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anonymous
 5 years ago
S.O.S: I need help with this: prove that if a+1 divides b and b divides b+3, then a=2 and b=3. Proofs I have covered are direct, contrapositive, and contradiction!!
anonymous
 5 years ago
S.O.S: I need help with this: prove that if a+1 divides b and b divides b+3, then a=2 and b=3. Proofs I have covered are direct, contrapositive, and contradiction!!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if a+1 divides b, this means that b = k *(a+1) for some integer k. if b divides b+3 this means that b+3 = mb for some integer. ok so far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right that is what I have so far

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets subtract the two equations

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(b+3)  b = mb  (a (k+1))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get 3 = mb  ak  a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is what I got too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there more info are a and b must be integers? well yes if we are discussing "divides"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok now substitute back

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0suppose a and b are positive integers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nevermind, thats circular to substitute back

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait I have a question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't b= (a+1)k? and b+3= bj?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have a different idea

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0add 3 to the first equation , so b+3 = 3 +(a+1)k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why do you have (b+3)b= mb(a(k+1))?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait you did that backwards

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isn't it suppose to be mb(a+1)k?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b= (a+1)k? and b+3= bj? this is false

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if a+1 divides b then b=(a+1)k?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0isnt the definition of divisiblity if a divides b then b= ak?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so we have bj3 = (a +1) k ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get bj3= (a+1)k?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok first step, b = (a+1)k , b+3 = bj

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you do that by substituting? i see what you did

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but we know that b = (a+1) k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0follow so far, i might have gone too fast

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b = (a+1)k , b+3 = bj , subtracting the former from the latter we get 3 = bj  (a+1)k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0add (a+1)k to both sides, subtract 3 from both sides you should get bj = (a+1)k ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok we know that (a+1)k = b, so by transitive rule we have b = bj  3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so b+3 = bj , so b divides b+3, which we already know, shoot

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok lets do some substitution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b = (a+1)k , b+3 = bj , thats the given. substitute (a+1)k for b in the second equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we get (a+1)k + 3 = (a+1)k*j

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0rewriting it as b= bj3 does that mean that we proved b=3 since that can be rewritten as b3=0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get that? i got b + 3 = bj

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bj3= (a+1) k and we know that b=(a+1)k

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which is b + 3 = bj , and we already know this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok then, ready for the solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah but I am not sure if that proves that b=3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it doesnt add any new information its not helpful

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b = (a+1)k , b+3 = bj , thats the given. substitute (a+1)k for b in the second equation so we get (a+1)k + 3 = (a+1)k*j

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good now divide both sides by a+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0just to make sure we are starting all over?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you get a+1)k + 3 = (a+1)k*j

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 3/(a+1) is an integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because kj and k are integers, and the difference of integers is always an integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or 3/(a+1) = k ( j1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is only one positive value that will make 3/(a+1) an integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is that wrong? I did it twice and that is what i got

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0a=2 will make 3/(a+1) = 3/3 = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we assumed, k, j, and a are positive initially

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we can prove this actually

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0, k, j, a, b are all positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we'll go back to show that k and j are positive, but lets just assume it is at the moment

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we have k + 3/ (a+1) = kj, agreed?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right but what does the k(j1) mean in this context?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well we dont need that. that just shows the left left is an integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but i want to make this airtight

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to avoid the negative cases

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k + 3/(a+1) = kj , agreed?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont subtract k from both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right hand side is positive, and left hand side is positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. so we would have 2 cases where k and j are even and then another where k and j are odd

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for this to be an equality, then we need kj to be a positive integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0err, i mean the left side to be a positive integer, since the right side is already a positive integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for this to be an equality we need the left side to be a positive integer , because the right side we know is a positive integer , (k , j are both positive integers so the product of two positive integers is another positive integer)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the right side is a positive integer, because k is positive, and j are positive. and we know pos. integer* pos. integer = pos. integer ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its an axiom, positive integers are closed under multiplication (or a theorem if youre doing peano math, etc)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then having k+ 3/(a+1)=kj and rewriting it by subtracting k to the other side : 3/(a+1) = k(j1) wouldn't that alright since k is a positive and j is a positive integer and subtracting 1 from a positive integer is still positive so the right side would be a positive integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, it isnt always positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we dont have to subtract k from both sides

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well, now that you mention it, since a > 0 , we assume, then 3/(a+1) is positive, so k(j1) has to be positive

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and since k>0 , j>0 , we have k>0 , j>1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the punchline is, the right hand side is a positive integer, so the left hand side must be as well

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we know that k(j1) is an integer, correct? since integers are closed under multiplication

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01. k(j1) is integer since integers are closed under multiplication 2. k(j1) is positive because it equals to 3/(a+1) , and we assumed a > 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01. k(j1) is integer since integers are closed under multiplication 2. k(j1) is positive because it equals to 3/(a+1) , and we assumed a > 0 3. 3/(a+1) must be a positive integer because it equals k(j1) which we have shown earlier is a positive integer 4. this only leaves one option for a, for a solution to exist. a = 2,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because if a = 1, we have 3/(1+1), not a positive integer if a = 2 , 3/(2+1), that works 3/3 = 1 a=3 , wont work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok I get this a little bit more but then how would we do this for b=3 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you just plug it back in

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then we can use our other formula

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean it won't work when a=1 or a=3 because we get fractions, which aren't integers?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0after substituting a = 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait shouldn't it be a=2 not 1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so substituting we get

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b = 3k for the first equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have a much shorter proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0substituting 3k for b into second equation we get

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01/ j1 = k, and we know that k > 0 and positive integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now we had before, b + 3 = bj, so b + 3 = b*2 so b = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is a shorter proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would we put this as 2 two cases or just a long direct proof?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that was a direct proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0here is a shorter direct proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that was excruciating long :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0b+3 = bj, so 3 = bj  b or 3 = b (j1), so 3 is divisible by b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so since b > 0 , there is only 2 options for b , b = 1 or b = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but b would be 0 if its 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0only 1 goes into 3 evenly and 3 goes into 3 evenly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you mean a = 0 if b = 1 , right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i was thinking about j

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have b = (a+1) k from second equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait thats false what you said

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if b = 1 , we have b+3 = bj , so 1 + 3 = 1(4) or j = 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and if b = 2 , we have b + 3 = b j , so 3 + 3 = 3(2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0err , if b = 3 i mean

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the second equation clinches the deal , b = (a+1)k , k = b / (a+1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k = 1/ (a+1) has only a = 0 solution if k is an integer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but we assumed that a > 0 , so contradiction. therefore b = 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the proof goes, you have two possible options in the beginning, b=3, or b = 1. But b=1 leads to a = 0 , which contradicts a >0. and if b = 3 , then a = 2 only solution, so thats fine

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the proof goes, a>0 , b>0 b divides b+3. so (b+3) / b is a positive integer. but (b+3)/b = 1 + 3/b This leaves only two options for b, either b = 1 , or b = 3. But b=1 leads to a = 0 from the second condition that a+1 divides b. This contradicts a >0. If b = 3 , then a = 2 works, 2+1  3 , and everyone is happy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha :) you are good at these. I Need so much practice. its hard to learn on you own and I thank you for your time and help :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your welcome i feel stupid now that i did such a long proof

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0always look for a trick, to avoid time loss
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