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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!

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  • $\begingroup$ "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".! $\endgroup$
    – amWhy
    Jan 11, 2013 at 4:19
  • $\begingroup$ @amWhy Hopefully its complete and valid, for reals this time. $\endgroup$
    – Leonardo
    Jan 11, 2013 at 4:23
  • $\begingroup$ Got it! Nice work! $\endgroup$
    – amWhy
    Jan 11, 2013 at 4:26
  • $\begingroup$ 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.) $\endgroup$
    – copper.hat
    Jan 11, 2013 at 7:25

1 Answer 1

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Your proof strategy is fine:

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 $A = B$

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  • $\begingroup$ 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. $\endgroup$
    – amWhy
    Jan 11, 2013 at 3:52
  • $\begingroup$ Yes, provided you show/state that for each $n \in A$, $n^2 < 17.$ Therefore, $n \in B$ $\endgroup$
    – amWhy
    Jan 11, 2013 at 4:00
  • $\begingroup$ 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. $\endgroup$
    – Leonardo
    Jan 11, 2013 at 4:15
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    $\begingroup$ Should it not be n belongs to B then n belongs to A in the second part? (4th line) $\endgroup$ Nov 5, 2015 at 8:19

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