anonymous
  • anonymous
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!!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
if 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
  • anonymous
right that is what I have so far
anonymous
  • anonymous
ok lets subtract the two equations

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anonymous
  • anonymous
(b+3) - b = mb - (a (k+1))
anonymous
  • anonymous
so subtract b+3-b?
anonymous
  • anonymous
yes
anonymous
  • anonymous
yes
anonymous
  • anonymous
system of equations
anonymous
  • anonymous
i get 3 = mb - ak - a
anonymous
  • anonymous
that is what I got too
anonymous
  • anonymous
is there more info are a and b must be integers? well yes if we are discussing "divides"
anonymous
  • anonymous
ok now substitute back
anonymous
  • anonymous
suppose a and b are positive integers
anonymous
  • anonymous
nevermind, thats circular to substitute back
anonymous
  • anonymous
wait I have a question
anonymous
  • anonymous
isn't b= (a+1)k? and b+3= bj?
anonymous
  • anonymous
i have a different idea
anonymous
  • anonymous
yes , thats correct
anonymous
  • anonymous
add 3 to the first equation , so b+3 = 3 +(a+1)k
anonymous
  • anonymous
so bj = 3 + (a+1)k
anonymous
  • anonymous
so bj - 3 = (a+1)k
anonymous
  • anonymous
why do you have (b+3)-b= mb-(a(k+1))?
anonymous
  • anonymous
wait you did that backwards
anonymous
  • anonymous
isn't it suppose to be mb-(a+1)k?
anonymous
  • anonymous
b= (a+1)k? and b+3= bj? this is false
anonymous
  • anonymous
if a+1 divides b then b=(a+1)k?
anonymous
  • anonymous
isnt the definition of divisiblity if a divides b then b= ak?
anonymous
  • anonymous
ok so we have bj-3 = (a +1) k ,
anonymous
  • anonymous
yes i misread
anonymous
  • anonymous
how did you get bj-3= (a+1)k?
anonymous
  • anonymous
a few steps
anonymous
  • anonymous
ok first step, b = (a+1)k , b+3 = bj
anonymous
  • anonymous
did you do that by substituting? i see what you did
anonymous
  • anonymous
3 = bj - (a+1)k
anonymous
  • anonymous
rearranging we get
anonymous
  • anonymous
(a+1)k = bj - 3
anonymous
  • anonymous
but we know that b = (a+1) k
anonymous
  • anonymous
so b = bj - 3
anonymous
  • anonymous
follow so far, i might have gone too fast
anonymous
  • anonymous
b = (a+1)k , b+3 = bj , subtracting the former from the latter we get 3 = bj - (a+1)k
anonymous
  • anonymous
add (a+1)k to both sides, subtract 3 from both sides you should get bj = (a+1)k ,
anonymous
  • anonymous
I got it so far
anonymous
  • anonymous
woops
anonymous
  • anonymous
bj - 3 = (a+1)k
anonymous
  • anonymous
ok we know that (a+1)k = b, so by transitive rule we have b = bj - 3
anonymous
  • anonymous
so b+3 = bj , so b divides b+3, which we already know, shoot
anonymous
  • anonymous
ok lets do some substitution
anonymous
  • anonymous
b = (a+1)k , b+3 = bj , thats the given. substitute (a+1)k for b in the second equation
anonymous
  • anonymous
so we get (a+1)k + 3 = (a+1)k*j
anonymous
  • anonymous
rewriting it as b= bj-3 does that mean that we proved b=3 since that can be rewritten as b-3=0?
anonymous
  • anonymous
how did you get that? i got b + 3 = bj
anonymous
  • anonymous
ok i have the answer
anonymous
  • anonymous
bj-3= (a+1) k and we know that b=(a+1)k
anonymous
  • anonymous
so b = bj - 3
anonymous
  • anonymous
which is b + 3 = bj , and we already know this
anonymous
  • anonymous
ok then, ready for the solution
anonymous
  • anonymous
yeah but I am not sure if that proves that b=3?
anonymous
  • anonymous
it doesnt add any new information its not helpful
anonymous
  • anonymous
so we abandon that
anonymous
  • anonymous
different approach
anonymous
  • anonymous
b = (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
  • anonymous
right
anonymous
  • anonymous
good now divide both sides by a+1
anonymous
  • anonymous
just to make sure we are starting all over?
anonymous
  • anonymous
yes
anonymous
  • anonymous
did you get a+1)k + 3 = (a+1)k*j
anonymous
  • anonymous
i got k+ 3/(a+1)= kj
anonymous
  • anonymous
so 3/(a+1) = kj - k
anonymous
  • anonymous
so 3/(a+1) is an integer
anonymous
  • anonymous
because kj and k are integers, and the difference of integers is always an integer
anonymous
  • anonymous
or 3/(a+1) = k ( j-1)
anonymous
  • anonymous
there is only one positive value that will make 3/(a+1) an integer
anonymous
  • anonymous
is that wrong? I did it twice and that is what i got
anonymous
  • anonymous
a=2 will make 3/(a+1) = 3/3 = 1
anonymous
  • anonymous
or backtrack
anonymous
  • anonymous
we assumed, k, j, and a are positive initially
anonymous
  • anonymous
we can prove this actually
anonymous
  • anonymous
, k, j, a, b are all positive
anonymous
  • anonymous
ok so far?
anonymous
  • anonymous
we'll go back to show that k and j are positive, but lets just assume it is at the moment
anonymous
  • anonymous
so we have k + 3/ (a+1) = kj, agreed?
anonymous
  • anonymous
right but what does the k(j-1) mean in this context?
anonymous
  • anonymous
well we dont need that. that just shows the left left is an integer
anonymous
  • anonymous
but i want to make this airtight
anonymous
  • anonymous
to avoid the negative cases
anonymous
  • anonymous
k + 3/(a+1) = kj , agreed?
anonymous
  • anonymous
dont subtract k from both sides
anonymous
  • anonymous
right hand side is positive, and left hand side is positive
anonymous
  • anonymous
Yes. so we would have 2 cases where k and j are even and then another where k and j are odd
anonymous
  • anonymous
dont need that
anonymous
  • anonymous
for this to be an equality, then we need kj to be a positive integer
anonymous
  • anonymous
err, i mean the left side to be a positive integer, since the right side is already a positive integer
anonymous
  • anonymous
why?
anonymous
  • anonymous
for 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
  • anonymous
the right side is a positive integer, because k is positive, and j are positive. and we know pos. integer* pos. integer = pos. integer ,
anonymous
  • anonymous
its an axiom, positive integers are closed under multiplication (or a theorem if youre doing peano math, etc)
anonymous
  • anonymous
so then having k+ 3/(a+1)=kj and rewriting it by subtracting k to the other side : 3/(a+1) = k(j-1) 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
  • anonymous
no, it isnt always positive
anonymous
  • anonymous
if j = 1, 1 -1 = 0
anonymous
  • anonymous
we dont have to subtract k from both sides
anonymous
  • anonymous
well, now that you mention it, since a > 0 , we assume, then 3/(a+1) is positive, so k(j-1) has to be positive
anonymous
  • anonymous
and since k>0 , j>0 , we have k>0 , j>1
anonymous
  • anonymous
ok we can do that
anonymous
  • anonymous
but the punchline is, the right hand side is a positive integer, so the left hand side must be as well
anonymous
  • anonymous
we know that k(j-1) is an integer, correct? since integers are closed under multiplication
anonymous
  • anonymous
1. k(j-1) is integer since integers are closed under multiplication 2. k(j-1) is positive because it equals to 3/(a+1) , and we assumed a > 0
anonymous
  • anonymous
ok so far?
anonymous
  • anonymous
you still there?
anonymous
  • anonymous
yes
anonymous
  • anonymous
yes im here
anonymous
  • anonymous
1. k(j-1) is integer since integers are closed under multiplication 2. k(j-1) 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(j-1) 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
  • anonymous
because 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
  • anonymous
Ok I get this a little bit more but then how would we do this for b=3 ?
anonymous
  • anonymous
then you just plug it back in
anonymous
  • anonymous
then we can use our other formula
anonymous
  • anonymous
you mean it won't work when a=1 or a=3 because we get fractions, which aren't integers?
anonymous
  • anonymous
right
anonymous
  • anonymous
which other formula?
anonymous
  • anonymous
3 = bj - 2k
anonymous
  • anonymous
after substituting a = 1
anonymous
  • anonymous
wait shouldn't it be a=2 not 1?
anonymous
  • anonymous
oh
anonymous
  • anonymous
ok
anonymous
  • anonymous
so substituting we get
anonymous
  • anonymous
b = 3k for the first equation
anonymous
  • anonymous
since b = (a+1)k
anonymous
  • anonymous
agreed?
anonymous
  • anonymous
you there?
anonymous
  • anonymous
gotcha
anonymous
  • anonymous
yes
anonymous
  • anonymous
i have a much shorter proof
anonymous
  • anonymous
but going with this
anonymous
  • anonymous
substituting 3k for b into second equation we get
anonymous
  • anonymous
3k + 3 = 3kj
anonymous
  • anonymous
right
anonymous
  • anonymous
so 3 = 3kj - 3k
anonymous
  • anonymous
3 = 3k ( j - 1)
anonymous
  • anonymous
yes
anonymous
  • anonymous
so 1 = k ( j-1)
anonymous
  • anonymous
yup
anonymous
  • anonymous
1/ j-1 = k, and we know that k > 0 and positive integer
anonymous
  • anonymous
so j = 2, k = 1
anonymous
  • anonymous
so now we had before, b + 3 = bj, so b + 3 = b*2 so b = 3
anonymous
  • anonymous
right
anonymous
  • anonymous
here is a shorter proof
anonymous
  • anonymous
would we put this as 2 two cases or just a long direct proof?
anonymous
  • anonymous
that was a direct proof
anonymous
  • anonymous
here is a shorter direct proof
anonymous
  • anonymous
that was excruciating long :)
anonymous
  • anonymous
b+3 = bj, so 3 = bj - b or 3 = b (j-1), so 3 is divisible by b
anonymous
  • anonymous
so since b > 0 , there is only 2 options for b , b = 1 or b = 3
anonymous
  • anonymous
ok so far?
anonymous
  • anonymous
yes but b would be 0 if its 1
anonymous
  • anonymous
only 1 goes into 3 evenly and 3 goes into 3 evenly
anonymous
  • anonymous
what do you mean?
anonymous
  • anonymous
oh nevermind
anonymous
  • anonymous
you mean a = 0 if b = 1 , right
anonymous
  • anonymous
i was thinking about j
anonymous
  • anonymous
we have b = (a+1) k from second equation
anonymous
  • anonymous
ok
anonymous
  • anonymous
wait thats false what you said
anonymous
  • anonymous
if b = 1 , we have b+3 = bj , so 1 + 3 = 1(4) or j = 4
anonymous
  • anonymous
and if b = 2 , we have b + 3 = b j , so 3 + 3 = 3(2)
anonymous
  • anonymous
err , if b = 3 i mean
anonymous
  • anonymous
the second equation clinches the deal , b = (a+1)k , k = b / (a+1)
anonymous
  • anonymous
k = 1/ (a+1) has only a = 0 solution if k is an integer
anonymous
  • anonymous
but we assumed that a > 0 , so contradiction. therefore b = 3
anonymous
  • anonymous
so 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
  • anonymous
so 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
  • anonymous
haha :) 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
  • anonymous
your welcome i feel stupid now that i did such a long proof
anonymous
  • anonymous
always look for a trick, to avoid time loss

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