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

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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
maybe another method
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
pick \(a=5,b=7\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
negation is \((\exists a,b\in N)(3a+4b\neq 2)\)
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
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).
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

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

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i like my proof better
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
oooh i thought it said \(\forall a,b\in \mathbb{N}\) did it change or can i not see straight?
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
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.
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
@satellite73 Bikes don't have a good reflector at night. That's why.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
it must be past my bed time
 2 years ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Sure it is.
 2 years ago

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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Mr.Math!
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
And by the way, that's the linear combination thingy
 one year ago
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