# How do I show that two sets are equal.

This is an ever so slightly modified version of a question from my book. My teacher went over this with me, but I would like an explanation that I can keep coming back to until I have this method figured out precisely.

$A = \{1, 2, 3, 4\}$

$B = \{ n | n \in \mathbb{Z}^+, n^2 < 17 \}$ where $\mathbb{Z}^+$ is the positive integers.

Show that $A = B$

I understand that to do this I must show for every $n$:

$n \in B \rightarrow n \in A$

and

$n \in A \rightarrow n \in B$

How to do that is still something I am not entirely sure how to do correctly, so I will give it my best and someone can show me errors or how to complete it.

If, $n \in B$ then $n^2 < 17$

$n < \sqrt{25}$, therefore $n < 5$

Since $n$ is a positive integer, $0 < n$

therefore, for every $n \in B, 0 < n < 5$, therefore $n \in A$

For every element $n$ such that $n \in A$, $n$ is one of 1, 2, 3, or 4

(I am probably taking too long to do this, but I am unsure how thorough it needs to be, so I will stop here to not waste time. What I would like to know is how can I write this quicker but still make valid statements and show that A = B more easily? I am also making statements that seem unnecessary or unconventional and perhaps even invalid from a logical perspective. I appreciate any help you can offer, the more the merrier!)

To finish the proof with help from amWhy:

$4$ is the largest element of $A$, and $4 < \sqrt{17}$

Assuming $n \in A$, $0 < n \leq 4$, $n^2 \leq 16 < 17$, therefore, $\forall n(n \in A \rightarrow n \in B)$

Therefore $A = B$, hurray!

• "Assuming $n \in A, 0 \lt n \leq 4, \;n^2 \leq 16 \lt 17$, therefore $\forall n (n \in A \rightarrow n\in B$). Then just conclude, "therefore, A = B".! Jan 11, 2013 at 4:19
• @amWhy Hopefully its complete and valid, for reals this time. Jan 11, 2013 at 4:23
• Got it! Nice work! Jan 11, 2013 at 4:26
• Another proof is to enumerate the elements of $B$ explicitly. This is made straightforward in this instance because it specifies $n \in \mathbb{Z}^+ = \{1,2,...\}$, and also the 'condition' $n^2 <17$ has the property that if it is not true for some number, then it is not true for larger numbers. (And personally I think it is very worthwhile to grind through details like you have above.) Jan 11, 2013 at 7:25

To show that sets $$A = B$$, one usually wants to show that $$A\subseteq B \text{ and}\; B \subseteq A$$, which means, equivalently $$(n \in A \rightarrow n \in B \;\text{ and}\;\; n \in A \rightarrow n \in B)$$

First part:

If, $$n \in B$$ then $$n^2 < 17$$, and $$n < \sqrt{25}$$, therefore $$n < 5$$

Replace this last statement above with:

$$n^2 \lt 17$$ which implies $$n \lt \sqrt{17}$$, therefore, $$n \leq 4$$.

Since $$n$$ is a positive integer, $$0 < n,\;$$ therefore, for every $$\;n \in B, 0 < n < 5$$.

For every element $$n$$ such that $$n \in A$$, $$n$$ is [one] of $$1, 2, 3, \,\text{or}\; 4$$

Can you proceed with the other direction? $$n \in A \rightarrow n \in B\;$$?

So for (II), We assume $$n \in A$$, $$n\in \{1, 2, 3, 4\}$$ and show that it follows that $$n \in B$$. It suffices to check that the largest element $$n \in A$$, $$n = 4$$, is such that $$4^2 \lt 17$$, then the square of the rest of the values in A must also be less than $$17$$, since $$\forall n \in A, n^2 \leq 4^2 = 16 \lt 17.$$ Hence, $$\forall n \in A, n \in B$$.

Part I and Part II show that $$A = B$$

• No, you need to also start with the assumption (Part II) $n \in A ...$ then ... ? (you want to show that $n\in A$ implies $n \in B$.) Once that's done, your posted proof, together with your proof of II, give you exactly what you need to prove A = B. Jan 11, 2013 at 3:52
• Yes, provided you show/state that for each $n \in A$, $n^2 < 17.$ Therefore, $n \in B$ Jan 11, 2013 at 4:00
• Thank you, this will take some getting used to but I think I am seeing the logic better, I should probably delete the previous comments though to save space. Jan 11, 2013 at 4:15
• Should it not be n belongs to B then n belongs to A in the second part? (4th line) Nov 5, 2015 at 8:19