Which equations show that the set of whole numbers is not closed under subtraction?

Choose all answers that are correct.

A.
1 – (–2) = 3

B.
1 – 2 = –1

C.
2 – 0 = 2

D.
2 – 4 = –2

Question

Which equations show that the set of whole numbers is not closed under subtraction?
 
Choose all answers that are correct.
 
 A.
1 – (–2) = 3
 
 B.
1 – 2 = –1
 
 C.
2 – 0 = 2
 
 D.
2 – 4 = –2

anonymous · Answer

Yes That is true

anonymous · Answer

But its Mulitple anwsers

anonymous · Answer

so there is more than 1 anwser

zepdrix · Answer

Cave, in order to show closure for a set, you need to start with two elements in the set.
(Keep in mind that negative numbers are NOT in the set of Whole numbers).

So you need to look for which one starts with two whole numbers,
then ends up with something that is not a whole number

zepdrix · Answer

It is not A.

anonymous · Answer

Its c

anonymous · Answer

and A

zepdrix · Answer

Notice that option A shows us doing subtraction with a negative number.
Negatives are not in the set of whole numbers.
You need to start with this :)
$\Large\rm whole~number -whole~number=$

daniel_chernioglo · Answer

True

anonymous · Answer

D

anonymous · Answer

and b

zepdrix · Answer

D sounds correct,
you start with two elements in the wholes,
then doing subtraction, you left the set of whole numbers.

And B also?
Yay good job \c:/

anonymous · Answer

so is B correct?

zepdrix · Answer

It's not C.
C does not show us leaving the set of whole numbers.
See how the result is still positive.

zepdrix · Answer

Daniel stop guessing D: oh my goodness lol

anonymous · Answer

So WHat is the second one Im confused!

daniel_chernioglo · Answer

so the answer has to end with negative?

zepdrix · Answer

How can you honestly not know which ones are correct?
Read my previous posts silly :O
Two of the options you choose were correct.... just read it

anonymous · Answer

Okay, Daniel was confusing me....

daniel_chernioglo · Answer

sorry guys!!!

anonymous · Answer

D and B

anonymous · Answer

Yes or no thats all i need to know

zepdrix · Answer

Yes daniel:$$\Large\rm whole~number -whole~number=not~whole~number$$$$\Large\rm positive-positive=negative$$That's what we were looking for ^
 yes cave :3

daniel_chernioglo · Answer

I just didnt get it b/c it sait: not closed under subtraction

anonymous · Answer

D and b thats Final

daniel_chernioglo · Answer

I get it now though

daniel_chernioglo · Answer

Yes its D and B

anonymous · Answer

could you help with another question plz

daniel_chernioglo · Answer

Ahhh i could try?

anonymous · Answer

For which operations is the set {0, 1} closed?
 
Choose all answers that are correct.
 
 A.
multiplication
 
 B.
division
 
 C.
addition
 
 D.
subtraction

daniel_chernioglo · Answer

let me think about it

anonymous · Answer

I think its division (a) and Multipilcation

anonymous · Answer

a and b

anonymous · Answer

@zepdrix need help plz

zepdrix · Answer

If we do this:$$\Large\rm 0-1$$Do we end up with a number which is in the set?

anonymous · Answer

Subtraction

zepdrix · Answer

I asked you a question

anonymous · Answer

me

zepdrix · Answer

I didn't ask if it was subtraction.
That's not the question I asked -_-

anonymous · Answer

No you wont

daniel_chernioglo · Answer

I know it A

zepdrix · Answer

Ok good, so it's not closed under subtraction :)
That takes away option D.

anonymous · Answer

yes

daniel_chernioglo · Answer

addition doesnt work either

zepdrix · Answer

$$\Large\rm \frac{1}{0}$$How bout division?
Do we end up with an element in the set if we do this?

anonymous · Answer

yes!

anonymous · Answer

oops no!

zepdrix · Answer

Ok good :)
 No division by zero in the land of math.
So division is also a no no.

anonymous · Answer

so its MultiPlication and addition its multi anwser question!

anonymous · Answer

Zeptrix I feel like you are my online teacher XD

anonymous · Answer

you help alot!

zepdrix · Answer

@jim_thompson5910 do we have to add two `different` elements to show closure under addition? I can't remember :(
Or can we add the same element to itself?

anonymous · Answer

can you help me with another question

daniel_chernioglo · Answer

i dont think that addition works

anonymous · Answer

Which sets of numbers are closed under division?
 
Choose all answers that are correct.
 
 A.
rational numbers
 
 B.
integers
 
 C.
{–1, 0, 1}
 
 D.
whole numbers

anonymous · Answer

let me fix that

jim_thompson5910 · Answer

They don't have to be different, but to get a counterexample, you have to use different numbers to subtract.

zepdrix · Answer

ah ok :)

zepdrix · Answer

So did you understand the last question cave? :O
$$\Large\rm 1+1$$We are not closed under addition, yah?

jim_thompson5910 · Answer

The whole numbers are closed under addition. There is no way to add two whole numbers to get a nonwhole number.

zepdrix · Answer

Yah we were on a second question :) The set containing 0 and 1.

anonymous · Answer

Which sets of numbers are closed under division?
 
Choose all answers that are correct.
 
 A.
rational numbers
 
 B.
integers
 
 C.
{–1, 0, 1}
 
 D.
whole numbers

anonymous · Answer

the question marks are a -

zzr0ck3r · Answer

what is the one element that would make it hard to divide by?

zzr0ck3r · Answer

what set does not contain this element?