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Is this well-defined in the rational numbers

  • Thread starter iNCREDiBLE
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  • #1
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Need help with proving:

Show that (a,b) + (c,d) = (a+c, b+d) is not well-defined in the rational numbers.
[Note: (a,b) + (c,d) = (ad+bc, bd) is well-defined because (a,b) is related to (c,d) when ad = bc.)]
 

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  • #2
Hurkyl
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Have you tried at all? Do you know what it means to fail to be well-defined?
 
  • #3
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Hurkyl said:
Have you tried at all? Do you know what it means to fail to be well-defined?
I know what it means. But is it enough with just stating an example?
 
  • #4
HallsofIvy
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When proving a general statement is not true, yes, it is sufficient to give a "counter-example": one case showing for which the statement is not true, thus proving it is not always true.

What you need to do is find an example of rational numbers (a,b),(a',b'),(c,d),(c',d') such that (a,b) and (a', b') are equivalent, (c,d) and (c',d') are equivalent, but (a+c,b+d) is not equivalent to (a'+c', b'+d').

(For those who are not clear on this, this is using a method of defining rational number from the integers by saying that two ordered pairs of integers, (a,b) and (a',b') (with second integer non-zero) are equivalent if and only if ab'= a'b. That is an equivalence relation and so partitions the set of all such pairs into equivalence classes. A rational number is such an equivalence class.)
 
  • #5
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HallsofIvy said:
When proving a general statement is not true, yes, it is sufficient to give a "counter-example": one case showing for which the statement is not true, thus proving it is not always true.

What you need to do is find an example of rational numbers (a,b),(a',b'),(c,d),(c',d') such that (a,b) and (a', b') are equivalent, (c,d) and (c',d') are equivalent, but (a+c,b+d) is not equivalent to (a'+c', b'+d').

(For those who are not clear on this, this is using a method of defining rational number from the integers by saying that two ordered pairs of integers, (a,b) and (a',b') (with second integer non-zero) are equivalent if and only if ab'= a'b. That is an equivalence relation and so partitions the set of all such pairs into equivalence classes. A rational number is such an equivalence class.)
That's just what I did, I wasn't just sure if that was enough.. :bugeye:
 
  • #6
HallsofIvy
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Out of curiosity then, what was your counter-example?
 
  • #7
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HallsofIvy said:
Out of curiosity then, what was your counter-example?
(1,2) ~ (1,2) and (1,3) ~ (2,6).
 

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