is the following statement true or false. prove?

charmedbeauty

1. The problem statement, all variables and given/known data

let a,b,c be three integers is a divides c and b divides c, then either a divides b or b divides a.


if a,b,c,d are real numbers with a<b and c<d, then ac<bd.



2. Relevant equations



3. The attempt at a solution

So for part 1

Since a/c and b/c

then, c=ak and c=bd for some integers k,d.

hence, ak=bd

so a/b=d/k

where d/k =l for some integer l.

so a/b=l... Therefore a/b. end of proof.

for part 2)


a<b and c<d (given).

let a= -5 and b=-4 satisfying a<b

and let c=-7 and d=-5 satisfying c<d

so then ac

= 35

and bd

= 20

so 35<20

but this is false

therefore ac<bd is false. end of proof.

Can Someone please tell me if I have done this correctly?

scurty

How do you know d/k is an integer? Try a few values out for a, b, and c and see how the corresponding d and k correlate. You should see the answer to the problem pretty quickly!

The second counterexample looks good!

charmedbeauty

So for the first part can I just use examples and then prove it false I guess I can, but I would like to know how to prove it algebraically without having to use real number, since most of the proofs I am doing at the moment involve this.

Ok so

a/c and b/c (given)

let c=3 and a=9 and b=12

so 9=3k for some positive integer k

and 12=3s for some positive integer s

but 12 does not divide 9 and 9 does not divide 12. therefore a/b or b/a is false.

but obviously I can make this statement true ie if k=6 and s=12... Although I guess this in not a "for some" statement... but I really don't like doing proofs this way.

scurty

Well, if a statement is sfalse, the only way to really prove it is to provide a counterexample.

In your example, 9 and 12 don't both divide 3. You have the definition backwards, a and b will be multiplied by the interger terms.