Question

Prove that {10k + 7} is a proper subset of {5m - 8}.

Answer 1

No idea where to start with this one. Usually the questions are of the form prove set A is a subset of set B, not with formulas...

Answer 2

what are k and m ??

Answer 3

Oh sorry, the full question is prove that \[\left\{ 10k + 7 | k \in Z \right\}\] is a proper subset of \[\left\{ 5m - 8 | m \in Z \right\}\].

Answer 4

What is your understanding of the problem ?

Answer 5

I want to show that the second set (5m-8) contains all of the elements in the first set, and others (definition of proper subset). I'm not sure how to go about that though. I have done similar problems like this, except in those the contents of the sets were explicitly defined, not defined as functions.

Answer 6

I think you need to rewrite one in terms of the other somehow? Is that on the right track?

Answer 7

yup. In my opinion, you need to find an "m" for any given "k".

Answer 8

So I need to show that every x that satisfies the first set also satisfies the second set. I'm not sure how to do that?

Answer 9

Lets go step by step:
To Prove: Every number in set1 is also present in set 2
=> for every number "10k+7" we have an "m" such that 10k+7=5m-8

Answer 10

does that make sense?

Answer 11

Yes.

Answer 12

I don't understand how you can prove it though.

Answer 13

just see the last equation i wrote. What does that tell you ?

Answer 14

10k+7 = 5m-8 ?

Answer 15

yup

Answer 16

So if 10k+7 = x, then 5m-8 also = x?

Answer 17

yup. only then the number will also be available in the set2 right ?

Answer 18

Is that all you need to prove it?

Answer 19

yes. But did you understand the reasoning behind this proof ?

Answer 20

Yes. By showing how a number in one set can't escape also being in the other set, you can prove that one is a subset of the other.

Answer 21

(I think). Haha. I find maths to be harder than any other subject for some reason.
Thanks for your help by the way :)

Answer 22

you are welcome :)