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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!!

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  1. anonymous
    • 5 years ago
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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?

  2. anonymous
    • 5 years ago
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    right that is what I have so far

  3. anonymous
    • 5 years ago
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    ok lets subtract the two equations

  4. anonymous
    • 5 years ago
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    (b+3) - b = mb - (a (k+1))

  5. anonymous
    • 5 years ago
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    so subtract b+3-b?

  6. anonymous
    • 5 years ago
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    yes

  7. anonymous
    • 5 years ago
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    yes

  8. anonymous
    • 5 years ago
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    system of equations

  9. anonymous
    • 5 years ago
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    i get 3 = mb - ak - a

  10. anonymous
    • 5 years ago
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    that is what I got too

  11. anonymous
    • 5 years ago
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    is there more info are a and b must be integers? well yes if we are discussing "divides"

  12. anonymous
    • 5 years ago
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    ok now substitute back

  13. anonymous
    • 5 years ago
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    suppose a and b are positive integers

  14. anonymous
    • 5 years ago
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    nevermind, thats circular to substitute back

  15. anonymous
    • 5 years ago
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    wait I have a question

  16. anonymous
    • 5 years ago
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    isn't b= (a+1)k? and b+3= bj?

  17. anonymous
    • 5 years ago
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    i have a different idea

  18. anonymous
    • 5 years ago
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    yes , thats correct

  19. anonymous
    • 5 years ago
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    add 3 to the first equation , so b+3 = 3 +(a+1)k

  20. anonymous
    • 5 years ago
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    so bj = 3 + (a+1)k

  21. anonymous
    • 5 years ago
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    so bj - 3 = (a+1)k

  22. anonymous
    • 5 years ago
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    why do you have (b+3)-b= mb-(a(k+1))?

  23. anonymous
    • 5 years ago
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    wait you did that backwards

  24. anonymous
    • 5 years ago
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    isn't it suppose to be mb-(a+1)k?

  25. anonymous
    • 5 years ago
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    b= (a+1)k? and b+3= bj? this is false

  26. anonymous
    • 5 years ago
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    if a+1 divides b then b=(a+1)k?

  27. anonymous
    • 5 years ago
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    isnt the definition of divisiblity if a divides b then b= ak?

  28. anonymous
    • 5 years ago
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    ok so we have bj-3 = (a +1) k ,

  29. anonymous
    • 5 years ago
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    yes i misread

  30. anonymous
    • 5 years ago
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    how did you get bj-3= (a+1)k?

  31. anonymous
    • 5 years ago
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    a few steps

  32. anonymous
    • 5 years ago
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    ok first step, b = (a+1)k , b+3 = bj

  33. anonymous
    • 5 years ago
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    did you do that by substituting? i see what you did

  34. anonymous
    • 5 years ago
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    3 = bj - (a+1)k

  35. anonymous
    • 5 years ago
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    rearranging we get

  36. anonymous
    • 5 years ago
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    (a+1)k = bj - 3

  37. anonymous
    • 5 years ago
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    but we know that b = (a+1) k

  38. anonymous
    • 5 years ago
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    so b = bj - 3

  39. anonymous
    • 5 years ago
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    follow so far, i might have gone too fast

  40. anonymous
    • 5 years ago
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    b = (a+1)k , b+3 = bj , subtracting the former from the latter we get 3 = bj - (a+1)k

  41. anonymous
    • 5 years ago
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    add (a+1)k to both sides, subtract 3 from both sides you should get bj = (a+1)k ,

  42. anonymous
    • 5 years ago
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    I got it so far

  43. anonymous
    • 5 years ago
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    woops

  44. anonymous
    • 5 years ago
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    bj - 3 = (a+1)k

  45. anonymous
    • 5 years ago
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    ok we know that (a+1)k = b, so by transitive rule we have b = bj - 3

  46. anonymous
    • 5 years ago
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    so b+3 = bj , so b divides b+3, which we already know, shoot

  47. anonymous
    • 5 years ago
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    ok lets do some substitution

  48. anonymous
    • 5 years ago
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    b = (a+1)k , b+3 = bj , thats the given. substitute (a+1)k for b in the second equation

  49. anonymous
    • 5 years ago
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    so we get (a+1)k + 3 = (a+1)k*j

  50. anonymous
    • 5 years ago
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    rewriting it as b= bj-3 does that mean that we proved b=3 since that can be rewritten as b-3=0?

  51. anonymous
    • 5 years ago
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    how did you get that? i got b + 3 = bj

  52. anonymous
    • 5 years ago
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    ok i have the answer

  53. anonymous
    • 5 years ago
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    bj-3= (a+1) k and we know that b=(a+1)k

  54. anonymous
    • 5 years ago
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    so b = bj - 3

  55. anonymous
    • 5 years ago
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    which is b + 3 = bj , and we already know this

  56. anonymous
    • 5 years ago
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    ok then, ready for the solution

  57. anonymous
    • 5 years ago
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    yeah but I am not sure if that proves that b=3?

  58. anonymous
    • 5 years ago
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    it doesnt add any new information its not helpful

  59. anonymous
    • 5 years ago
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    so we abandon that

  60. anonymous
    • 5 years ago
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    different approach

  61. anonymous
    • 5 years ago
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    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

  62. anonymous
    • 5 years ago
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    right

  63. anonymous
    • 5 years ago
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    good now divide both sides by a+1

  64. anonymous
    • 5 years ago
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    just to make sure we are starting all over?

  65. anonymous
    • 5 years ago
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    yes

  66. anonymous
    • 5 years ago
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    did you get a+1)k + 3 = (a+1)k*j

  67. anonymous
    • 5 years ago
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    i got k+ 3/(a+1)= kj

  68. anonymous
    • 5 years ago
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    so 3/(a+1) = kj - k

  69. anonymous
    • 5 years ago
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    so 3/(a+1) is an integer

  70. anonymous
    • 5 years ago
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    because kj and k are integers, and the difference of integers is always an integer

  71. anonymous
    • 5 years ago
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    or 3/(a+1) = k ( j-1)

  72. anonymous
    • 5 years ago
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    there is only one positive value that will make 3/(a+1) an integer

  73. anonymous
    • 5 years ago
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    is that wrong? I did it twice and that is what i got

  74. anonymous
    • 5 years ago
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    a=2 will make 3/(a+1) = 3/3 = 1

  75. anonymous
    • 5 years ago
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    or backtrack

  76. anonymous
    • 5 years ago
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    we assumed, k, j, and a are positive initially

  77. anonymous
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    we can prove this actually

  78. anonymous
    • 5 years ago
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    , k, j, a, b are all positive

  79. anonymous
    • 5 years ago
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    ok so far?

  80. anonymous
    • 5 years ago
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    we'll go back to show that k and j are positive, but lets just assume it is at the moment

  81. anonymous
    • 5 years ago
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    so we have k + 3/ (a+1) = kj, agreed?

  82. anonymous
    • 5 years ago
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    right but what does the k(j-1) mean in this context?

  83. anonymous
    • 5 years ago
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    well we dont need that. that just shows the left left is an integer

  84. anonymous
    • 5 years ago
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    but i want to make this airtight

  85. anonymous
    • 5 years ago
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    to avoid the negative cases

  86. anonymous
    • 5 years ago
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    k + 3/(a+1) = kj , agreed?

  87. anonymous
    • 5 years ago
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    dont subtract k from both sides

  88. anonymous
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    right hand side is positive, and left hand side is positive

  89. anonymous
    • 5 years ago
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    Yes. so we would have 2 cases where k and j are even and then another where k and j are odd

  90. anonymous
    • 5 years ago
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    dont need that

  91. anonymous
    • 5 years ago
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    for this to be an equality, then we need kj to be a positive integer

  92. anonymous
    • 5 years ago
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    err, i mean the left side to be a positive integer, since the right side is already a positive integer

  93. anonymous
    • 5 years ago
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    why?

  94. anonymous
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    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)

  95. anonymous
    • 5 years ago
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    the right side is a positive integer, because k is positive, and j are positive. and we know pos. integer* pos. integer = pos. integer ,

  96. anonymous
    • 5 years ago
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    its an axiom, positive integers are closed under multiplication (or a theorem if youre doing peano math, etc)

  97. anonymous
    • 5 years ago
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    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

  98. anonymous
    • 5 years ago
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    no, it isnt always positive

  99. anonymous
    • 5 years ago
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    if j = 1, 1 -1 = 0

  100. anonymous
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    we dont have to subtract k from both sides

  101. anonymous
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    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

  102. anonymous
    • 5 years ago
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    and since k>0 , j>0 , we have k>0 , j>1

  103. anonymous
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    ok we can do that

  104. anonymous
    • 5 years ago
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    but the punchline is, the right hand side is a positive integer, so the left hand side must be as well

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

  106. anonymous
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    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

  107. anonymous
    • 5 years ago
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    ok so far?

  108. anonymous
    • 5 years ago
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    you still there?

  109. anonymous
    • 5 years ago
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    yes

  110. anonymous
    • 5 years ago
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    yes im here

  111. anonymous
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    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,

  112. anonymous
    • 5 years ago
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    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

  113. anonymous
    • 5 years ago
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    Ok I get this a little bit more but then how would we do this for b=3 ?

  114. anonymous
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    then you just plug it back in

  115. anonymous
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    then we can use our other formula

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

  117. anonymous
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    right

  118. anonymous
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    which other formula?

  119. anonymous
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    3 = bj - 2k

  120. anonymous
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    after substituting a = 1

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

  122. anonymous
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    oh

  123. anonymous
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    ok

  124. anonymous
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    so substituting we get

  125. anonymous
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    b = 3k for the first equation

  126. anonymous
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    since b = (a+1)k

  127. anonymous
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    agreed?

  128. anonymous
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    you there?

  129. anonymous
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    gotcha

  130. anonymous
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    yes

  131. anonymous
    • 5 years ago
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    i have a much shorter proof

  132. anonymous
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    but going with this

  133. anonymous
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    substituting 3k for b into second equation we get

  134. anonymous
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    3k + 3 = 3kj

  135. anonymous
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    right

  136. anonymous
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    so 3 = 3kj - 3k

  137. anonymous
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    3 = 3k ( j - 1)

  138. anonymous
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    yes

  139. anonymous
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    so 1 = k ( j-1)

  140. anonymous
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    yup

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

  142. anonymous
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    so j = 2, k = 1

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

  144. anonymous
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    right

  145. anonymous
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    here is a shorter proof

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

  147. anonymous
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    that was a direct proof

  148. anonymous
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    here is a shorter direct proof

  149. anonymous
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    that was excruciating long :)

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

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

  152. anonymous
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    ok so far?

  153. anonymous
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    yes but b would be 0 if its 1

  154. anonymous
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    only 1 goes into 3 evenly and 3 goes into 3 evenly

  155. anonymous
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    what do you mean?

  156. anonymous
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    oh nevermind

  157. anonymous
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    you mean a = 0 if b = 1 , right

  158. anonymous
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    i was thinking about j

  159. anonymous
    • 5 years ago
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    we have b = (a+1) k from second equation

  160. anonymous
    • 5 years ago
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    ok

  161. anonymous
    • 5 years ago
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    wait thats false what you said

  162. anonymous
    • 5 years ago
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    if b = 1 , we have b+3 = bj , so 1 + 3 = 1(4) or j = 4

  163. anonymous
    • 5 years ago
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    and if b = 2 , we have b + 3 = b j , so 3 + 3 = 3(2)

  164. anonymous
    • 5 years ago
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    err , if b = 3 i mean

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

  166. anonymous
    • 5 years ago
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    k = 1/ (a+1) has only a = 0 solution if k is an integer

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

  168. anonymous
    • 5 years ago
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    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

  169. anonymous
    • 5 years ago
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    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

  170. anonymous
    • 5 years ago
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    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 :)

  171. anonymous
    • 5 years ago
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    your welcome i feel stupid now that i did such a long proof

  172. anonymous
    • 5 years ago
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    always look for a trick, to avoid time loss

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