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