Proof must be integer or irrational?

[hackboiz29]

Homework Statement

Suppose a, b ε Z. Prove that any solution to the equation x^3 +ax+b = 0 must either be an integer, or else be irrational.

Homework Equations

Not sure if this is right but x = m / n where m divides b and n divides 1

The Attempt at a Solution

So far i think i got x = m / n where m divides b and n divides 1 but i don't know where to go from there and i am kind of not sure if that's even right. There's probably some way to factor this right? I am trying to do this by contradiction so i assume that x = a/b where a,b belong to N


One More problem...

Homework Statement

Prove that log2n : n ε N consists entirely of integers and irrational numbers. (it's base 2 n not log(2n))

Homework Equations

log2n = a / b. Trying to do a proof by contradiction so i started off by assuming log2n = a / b where a,b E N.

The Attempt at a Solution

As i said I'm doing a proof by contradiction. So far i have come up with log2n = a / b. I don't really know where to go from here. Rearranging the equation doesn't really help. Maybe use some laws for logarithms to write them differently? :/

 

[HallsofIvy]

Homework Statement

Suppose a, b ε Z. Prove that any solution to the equation x^3 +ax+b = 0 must either be an integer, or else be irrational.

Homework Equations

Not sure if this is right but x = m / n where m divides b and n divides 1

IF you allowed to use this, then surely you see that the only integers that divide 1 are 1 and -1! So if x is rational it is either m/1 or m/-1 for some integer m.

The Attempt at a Solution

So far i think i got x = m / n where m divides b and n divides 1 but i don't know where to go from there and i am kind of not sure if that's even right. There's probably some way to factor this right? I am trying to do this by contradiction so i assume that x = a/b where a,b belong to N


One More problem...

Homework Statement

Prove that log2n : n ε N consists entirely of integers and irrational numbers. (it's base 2 n not log(2n))

Homework Equations

log2n = a / b. Trying to do a proof by contradiction so i started off by assuming log2n = a / b where a,b E N.

The Attempt at a Solution

As i said I'm doing a proof by contradiction. So far i have come up with log2n = a / b. I don't really know where to go from here. Rearranging the equation doesn't really help. Maybe use some laws for logarithms to write them differently? :/

 

[hackboiz29]

Oh... wow okay so x could = m or -m both of which are integers. That makes sense. I don't understand how to prove that any other solution would be irrational :/

 

[willem2]

Oh... wow okay so x could = m or -m both of which are integers. That makes sense. I don't understand how to prove that any other solution would be irrational :/

what you used is called the rational root theorem, giving all possible values of rational roots, so any other roots must be ...

 

[HallsofIvy]

That proves that any rational root must be an integer. Of course, any other roots, that is, any root that is not rational, is irrational, by definition.

But I still wonder if you are allowed to use that. It seems too easy.

 

[hackboiz29]

My teacher told us to use that because we haven't really learned a lot yet. Thanks for your help! :)