Question

Which sets of numbers are closed under subtraction?

Choose all answers that are correct.

A.
odd natural numbers

B.
rational numbers

C.
{0, 1}

D.
{0, 1, 2}

Answer 1

@AlexandervonHumboldt2

Answer 2

@chosenmatt

Answer 3

@freckles

Answer 4

well didn't we already talk about the set {0,1} under subtraction?
and we didn't select that choice when your question was asking for which operations the set is closed..

anyways

let's think about the last set since it is a finite set {0,1,2}
is 1-2 in {0,1,2}?

Answer 5

yup

Answer 6

1-2 is -1 right?

Answer 7

is -1 a member of the set {0,1,2}?

Answer 8

no

Answer 9

so since 1-2 is no in {0,1,2} then the set {0,1,2} is not closed under subtraction...

now
rational numbers
rational numbers look like a/b or c/d
where a,b,c,d are integers (and b and d not equal to 0)
\[\frac{a}{b}-\frac{c}{d}\]
when you subtract these two fractions
do you still get a rational number?

Answer 10

try combining the fractions

Answer 11

yes you still get a rational #

Answer 12

ok then that would make the rationals closed under subtraction

---
now odd natural numbers...
odd natural numbers={1,3,5,7,9,11,13,...}
try to see if you can think of two numbers that you can subtract from this set and that result is not in the set {1,3,5,7,9,11,13,...}

Answer 13

13 and 1

Answer 14

13-1 =12

Answer 15

sounds great :)
also
2k+1 and 2m+1 are odd numbers where k and m are integers
(2k+1)-(2m+1)
2k+1-2m-1
2k+2m+1-1
2k+2m+0
2k+2m
2(k+m)
and an even number is not going to be in any set of odd numbers

Answer 16

so you only had one choice to go with here

Answer 17

rational, right?

Answer 18

yep

Answer 19

thx

Answer 20

\[\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd} \in \mathbb{Q}\]
subce ad-cb is an integer
and bd is also in integer
and a rational number is a integer/integer
(where the bottom doesn't equal 0 of course)