Is the following statement true or false. prove?

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Homework Help Overview

The discussion revolves around two mathematical statements involving integers and real numbers. The first statement concerns divisibility among integers, while the second involves inequalities with real numbers. Participants are exploring the validity of these statements and attempting to provide proofs or counterexamples.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to prove or disprove the statements through examples and algebraic reasoning. Questions are raised about the validity of certain assumptions, particularly regarding the nature of integers and their divisibility. Counterexamples are being discussed to challenge the original claims.

Discussion Status

The discussion is active, with participants providing examples and questioning the reasoning behind the proofs. Some participants express a desire to find algebraic proofs rather than relying solely on examples. There is recognition of the need for clarity in definitions and the conditions under which the statements hold.

Contextual Notes

Participants are navigating the constraints of proving statements without relying on real numbers, indicating a preference for integer-based proofs. There is also a mention of the definitions of divisibility and the conditions under which the statements can be considered true or false.

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Homework Statement



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.



Homework Equations





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?
 
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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!
 
scurty said:
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!

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.
 
charmedbeauty said:
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.

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.
 

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