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UnkleRhaukus

how can i disprove the following statement \[(\exists a ,b \in \mathbb N)(3a + 4b = 2)\]

  • one year ago
  • one year ago

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  1. UnkleRhaukus
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    maybe another method

    • one year ago
  2. satellite73
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    pick \(a=5,b=7\)

    • one year ago
  3. satellite73
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    negation is \((\exists a,b\in N)(3a+4b\neq 2)\)

    • one year ago
  4. ParthKohli
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    Ohhhh.

    • one year ago
  5. ParthKohli
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    Assuming the statement is true,\[\rm b = {2 - 3a \over 4}\]so\[\rm b = {1 \over 2} - {3a \over 4}\]Therefore, if \(\rm a\) is a natural number, then \(\rm b \) isn't (try it - picking 1 will get ya some idiotic negative fraction).

    • one year ago
  6. UnkleRhaukus
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    oh,

    • one year ago
  7. ParthKohli
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    Let's do the magic for \(\rm a\) as well.\[\rm a = {2 - 4b \over 3}\]\[\rm a = {2 \over 3} - {4b \over 3}\]If \(\rm b\) is a natural number, then you get a negative fraction and therefore \(\rm a\) is not.

    • one year ago
  8. ParthKohli
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    We've proved that both \(\rm a\) and \(\rm b\) don't belong to the set of natural numbers. QED

    • one year ago
  9. UnkleRhaukus
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    so i assume that \(a\) is a natural number and show that if it is; \(b\) cannot be a natural

    • one year ago
  10. satellite73
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    i like my proof better

    • one year ago
  11. satellite73
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    oooh i thought it said \(\forall a,b\in \mathbb{N}\) did it change or can i not see straight?

    • one year ago
  12. ParthKohli
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    Basically, your statement says that both a and b must be natural. I've shown that in both cases, one must be a negative fraction - not belonging to the set of natural numbers.

    • one year ago
  13. ParthKohli
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    @satellite73 Bikes don't have a good reflector at night. That's why.

    • one year ago
  14. satellite73
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    it must be past my bed time

    • one year ago
  15. ParthKohli
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    Sure it is.

    • one year ago
  16. Mr.Math
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    Or just note that \(3a+4b\ge7\) \( \forall a,b \in \mathbb{N}\). Then the statement is false since \(2<7\).

    • one year ago
  17. ParthKohli
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    Mr.Math!

    • one year ago
  18. ParthKohli
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    And by the way, that's the linear combination thingy

    • one year ago
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