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This is found in the book the instructor is using, An Introduction to Proof through Real Analysis by Daniel J. Madden and Jason A. Aubrey The University of Arizona Tucson, Arizona, USA

Here is one way to define $A$ in set-builder notation: $$A = \{x | x \text{ is an even integer}\} \text. \tag{9.1}$$ In general, set-builder notation takes the form $$S = \{x | P(x) \text { is true}\} \text, \tag{9.2}$$ where $P(x)$ is some mathematical statement. We read this definition of the set $S$ as "$S$ is the set of all $x$ such that $P(x)$ is true". So

  • $s \in S$ if and only if $P(s)$ is true;
  • $s \notin S$ if and only if $P(s)$ is false.

There are some common variations on set-builder notation that you will see. For example, people will often use a colon ":" in place of the bar "|". That is fine; the idea is the same. Sometimes, another condition on elements of a set is slipped in before the "such that" symbol by limiting elements to members of a larger set. For example, we could have defined the set of even integers as this: $$A = \{x \in \mathbb Z | x \text { is even}\} \text. \tag{9.3}$$ Besides using English or set-builder notation to define sets, we can define sets by simply listing their elements. For example, we can write $$A = \{\dots, −6, −4, −2, 0, 2, 4, 6, \dots\} \tag{9.4}$$ to define the set of all even integers. But this really only works when the set is small enough that all of its elements can be reasonably listed or when the pattern is strong enough to be recognized. For example, we could write $$B = \{n \in \mathbb N | 2 \le n \le 5\} \text { or } B = \{2, 3, 4, 5\} \text. \tag{9.5}$$

This is all the information from the book that I could find that is pertaining to the purposed question. Please Help Me!

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    $\begingroup$ Welcome to Mathematics Stack Exchange. Can you show $n$ is even $\iff n-1$ is odd? $\endgroup$ – J. W. Tanner Sep 30 '20 at 1:02
  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Sep 30 '20 at 1:02
  • $\begingroup$ Um... do you have a question? Any question? You quote something. To a experienced person everything you quote is straightforward and fairly simple. But obviously its not or you wouldn't ask anything. So it's up to you to tell us where you have difficulty and what you want clarified. $\endgroup$ – fleablood Sep 30 '20 at 2:36
  • $\begingroup$ The point of this is to get you familiar with set builder notation. What is actually being proven is fairly simple: The integers that are even are precisely the numbers that are one more than an odd number. this is obvious. If $n = 2k$ is even then $n-1 = 2k -1$ is odd. That's it we are done. What the excercise though is to put it in terms of set builder notation. And ... I think the text you quoted does a fine job explaining that. $\endgroup$ – fleablood Sep 30 '20 at 2:41
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To prove "set A= set B" you prove "A is a subset of B" and "B is a subset of A". And to prove "A is a subset of B" start "if x is in A" and then use the definitions of A and B to conclude "then x is in B"

Here that means we want to prove "if x is even then x- 1 is odd" and "if x- 1 is odd then x is even".

Of course, we need to use the fact that any even number is of the form 2k for some integer k and any odd number is of the form 2k+ 1 for some integer k.

So "If x is even then there exist an integer k such that x= 2k. Then x- 1= 2k- 1= 2k- 2+ 1= 2(k-1)+ 1 so x-1 is odd."

And "if x- 1 is odd then there exist an integer k such that x- 1= 2k+ 1. Then x= 2k+ 2= 2(k+ 1) so x is even."

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$\{n\in \mathbb Z| n$ is even $\}=$

$\{n\in \mathbb Z| n$ is divisible by $2\}=$

$\{n\in \mathbb Z|$ there exists an integer $k$ so that $n = 2k\}=$

$\{n\in \mathbb Z|$ there exists an integer $k$ so that $n-1 = 2k-1\}=$

$\{n\in \mathbb Z| n-1$ is odd$\}$.

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In order to show equality of sets, you need to show two-way containment; that is, show that $A\subseteq B$ and $A\supseteq B$, which is equivalent to saying that $x\in A \Leftrightarrow x\in B$. Sometimes it's best to show each direction independently, but in this case, I would recommend a chain of biconditionals, as it's quicker.

Let $A=\{x\in\mathbb{Z} | x\text{ is even}\}$ and $B=\{x\in\mathbb{Z} | x-1\text{ is odd}\}$. Then we have the following chain of biconditionals:

$x\in A\Leftrightarrow x$ is even $\Leftrightarrow x=2k$ for some $k\in\mathbb{Z} \Leftrightarrow x-1=2k-1\Leftrightarrow x-1$ is odd $\Leftrightarrow x\in B$.

Therefore, $A=B$.

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