Problem with understanding closure property in binary operation

[chijioke]

Please take a look at this problem.

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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]

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I decided to go for (d):

Suppose a=1,b=2 then

[imath]\qquad 1-2=-1[/imath]

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

[imath]\qquad -1 \cancel{\epsilon} N[/imath]

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

[imath]\qquad 2-1=1[/imath]

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

[imath]\qquad 1 \in N[/imath]

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

 

[Dr.Peterson]

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?

 

[blamocur]

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 ?

 

[lev888]

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
[math]1-2=-1[/math]The set N is not closed under the given binary operations [math]\because[/math][math]-1 \cancel{\epsilon} N[/math]But on the other hand, if a=2, b=1 then
[math]2-1=1[/math]I can now say that the set N, is closed [math]\becuase[/math][math]1 \epsilon N[/math]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?

 

[chijioke]

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].

Please help me by defining [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.

 

[Dr.Peterson]

No, I don't.

Please help me by defining [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.

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):

Absolute Value

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.)

 

[stapel]

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

[imath]\qquad 1-2=-1[/imath]

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

[imath]\qquad -1 \cancel{\epsilon} N[/imath]

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

[imath]\qquad 2-1=1[/imath]

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

[imath]\qquad 1 \in N[/imath]

---

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.

 

[chijioke]

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.

 

[lev888]

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?