# Problem with understanding closure property in binary operation

#### chijioke

##### Full Member
Please take a look at this problem.

1. Determine whether or not, each of the following sets is closed under the given operations defined by:

[imath]\qquad \textrm{(a) } a * b = 4(a + b),\; a,\,b \in \mathbb{R}[/imath]

[imath]\qquad \textrm{(b) } p \nabla q = \dfrac{pq}{5},\; p,\,q \in \mathbb{R}[/imath]

[imath]\qquad \textrm{(c) } x \Omicron y = x + y + \dfrac{xy}{3},\; x,\,y \in \mathbb{Q}[/imath]

[imath]\qquad \textrm{(d) } a \nabla b = \vert a - b \vert,\; a,\,b \in\mathbb{N}[/imath]

I decided to go for (d):

Suppose a=1,b=2 then

The set N is not closed under the given binary operations because:

But on the other hand, if a=2, b=1 then:

I can now say that the set N, is closed because:

What do you have to say about my idea on this problem?

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Please take a look at this problem.
View attachment 36145
View attachment 36146
I decided to go for (d)
Suppose a=1,b=2 then [imath]1-2=-1[/imath]
The set N is not closed under the given binary operations [imath]\because[/imath] [imath]-1 \cancel{\epsilon} N[/imath]
But on the other hand, if a=2, b=1 then [imath]2-1=1[/imath]
I can now say that the set N, is closed [imath]\because[/imath] [imath]1 \epsilon N[/imath]
What do you have to say about my idea on this problem?
Do you know what the absolute value means? The operation is not [imath]a-b[/imath], but [imath]|a-b|[/imath].

Also, how is "natural number" defined in this book? Do they include 0 or not?

I decided to go for (d)
Do you have to pick only one of them?

Also, while [imath]1-2=-1 \notin N[/imath], you forgot to use absolute value, i.e. [imath]|1-2| = 1 \in N[/imath].

Which defintion of [imath]N[/imath] is used in your class/book ?

Please take a look at this problem.
View attachment 36145
View attachment 36146
I decided to go for (d)
Suppose a=1,b=2 then
$1-2=-1$The set N is not closed under the given binary operations $\because$$-1 \cancel{\epsilon} N$But on the other hand, if a=2, b=1 then
$2-1=1$I can now say that the set N, is closed $\becuase$$1 \epsilon N$What do you have to say about my idea on this problem?
1. Please post the definition of the closure property.
2. in (d), do you know what the 2 vertical lines mean?

Do you know what the absolute value means?

No, I don't.

The operation is not [imath]a-b[/imath], but [imath]|a-b|[/imath].

I thinking it means 'side' something... when dealing with geometry. But in this context, I can't tell what it is.

Also, how is "natural number" defined in this book? Do they include 0 or not?

I understand natural numbers as set N of numbers used for counting excluding 0. That is [imath]\left\{ 1,2,3,4, \cdots \right\}[/imath]
I don't know why the smaller brace stubbornly choose to remain inside depite all my efforts in trying to remove it.

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No, I don't.

I thinking it means 'side' something... when dealing with geometry. But in this context, I can't tell what it is.

I understand natural numbers as set N of numbers used for counting excluding 0. That is [imath]\{ 1,2,3,4, \cdots \}[/imath]
I don't know why the smaller brace stubbornly choose to remain inside depite all my efforts in trying to remove it.
Here is information about |x|, the absolute value (also called, in some places, the modulus):

Now you should be able to answer the question. (But you may want to check how the book defines "natural number"; the usage varies, even within one country, so they may not agree with your own understanding.)

1. Determine whether or not, each of the following sets is closed under the given operations defined by:

[imath]\qquad \textrm{(d) } a \nabla b = \vert a - b \vert,\; a,\,b \in\mathbb{N}[/imath]

I decided to go for (d):

Suppose a=1,b=2 then

The set N is not closed under the given binary operations because:

But on the other hand, if a=2, b=1 then:

I can now say that the set N, is closed because:

What do you have to say about my idea on this problem?
(a) Study up on absolute values (which should have been covered when you were back in pre-algebra), and re-attempt part (d) of the exercise.

(b) You have been given various relations, and, with respect to each, have been tasked with (i) disproving the closed-ness of the given operation (generally with a counter-example) or else (ii) proving that the given operation is closed "in full generality".

To disprove a rule which claims always to be true, one need find only one counter-example. If the rule is not true in one spot, then it certainly isn't true "everywhere".

To prove a rule "in full generality", one cannot got number-by-number. One cannot say, based on one place where the rule works, that then rule then works "everywhere". You have to *prove* that it's true everywhere. Since there are infinitely-many natural numbers, rational numbers, and real numbers, obviously you cannot prove that the rule works everywhere, because you cannot possible *test* everywhere. Instead, you'll have to use symbols (that is, variables) and logic to do a proof.

Here is information about |x|, the absolute value (also called, in some places, the modulus):
(a) Study up on absolute values (which should have been covered when you were back in pre-algebra), and re-attempt part (d) of the exercise.
Now I understand.
1. Please post the definition of the closure property.
Closure property says that if I carry operations (addition, subtraction, multiplication and division ) using any two members of a chosen set of numbers, the result obtained will be a member of the set of numbers chosen.

Now I understand.

Closure property says that if I carry operations (addition, subtraction, multiplication and division ) using any two members of a chosen set of numbers, the result obtained will be a member of the set of numbers chosen.
Is anything still not clear? Can you correct your solution?