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

1. UnkleRhaukus

maybe another method

2. satellite73

pick $$a=5,b=7$$

3. satellite73

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

4. ParthKohli

Ohhhh.

5. ParthKohli

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).

6. UnkleRhaukus

oh,

7. ParthKohli

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.

8. ParthKohli

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

9. UnkleRhaukus

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

10. satellite73

i like my proof better

11. satellite73

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

12. ParthKohli

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.

13. ParthKohli

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

14. satellite73

it must be past my bed time

15. ParthKohli

Sure it is.

16. Mr.Math

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

17. ParthKohli

Mr.Math!

18. ParthKohli

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