I don't understand why this polynomial can't have rational but not integer solutions
#Clear a small doubt about this problem
hardy pollen
gaunt cobaltBOT
I don't understand why this polynomial can't have rational but not integer solut...
- Do not ping the Moderators, unless someone is breaking the rules.
- Do not ping the Helper Moderators, unless there is a conflict between helpers.
- Do not ping other members randomly for help.
- Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
- Wait patiently for a helper to come along.
- If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:
+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:
tepid cape
I don't understand why this polynomial can't have rational but not integer solut...
It actually seems false?
Suppose the a's all had a common factor of q^n, for instance.
Then it would be trivial to construct from that a polynomial with integer coefficients and rational roots.
hardy pollen
It actually seems false?
Are you saying that it seems false or asking me
hardy pollen
Then it would be trivial to construct from that a polynomial with integer coeffi...
I also though this, what if I construct a polynomial like
(x-5/6)(x+3/4)
tepid cape
I also though this, what if I construct a polynomial like
(x-5/6)(x+3/4)
That doesn't have integer coefficients.
hardy pollen
That doesn't have integer coefficients.
Find the solutions of 12x^2-28x+15
low osprey
so the claim is if a root is rational, then it must be an integer?
hardy pollen
so the claim is if a root is rational, then it must be an integer?
Ye
hardy pollen
Find the solutions of 12x^2-28x+15
But if you solve this you get 5/6 and 3/2
tepid cape
Find the solutions of 12x^2-28x+15
That doesn't have lead coefficient 1.
hardy pollen
That doesn't have lead coefficient 1.
Oh💀
I was feeling like I was missing something but didn't understand what it was
tepid cape
No, I see it clearly now, and it's a true statement.
Like I said, it's a corollary to the rational root theorem.
Did you already do Problem 16?
hardy pollen
Did you already do Problem 16?
Yeah, problem 16 is pretty much basic number theory
tepid cape
Show me.
hardy pollen
tepid cape
Ah. The fundamental theorem of arithmetic.
Okay, so then the natural extension to algebra would seem to me to be to consider the polynomial as a product of linear terms.
Times some irreducible polynomial.
Perhaps.
Though per the fundamental theorem of algebra there's no such thing as an irreducible polynomial in complex numbers.
low osprey
ah I also see it now, it's a consequence of Gauss lemma for polynomials
hardy pollen
Though per the fundamental theorem of algebra there's no such thing as an irredu...
Is it an abstract algebra concept
tepid cape
ah I also see it now, it's a consequence of Gauss lemma for polynomials
Or, y'know, the rational root theorem. Like I said.
tepid cape
Is it an abstract algebra concept
No. The fundamental theorem of algebra states that a polynomial of degree n with complex coefficients has exactly n complex roots up to multiplicity.
placid trenchBOT
@hardy pollen
hardy pollen
+close
placid trenchBOT
Thank you for your feedback! Techie Literate has been awarded 1 
. They now have 1054 . They have 2 daily left for today.
placid trenchBOT
Thank you for your feedback! aL has been awarded 1 . They now have 922 . They have 2 daily left for today.