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UnkleRhaukus
Group Title
how can i disprove the following statement
\[(\exists a ,b \in \mathbb N)(3a + 4b = 2)\]
 one year ago
 one year ago
UnkleRhaukus Group Title
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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UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
maybe another method
 one year ago

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

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i like my proof better
 one year 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?
 one year 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.
 one year 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.
 one year ago

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

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Sure it is.
 one year 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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