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lgbasallote Group Title

Prove that if n is an integer then 3n + 2 is even, then n is even

  • 2 years ago
  • 2 years ago

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  1. mathslover Group Title
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    we have to prove two things? : 1) 3n+2 is even , 2) n is even

    • 2 years ago
  2. vf321 Group Title
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    no, it's prove n is even GIVEN 3n + 2 is even.

    • 2 years ago
  3. mathslover Group Title
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    ok , thanks

    • 2 years ago
  4. vf321 Group Title
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    If 3n + 2 is even, then 3n is even

    • 2 years ago
  5. mathslover Group Title
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    if 3n is even then n is even.

    • 2 years ago
  6. sara12345 Group Title
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    Given : 3n+2 is even To prove : n is even 3n+2 = 2k 3n = 2k-2 3n = 2(k-1)

    • 2 years ago
  7. hartnn Group Title
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    not true for n=1 ? 3+2=5<---not even

    • 2 years ago
  8. vf321 Group Title
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    @hartnn GIVEN that 3+2 is even, n is even

    • 2 years ago
  9. mathslover Group Title
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    IF 3n+2 is even , prove that is even. @hartnn

    • 2 years ago
  10. vf321 Group Title
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    But it's not so no guarentee is made.

    • 2 years ago
  11. hartnn Group Title
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    lets clear it from @lgbasallote what the exact question is...

    • 2 years ago
  12. mathslover Group Title
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    @lgbasallote ?

    • 2 years ago
  13. JamesWolf Group Title
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    Just out of interest whats the general way to prove something is even, divide by 2?

    • 2 years ago
  14. sara12345 Group Title
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    yes

    • 2 years ago
  15. sara12345 Group Title
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    if we prove a number is of form 2k that proves it is even

    • 2 years ago
  16. JamesWolf Group Title
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    right I see

    • 2 years ago
  17. vf321 Group Title
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    btw any1 here mind taking a look at my question? http://openstudy.com/updates/5078b9d7e4b02f109be44f95

    • 2 years ago
  18. lgbasallote Group Title
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    sorry i just came back....anyway...i'd llike to see how proving by contradiction is done.. direct proof is too easy

    • 2 years ago
  19. hartnn Group Title
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    but whats the question ?

    • 2 years ago
  20. lgbasallote Group Title
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    the statement in the blue box

    • 2 years ago
  21. lgbasallote Group Title
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    forgot to mention by the way....that the direct proof done here was wrong

    • 2 years ago
  22. lgbasallote Group Title
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    they took n as the condition... 3n + 2 is supposed to be the condition

    • 2 years ago
  23. JamesWolf Group Title
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    so 3n + 2 = (a different n)?

    • 2 years ago
  24. hartnn Group Title
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    if 3n+2 is even , n is even. thats the question and u need to prove that using contradiction, right ?

    • 2 years ago
  25. lgbasallote Group Title
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    if you use direct proof... it should be 3n + 2 = 2x then prove n is even

    • 2 years ago
  26. lgbasallote Group Title
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    but like i said...should be contradiction though

    • 2 years ago
  27. vf321 Group Title
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    well okay thats easy enough, too

    • 2 years ago
  28. lgbasallote Group Title
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    hmmm then let's see you try

    • 2 years ago
  29. vf321 Group Title
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    All you have to say is that assume that given 3n + 2 is even, n is not even Then n has to be odd. Thus, 3n + 2 can be represented as 2k + 1 for some k

    • 2 years ago
  30. sara12345 Group Title
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    Given : 3n+2 is even To prove : n is even to prove by contradiction, lets assume the opposite - lets assume n is odd, 3n+2 = 2k 3n = 2k-2 3n = 2(k-1) so we got the right side as even number, but we assumed n is odd, so left 3n becomes odd - contradiction

    • 2 years ago
  31. vf321 Group Title
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    @sara12345 you cannot say "assume that n is odd" and then equate it to 2k

    • 2 years ago
  32. lgbasallote Group Title
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    @sara12345 how is that contradiction

    • 2 years ago
  33. sara12345 Group Title
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    we do that always in proof by contradiction

    • 2 years ago
  34. sara12345 Group Title
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    RHS = even , LHS = odd => contradiction

    • 2 years ago
  35. lgbasallote Group Title
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    i was referring to your solution

    • 2 years ago
  36. vf321 Group Title
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    How can you prove that it equals 2k, though?

    • 2 years ago
  37. JamesWolf Group Title
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    if 3n + 2 is odd, then it can be expressed 3n + 2 = 2k + 1. solving for n gives \[n = \frac{2k}{3} - 1\] substituting back for n gives \[3 (\times \frac{2k}{3} - 1) + 2 = 2k + 1so\] so \[2k - 1 = 2k + 1 \] which is absurd

    • 2 years ago
  38. lgbasallote Group Title
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    hmm seems legit

    • 2 years ago
  39. lgbasallote Group Title
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    so in proof by contradiction...you still substitute back huh

    • 2 years ago
  40. vf321 Group Title
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    As @JamesWolf showed for you,

    • 2 years ago
  41. vf321 Group Title
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    you take the contradiction, show that it is not internally consistent, and therefore it cannot be true.

    • 2 years ago
  42. JamesWolf Group Title
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    no its not legit

    • 2 years ago
  43. JamesWolf Group Title
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    ive been an idiot int he first step

    • 2 years ago
  44. vf321 Group Title
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    @JamesWolf no, you inadvertently put up my proof.

    • 2 years ago
  45. sara12345 Group Title
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    in proof by contradiction, we assum e the opposite of what we need to prove as true, and proceed, not the opposite of given conditions

    • 2 years ago
  46. lgbasallote Group Title
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    if you assume n is odd and 3n + 2 is even... 3(2k + 1) + 2 6k + 3 + 2 6k + 2 + 3 2(3k + 1) + 3 so even + odd would be odd...so contradiction i suppose that works as well

    • 2 years ago
  47. vf321 Group Title
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    @sara12345 yes you're right, sorry.

    • 2 years ago
  48. JamesWolf Group Title
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    yes its right actually i managed to not divide by 3 at the start, but luckily i messed up by not multiplying - 1by 3 at the end

    • 2 years ago
  49. lgbasallote Group Title
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    oh so you take the negation of q then proceed from there?

    • 2 years ago
  50. lgbasallote Group Title
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    anyway...is my proof right?

    • 2 years ago
  51. sara12345 Group Title
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    lgba ur proof is more correct as it shows the assumption n is odd as well by letting n =2k+1

    • 2 years ago
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