# Please help me verify the following set equation from The Essence of Discrete Mathematics by Neville Dean

LaTeX markup suggested format: $\{y\in N_1 | y\le 4 \land (\exists x\in Z| x< 4 \land 6=yx\}$

Author's original post: {y:∈ N1 | y ≤ 4 Ʌ (∃x:∈Z | x < 4 . 6=y*x)}

The first part of the equation states that y is a natural number except zero and it is less than or equal to 4. $N_1$ is all natural numbers except zero. $y \le 4$ means that $y = 1, 2, 3, 4$

The second part of the equation after the AND ($\land$) states: there exists ($\exists$) a variable $x$ that is an integer (Z) such that $x$ is less than 4, which include: (...-2, -1, 0, 1, 2, 3)

The final calculation 6 = y * x yields the following combinations:

y    x    Truth Value
1    6     F (since x must be less than 4)
2    3     T
3    2     T
4    1.5   F (since x is not an integer)


No negative numbers in the set of $x$ need to be evaluated since the value must be 6 and $y$ cannot be negative. Therefore, I believe the set enumeration is: $\{2, 3\}$ However, the book says that the answer is $\{1, 2, 3\}$.

Please tell me if I'm right or the book is right.

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The LaTeX markup you provided is interesting. Thanks for opening my eyes to it. I'm new to this and am now reading about this document markup language. Is it true that the end of the equation shown in LaTeX markup: $x< 4 \land 6=yx\}$ is equivalent to what I originally posted? x < 4 . 6=y*x)}. The period, which admittedly falls short, was supposed to be a large dot to denote a set replacement. – Ethan Jan 12 '12 at 19:24
What's a "set replacement"? The version with $\land$ makes sense to me; I don't know your dot notation; but there are lots of different ways of writing sets, so yours may well be in use, too. – joriki Jan 12 '12 at 19:36
I'm a beginner at this and, being a novice, I sure wish there were standards around notation in Discrete Mathematics. The author of the book I'm reading says that an extension of set comprehension is to follow the declaration and predicate by a formula. That's what appears after the dot. Thanks again for your comments. – Ethan Jan 12 '12 at 19:46

The way you've written down your analysis is slightly confusing because what you list under "Truth Value" is the truth value of the entire parenthesized expression after the "and", which quantifies over $x$, yet each row contains a value of $x$, which looks as if you're substituting that value of $x$ into some expression containing $x$ as a free variable to get that truth value. A somewhat clearer way to express this would be: In each row, for the given value of $y$, there is only one rational number $x$ for which $6=y\cdot x$. For the parenthesized expression to be true, this rational number would have to be an integer less than $4$. Thus we can calculate this $x$ and then deduce the truth value of the parenthesized expression from whether this $x$ is an integer less than $4$. I know this is roughly what you meant; I'm just trying to help you present your thoughts more precisely.
Would it be wrong to recast this solely with integers rather than introducing rational numbers? "In each row, for the given value of $y$, there is only one integer $x$ for which $6=y\cdot x$. For the parenthesized expression to be true, this integer would have to be less than $4$. Thus we can calculate this $x$ and then deduce the truth value of the parenthesized expression from whether this $x$ is an integer less than $4$." – Ethan Jan 12 '12 at 19:41
@Ethan: That's not quite right; as your table shows, there isn't always an integer $x$ for which $6=y\cdot x$. – joriki Jan 12 '12 at 20:09
You could start out with "In each row, for the given value of $y$, there is at most one integer $x$...", but when you get to "we can calculate this $x$", you'd have to rewrite it more substantially, since you can't calculate such an $x$ without the rational numbers. You might say something like "we can find this $x$ if there is one and then deduce the truth value of the parenthesized expression from whether such an $x$ exists and is less than $4$". No need to repeat "integer" in the end if you've introduced $x$ as an integer. – joriki Jan 12 '12 at 20:10