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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.
Conjectures I know its easy but can someone help before monday
ad absurdum
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.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.
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.
Steve
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).
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).