Question

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

Answer 1

maybe another method

Answer 2

pick \(a=5,b=7\)

Answer 3

negation is \((\exists a,b\in N)(3a+4b\neq 2)\)

Answer 4

Ohhhh.

Answer 5

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).

Answer 6

oh,

Answer 7

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.

Answer 8

We've proved that both \(\rm a\) and \(\rm b\) don't belong to the set of natural numbers.
QED

Answer 9

so i assume that \(a\) is a natural number and show that if it is; \(b\) cannot be a natural

Answer 10

i like my proof better

Answer 11

oooh i thought it said \(\forall a,b\in \mathbb{N}\) did it change or can i not see straight?

Answer 12

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.

Answer 13

@satellite73 Bikes don't have a good reflector at night.
That's why.

Answer 14

it must be past my bed time

Answer 15

Sure it is.

Answer 16

Or just note that \(3a+4b\ge7\) \( \forall a,b \in \mathbb{N}\). Then the statement is false since \(2<7\).

Answer 17

Mr.Math!

Answer 18

And by the way, that's the linear combination thingy