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

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UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0maybe another method

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0negation is \((\exists a,b\in N)(3a+4b\neq 2)\)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Assuming 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).

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Let'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
 3 years ago
Best ResponseYou've already chosen the best response.1We've proved that both \(\rm a\) and \(\rm b\) don't belong to the set of natural numbers. QED

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0so i assume that \(a\) is a natural number and show that if it is; \(b\) cannot be a natural

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i like my proof better

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oooh i thought it said \(\forall a,b\in \mathbb{N}\) did it change or can i not see straight?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Basically, 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
 3 years ago
Best ResponseYou've already chosen the best response.1@satellite73 Bikes don't have a good reflector at night. That's why.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it must be past my bed time

Mr.Math
 3 years ago
Best ResponseYou've already chosen the best response.2Or just note that \(3a+4b\ge7\) \( \forall a,b \in \mathbb{N}\). Then the statement is false since \(2<7\).

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1And by the way, that's the linear combination thingy
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