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Proof must be integer or irrational?

  • Thread starter hackboiz29
  • Start date
  • #1

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? :/
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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955

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? :/
 
  • #3
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 :/
 
  • #4
1,953
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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 .....
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
41,833
955
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.
 
  • #6
My teacher told us to use that because we haven't really learned a lot yet. Thanks for your help! :)
 

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