UnkleRhaukus
  • UnkleRhaukus
how can i disprove the following statement \[(\exists a ,b \in \mathbb N)(3a + 4b = 2)\]
Meta-math
jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
maybe another method
anonymous
  • anonymous
pick \(a=5,b=7\)
anonymous
  • anonymous
negation is \((\exists a,b\in N)(3a+4b\neq 2)\)

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ParthKohli
  • ParthKohli
Ohhhh.
ParthKohli
  • ParthKohli
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).
UnkleRhaukus
  • UnkleRhaukus
oh,
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
We've proved that both \(\rm a\) and \(\rm b\) don't belong to the set of natural numbers. QED
UnkleRhaukus
  • UnkleRhaukus
so i assume that \(a\) is a natural number and show that if it is; \(b\) cannot be a natural
anonymous
  • anonymous
i like my proof better
anonymous
  • anonymous
oooh i thought it said \(\forall a,b\in \mathbb{N}\) did it change or can i not see straight?
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
@satellite73 Bikes don't have a good reflector at night. That's why.
anonymous
  • anonymous
it must be past my bed time
ParthKohli
  • ParthKohli
Sure it is.
Mr.Math
  • Mr.Math
Or just note that \(3a+4b\ge7\) \( \forall a,b \in \mathbb{N}\). Then the statement is false since \(2<7\).
ParthKohli
  • ParthKohli
Mr.Math!
ParthKohli
  • ParthKohli
And by the way, that's the linear combination thingy

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