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.__

Started by AM4R, Feb 17 2007 05:10 PM

3 replies to this topic

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.

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

::::::/\M/\R::::::

Posted 17 February 2007 - 06:33 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

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

For disproofs a counter-example is always more than enough. Don't go looking for anything fancy, there will be a straightforward counter-example.

HMFC - Founded 1874, beefing the Cabbage since 1875

Posted 18 February 2007 - 04:59 PM

thanks ad, easy as that

::::::/\M/\R::::::

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

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