Question

Assume sqrt(5) is irrational. By using the definition of a rational number, give a proof ny contradiction that for any rational number q, the expression 3q+sqrt(5) is irrational

Answer 1

WHALECOME TO OPENSTUDY

Answer 2

thanks

Answer 3

lol

Answer 4

please, what is q?

Answer 5

@addishuns

Answer 6

i believe it is just any random number, the question does not.

Answer 7

say what q is*

Answer 8

I put the whole question, maybe that will help

Answer 9

@addishuns Can I suppose q=1?

Answer 10

i believe so

Answer 11

No, because then you just showed that it is true for q=1. @Michele_Laino

You need to show that \(3q+\sqrt5\) is irrational for all \(q\in\mathbb{Q}\) (all rational).

Answer 12

so q=a/b where a,b are integers (not equal to zero)
\[\frac{3a}{b}+\sqrt{5}\]
Now since you are asking for a proof by contradiction
assume \[\frac{3a}{b}+\sqrt{5}=\frac{m}{n}\]
where m and n are integers (n not equal to zero)
solve for sqrt(5)
and see if you can conclude anything from the equation

Answer 13

\[\sqrt{5} = mb - 3an / bn\]

Answer 14

Right, what do you noticed about \(\dfrac{mb-3an}{bn}\)?

*cough* it is rational *cough*

Answer 15

Of course, so it is a contradiction and proves the original statement.

Answer 16

nice job guys