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

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- 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

right that is what I have so far

- anonymous

ok lets subtract the two equations

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- anonymous

(b+3) - b = mb - (a (k+1))

- anonymous

so subtract b+3-b?

- anonymous

yes

- anonymous

yes

- anonymous

system of equations

- anonymous

i get 3 = mb - ak - a

- anonymous

that is what I got too

- anonymous

is there more info
are a and b must be integers?
well yes if we are discussing "divides"

- anonymous

ok now substitute back

- anonymous

suppose a and b are positive integers

- anonymous

nevermind, thats circular to substitute back

- anonymous

wait I have a question

- anonymous

isn't b= (a+1)k? and b+3= bj?

- anonymous

i have a different idea

- anonymous

yes , thats correct

- anonymous

add 3 to the first equation , so b+3 = 3 +(a+1)k

- anonymous

so bj = 3 + (a+1)k

- anonymous

so bj - 3 = (a+1)k

- anonymous

why do you have (b+3)-b= mb-(a(k+1))?

- anonymous

wait you did that backwards

- anonymous

isn't it suppose to be mb-(a+1)k?

- anonymous

b= (a+1)k? and b+3= bj? this is false

- anonymous

if a+1 divides b then b=(a+1)k?

- anonymous

isnt the definition of divisiblity if a divides b then b= ak?

- anonymous

ok so we have bj-3 = (a +1) k ,

- anonymous

yes i misread

- anonymous

how did you get bj-3= (a+1)k?

- anonymous

a few steps

- anonymous

ok first step, b = (a+1)k , b+3 = bj

- anonymous

did you do that by substituting?
i see what you did

- anonymous

3 = bj - (a+1)k

- anonymous

rearranging we get

- anonymous

(a+1)k = bj - 3

- anonymous

but we know that b = (a+1) k

- anonymous

so b = bj - 3

- anonymous

follow so far, i might have gone too fast

- anonymous

b = (a+1)k , b+3 = bj , subtracting the former from the latter we get
3 = bj - (a+1)k

- anonymous

add (a+1)k to both sides, subtract 3 from both sides
you should get bj = (a+1)k ,

- anonymous

I got it so far

- anonymous

woops

- anonymous

bj - 3 = (a+1)k

- anonymous

ok we know that (a+1)k = b, so by transitive rule we have
b = bj - 3

- anonymous

so b+3 = bj , so b divides b+3, which we already know, shoot

- anonymous

ok lets do some substitution

- anonymous

b = (a+1)k , b+3 = bj ,
thats the given.
substitute (a+1)k for b in the second equation

- anonymous

so we get (a+1)k + 3 = (a+1)k*j

- 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

how did you get that? i got b + 3 = bj

- anonymous

ok i have the answer

- anonymous

bj-3= (a+1) k and we know that b=(a+1)k

- anonymous

so b = bj - 3

- anonymous

which is b + 3 = bj , and we already know this

- anonymous

ok then, ready for the solution

- anonymous

yeah but I am not sure if that proves that b=3?

- anonymous

it doesnt add any new information
its not helpful

- anonymous

so we abandon that

- anonymous

different approach

- 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

right

- anonymous

good
now divide both sides by a+1

- anonymous

just to make sure we are starting all over?

- anonymous

yes

- anonymous

did you get a+1)k + 3 = (a+1)k*j

- anonymous

i got k+ 3/(a+1)= kj

- anonymous

so 3/(a+1) = kj - k

- anonymous

so 3/(a+1) is an integer

- anonymous

because kj and k are integers, and the difference of integers is always an integer

- anonymous

or 3/(a+1) = k ( j-1)

- anonymous

there is only one positive value that will make 3/(a+1) an integer

- anonymous

is that wrong? I did it twice and that is what i got

- anonymous

a=2 will make 3/(a+1) = 3/3 = 1

- anonymous

or backtrack

- anonymous

we assumed, k, j, and a are positive initially

- anonymous

we can prove this actually

- anonymous

, k, j, a, b are all positive

- anonymous

ok so far?

- anonymous

we'll go back to show that k and j are positive, but lets just assume it is at the moment

- anonymous

so we have k + 3/ (a+1) = kj, agreed?

- anonymous

right but what does the k(j-1) mean in this context?

- anonymous

well we dont need that. that just shows the left left is an integer

- anonymous

but i want to make this airtight

- anonymous

to avoid the negative cases

- anonymous

k + 3/(a+1) = kj , agreed?

- anonymous

dont subtract k from both sides

- anonymous

right hand side is positive, and left hand side is positive

- anonymous

Yes. so we would have 2 cases where k and j are even and then another where k and j are odd

- anonymous

dont need that

- anonymous

for this to be an equality, then we need kj to be a positive integer

- anonymous

err, i mean the left side to be a positive integer, since the right side is already a positive integer

- anonymous

why?

- 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

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

its an axiom, positive integers are closed under multiplication (or a theorem if youre doing peano math, etc)

- 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

no, it isnt always positive

- anonymous

if j = 1, 1 -1 = 0

- anonymous

we dont have to subtract k from both sides

- 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

and since k>0 , j>0 , we have k>0 , j>1

- anonymous

ok we can do that

- anonymous

but the punchline is, the right hand side is a positive integer, so the left hand side must be as well

- anonymous

we know that k(j-1) is an integer, correct? since integers are closed under multiplication

- 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

ok so far?

- anonymous

you still there?

- anonymous

yes

- anonymous

yes im here

- 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

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

Ok I get this a little bit more but then how would we do this for b=3 ?

- anonymous

then you just plug it back in

- anonymous

then we can use our other formula

- anonymous

you mean it won't work when a=1 or a=3 because we get fractions, which aren't integers?

- anonymous

right

- anonymous

which other formula?

- anonymous

3 = bj - 2k

- anonymous

after substituting a = 1

- anonymous

wait shouldn't it be a=2 not 1?

- anonymous

oh

- anonymous

ok

- anonymous

so substituting we get

- anonymous

b = 3k for the first equation

- anonymous

since b = (a+1)k

- anonymous

agreed?

- anonymous

you there?

- anonymous

gotcha

- anonymous

yes

- anonymous

i have a much shorter proof

- anonymous

but going with this

- anonymous

substituting 3k for b into second equation we get

- anonymous

3k + 3 = 3kj

- anonymous

right

- anonymous

so 3 = 3kj - 3k

- anonymous

3 = 3k ( j - 1)

- anonymous

yes

- anonymous

so 1 = k ( j-1)

- anonymous

yup

- anonymous

1/ j-1 = k, and we know that k > 0 and positive integer

- anonymous

so j = 2, k = 1

- anonymous

so now we had before, b + 3 = bj,
so b + 3 = b*2
so b = 3

- anonymous

right

- anonymous

here is a shorter proof

- anonymous

would we put this as 2 two cases or just a long direct proof?

- anonymous

that was a direct proof

- anonymous

here is a shorter direct proof

- anonymous

that was excruciating long :)

- anonymous

b+3 = bj, so 3 = bj - b
or 3 = b (j-1), so 3 is divisible by b

- anonymous

so since b > 0 , there is only 2 options for b , b = 1 or b = 3

- anonymous

ok so far?

- anonymous

yes but b would be 0 if its 1

- anonymous

only 1 goes into 3 evenly and 3 goes into 3 evenly

- anonymous

what do you mean?

- anonymous

oh nevermind

- anonymous

you mean a = 0 if b = 1
, right

- anonymous

i was thinking about j

- anonymous

we have b = (a+1) k from second equation

- anonymous

ok

- anonymous

wait thats false what you said

- anonymous

if b = 1 , we have b+3 = bj , so 1 + 3 = 1(4) or j = 4

- anonymous

and if b = 2 , we have b + 3 = b j , so 3 + 3 = 3(2)

- anonymous

err
, if b = 3 i mean

- anonymous

the second equation clinches the deal , b = (a+1)k , k = b / (a+1)

- anonymous

k = 1/ (a+1) has only a = 0 solution if k is an integer

- anonymous

but we assumed that a > 0 , so contradiction. therefore b = 3

- 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

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

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

your welcome
i feel stupid now that i did such a long proof

- anonymous

always look for a trick, to avoid time loss

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