Can Positive Integers Prove These Division Properties?

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SUMMARY

The discussion focuses on proving two division properties involving positive integers a, b, c, and d. The first property states that if a divides b and c divides d, then ac divides bd. The second property establishes the equivalence between a dividing b and ac dividing bc. Participants clarify the notation used in mathematical expressions, emphasizing the distinction between divisibility and fractional representation. The conversation highlights the importance of precise language in mathematical proofs.

PREREQUISITES
  • Understanding of divisibility notation (e.g., a|b)
  • Familiarity with basic algebraic manipulation
  • Knowledge of equivalence relations in mathematics
  • Experience with proofs in number theory
NEXT STEPS
  • Study the properties of divisibility in number theory
  • Learn about equivalence relations and their applications
  • Explore algebraic proofs involving integers
  • Review common mathematical notation and its meanings
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Mathematics students, educators, and anyone interested in number theory and mathematical proofs will benefit from this discussion.

Cyborg31
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Homework Statement


If a, b < c, and d are positive integers, prove the following inferences.

1. a|b \wedge c|d \rightarrow ac|bd
2. a|b <=> ac|bc


Homework Equations





The Attempt at a Solution



1.

a|b = x, then b = ax

c|d = y, then d = cy

bd = axcy

thus ac|bd = ac|axcy, and ac|axcy = xy

therefore ac|bd = xy if a|b = x and c|d = y

2.

c|ac = a and c|bc = b

so c|(ac|bc) = a|b
 
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Cyborg31 said:

Homework Statement


If a, b < c, and d are positive integers, prove the following inferences.

1. a|b \wedge c|d \rightarrow ac|bd
2. a|b <=> ac|bc


Homework Equations





The Attempt at a Solution



1.

a|b = x, then b = ax

c|d = y, then d = cy
Yes, and these are the "relevant equations" above

bd = axcy

thus ac|bd = ac|axcy, and ac|axcy = xy
I think it is simpler and clearer to write "bd= (ac)(xy) so ac|bd".

therefore ac|bd = xy if a|b = x and c|d = y

2.

c|ac = a and c|bc = b
This makes no sense "c|ac" is the statement "c divides ac" and is not equal to anything. You mean to say ac/c= a and bc/b= b.

so c|(ac|bc) = a|b
You want to prove "a|b <=> ac|bc. That's and "if and only if" statement and must be proved both ways:

1) if a|b then b= ax for some x. bc= axc ...

2) if ac|bc then bc= acy for some y...
 
This makes no sense "c|ac" is the statement "c divides ac" and is not equal to anything. You mean to say ac/c= a and bc/b= b.

I know, that's what I meant "c divides ac". a|b = b/a right? Doesn't it follow that c|ac = ac/c?
 
No, the collection of symbols a|b means "b is a multiple of a, or equivalently, a divides b without remainder". The symbols b/a represent a (rational in this case) number.
 
For a|b <=> ac|bc

a|b = x
ac|bc = y

b = ax
bc = acy

bc/c = acy/c => b = ay

If b = ax and b = ay then x = y

a|b = x <=> a|b = y therefore a|b <=> ac|bc

Is this correct?

<=> is equivalence, not <->.
 
Please stop using the symbol a|b to mean the same as b/a. They are different.
 
Uh ok... what's wrong with what I did? I did c|ac = a, before and that apparently that's wrong so I used bc/c = b, this time. How else do I cancel out the c?

Should it be c|bc = c|acy => b = ay ?

Is my solution correct or wrong?
 
A vertical bar means "http://mathworld.wolfram.com/Divides.html" ." matt grime is just making a point about semantics and the use of symbols.

It's not the same as the "fraction bar."
 
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