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  Conjectures - HSN forum
  
  # Conjectures
  
  3 replies to this topic
  
  ### #1AM4R
  
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  Posted 17 February 2007 - 05:10 PM
  
  Right this will probably be in my nab and i dont have a clue what the question is asking me to do and what numbers I am suppoesed to use.
  
  Can someone help me and take me through how im meant to answer it.
  
  Thanks
  
  Disprove the conjecture:
  bc is divisible by a => either b is divisible by a or c is divisible by a.
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  Posted 17 February 2007 - 06:33 PM
  
  QUOTE(AM4R @ Feb 17 2007, 05:10 PM)
  Right this will probably be in my nab and i dont have a clue what the question is asking me to do and what numbers I am suppoesed to use.
  
  Can someone help me and take me through how im meant to answer it.
  
  Thanks
  
  Disprove the conjecture:
  bc is divisible by a => either b is divisible by a or c is divisible by a.
  Well all this statement in the question reall is "suppose we had two numbers and multiplied them together and the product was divisible by some number a; it follows that one of the two numbers we multiplied together must be divisible by a". You're told to disprove this, so you've got to show it to be false.
  
  Let a=4,b=2,c=2. Put these into the statement giving:
  
  "2*2 is divisible by 4 => either 2 is divisible by 4 or 2 is divisible by 4"
  
  Now obviously 2*2=4, which is divisible by 4, but 2 on it's own isn't. Thus the statement above is pish so you can see that the statement isn't true.
  
  For disproofs a counter-example is always more than enough. Don't go looking for anything fancy, there will be a straightforward counter-example.
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  ### #3AM4R
  
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  Posted 18 February 2007 - 04:59 PM
  
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  ### #4Steve
  
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  Posted 18 February 2007 - 05:15 PM
  
  Note that the reason this disproved the statement is that it implicity said that "(For all numbers a,b, and c,) bc is divisible by a => either b is divisible by a or c is divisible by a."
  
  If the statement had said "There are numbers a,b, and c such that bc is divisible by a => either b is divisible by a or c is divisible by a", then this would be true, e.g. a=3, b=6, c=5.
  
  You can only use a "counterexample" (as ad absurdum did) to disprove a conjecture which is about all "somethings". In particular, finding a counterexample is not enough to disprove my statement above, since it doesn't say "for all".
  
  I wish there was more stuff like this in AH maths.
  
  (By the way, the statement is true if you insist that a is prime. Indeed you can take this as a definition of what it means to be prime, as in the subject of Ring Theory).
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