Elements in Sets: Check/Confirm Answers

[Math100]

Homework Statement

Write each of the following sets by listing their elements between braces.

Relevant Equations

None.

Can anyone please check/confirm my answers if they are correct or not? I boxed around my answers just to be clear and understanding. Thank you.

 

[Mark44]

Homework Statement:: Write each of the following sets by listing their elements between braces.
Relevant Equations:: None.

Can anyone please check/confirm my answers if they are correct or not? I boxed around my answers just to be clear and understanding. Thank you.

They look fine.

 

[Math100]

So my answers are correct?

 

[PeroK]

So my answers are correct?

Yes, although any thoughts on whether the second question is valid?

 

[FactChecker]

It looks ok. For (2), some people consider 0 to be a natural number.

 

[PeroK]

It depends whether the notation, $\{x \in \mathbb N: \dots \}$ implies that $\mathbb N$ is the universal set under consideration. In that case, $-2 \notin \mathbb N$ and the comparison $-2 < x$ is not valid.

Or, if we consider that in all cases $\mathbb N \subset \mathbb Z$ and that it's valid to talk about $-2$ even when nominally restricting our attention to $\mathbb N$, then it's fine.

I'm not saying one way or the other, but the question just didn't look right to me.

 

[Math100]

Thank you guys for the help! I really appreciate it!

 

[PeroK]

Thank you guys for the help! I really appreciate it!

What do you think? Is the condition $-2 < x$ valid for $x \in \mathbb N$?

 

[Math100]

What do you think? Is the condition $-2 < x$ valid for $x \in \mathbb N$?

Yes.

 

[haruspex]

It depends whether the notation, $\{x \in \mathbb N: \dots \}$ implies that $\mathbb N$ is the universal set under consideration. In that case, $-2 \notin \mathbb N$ and the comparison $-2 < x$ is not valid.

By that reasoning, it would not be possible to say whether "1<x" is valid, since we could take the 1 as an element of that $\mathbb Z$ or $\mathbb R$.
Similarly, I could not write x-1 since that is shorthand for x+(-1).

Seems more reasonable to apply the programming language concept of type coercion. An element of $\mathbb N$ can be 'elevated' to $\mathbb Z$ etc. as necessary to make the operation valid.

Whether the result can be demoted to conform to the target variable type is another matter.

 

[PeroK]

By that reasoning, it would not be possible to say whether "1<x" is valid, since we could take the 1 as an element of that $\mathbb Z$ or $\mathbb R$.
Similarly, I could not write x-1 since that is shorthand for x+(-1).

Seems more reasonable to apply the programming language concept of type coercion. An element of $\mathbb N$ can be 'elevated' to $\mathbb Z$ etc. as necessary to make the operation valid.

Whether the result can be demoted to conform to the target variable type is another matter.

Usually $\mathbb R$ is implied at the universal set and $\mathbb Z \subset \mathbb R$.

The point about the question is that it explicitly defines the universal set as $\mathbb N$ and then talks about $-2$, which is not defined within $\mathbb N$.

Anyway, my main point is that it's definitely worth thinking about if you want to study pure maths,

 

[PeroK]

Similarly, I could not write x-1 since that is shorthand for x+(-1).

If you are dealing with natural numbers, that can't be right, as natural numbers have no additive inverse. Instead $n - m$ is defined to be the natural number $k$ such that $m + k = n$.

In general, $n - m$ is not well defined for all pairs of natural numbers. That IS important.