x}))\}$

then I transform $M_2$, such that

$M_2=\{l\in\mathbb{N}\ |\ \forall y(\lnot (y\in\mathbb{N})\ \lor (y\leq l))\}$

Now, I don't know what to do. What is the next step?

I know that, to prove the equality of two sets, I must proof that the first set is contained in the second set and vice versa. But how do I do that given the sets I have?

  • 1
    $\begingroup$ Hint: if all else fails, try a few simple examples: e.g., check whether $i \in M_1$ and $i \in M_2$ for small $i$, like $0$, $1$, $2$, ... If you find a discrepancy the two sets are not equal. $\endgroup$ – Rob Arthan Oct 28 '16 at 19:43
  • 2
    $\begingroup$ You're right. I missed the "or disprove" part, and assumed that the claim was true. I saw that the bottom one was $\emptyset$ so assumed the top was as well without actually checking. I call this the "textbook infallibility principle" but when I was in school my professors always rejected the proofs I wrote that invoked it. $\endgroup$ – Gabriel Burns Oct 28 '16 at 19:47

$2^2 + 4\times 2 + 4 = 16 = 2\times 6 + 4$ therefore $2\in M_1$.

Assume $l\in \Bbb N$. Then $l+1 \gt l$, thus $l \notin M_2$, therefore $M_2 = \emptyset$.

Thus $M_1 \ne M_2$.


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