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OpenStudy (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!!

OpenStudy (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?

OpenStudy (anonymous):

right that is what I have so far

OpenStudy (anonymous):

ok lets subtract the two equations

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so subtract b+3-b?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

system of equations

OpenStudy (anonymous):

i get 3 = mb - ak - a

OpenStudy (anonymous):

that is what I got too

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok now substitute back

OpenStudy (anonymous):

suppose a and b are positive integers

OpenStudy (anonymous):

nevermind, thats circular to substitute back

OpenStudy (anonymous):

wait I have a question

OpenStudy (anonymous):

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

OpenStudy (anonymous):

i have a different idea

OpenStudy (anonymous):

yes , thats correct

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so bj = 3 + (a+1)k

OpenStudy (anonymous):

so bj - 3 = (a+1)k

OpenStudy (anonymous):

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

OpenStudy (anonymous):

wait you did that backwards

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yes i misread

OpenStudy (anonymous):

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

OpenStudy (anonymous):

a few steps

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

3 = bj - (a+1)k

OpenStudy (anonymous):

rearranging we get

OpenStudy (anonymous):

(a+1)k = bj - 3

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so b = bj - 3

OpenStudy (anonymous):

follow so far, i might have gone too fast

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I got it so far

OpenStudy (anonymous):

woops

OpenStudy (anonymous):

bj - 3 = (a+1)k

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok lets do some substitution

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

rewriting it as b= bj-3 does that mean that we proved b=3 since that can be rewritten as b-3=0?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok i have the answer

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so b = bj - 3

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok then, ready for the solution

OpenStudy (anonymous):

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

OpenStudy (anonymous):

it doesnt add any new information its not helpful

OpenStudy (anonymous):

so we abandon that

OpenStudy (anonymous):

different approach

OpenStudy (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

OpenStudy (anonymous):

right

OpenStudy (anonymous):

good now divide both sides by a+1

OpenStudy (anonymous):

just to make sure we are starting all over?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so 3/(a+1) is an integer

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

or backtrack

OpenStudy (anonymous):

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

OpenStudy (anonymous):

we can prove this actually

OpenStudy (anonymous):

, k, j, a, b are all positive

OpenStudy (anonymous):

ok so far?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

but i want to make this airtight

OpenStudy (anonymous):

to avoid the negative cases

OpenStudy (anonymous):

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

OpenStudy (anonymous):

dont subtract k from both sides

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

dont need that

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

why?

OpenStudy (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)

OpenStudy (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 ,

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

no, it isnt always positive

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