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

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maybe another method
pick \(a=5,b=7\)
negation is \((\exists a,b\in N)(3a+4b\neq 2)\)

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

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