Discrete Proof

[Startbucks]

For some reason this proof is irritating me.

Prove the following:

For all integers $\displaystyle b$,$\displaystyle c$ prove that if $\displaystyle r$ is a rational sol'n of $\displaystyle x^2 + bx + c = 0$, then $\displaystyle r$ is an integer. HINT: Use quad. formula $\displaystyle \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

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So... we know r = a/b ... that's a start right?! x = quad. form. We know the numerator has to be rational... and, the numerator has to be even if the roots are going to be integers... then im stuck.

 

[daon]

Can you use the rational zeros thm? Rational solutions will be of the form p/q where p|c and q|1. Thus q = +/-1.

 

[Startbucks]

Here's my attempt:

We know if $\displaystyle r$ is a rational root, then that would have to be a factor +/- of $\displaystyle c$. Since $\displaystyle c$ is an integer, this would make it so that $\displaystyle r$ has to be an integer. Thus, we have shown it is a rational.

The roots $\displaystyle r_1$ and $\displaystyle r_2$ are given by: $\displaystyle r_{1,2} = \frac {-b \pm \sqrt{b^2 - 4c}}2$For this to be rational, we need $\displaystyle -b \pm \sqrt{b^2 - 4c} = k$ where $\displaystyle k$ is an integer

So now I need to show $\displaystyle k$ is even which will then mean dividing $\displaystyle k$ by $\displaystyle 2$ (hence getting $\displaystyle r_{1,2}$) will give an integer which will conclude the proof. How do I show it's even though?

Thanks for the help daon.

 

[daon]

If you;re using the rational zero's thm, you're done before you started. Since q=+/- 1. r = +/- p where p|c.

Alternatively where you left off, you can break it into whether or not s=b^2 +/- 4c is even or odd.

If s (being a perfect square) is even then b is even, and sqrt(s) is also. Then you have b+/-sqrt(s) is even and hence r is an integer.

If s is odd then b is also odd (since b^2 +/- 4c is odd => b^2 is odd => b is odd). Thus sqrt(s) is odd. b+/-sqrt(s) is the sum/difference of 2 odds which is even. Hence r is an integer.

edit: You may need lemmas to show that the parity of m^2 and m are the same for all positive integers m.

 

[PAULK]

Here's my attempt:

We know if $\displaystyle r$ is a rational root, then that would have to be a factor +/- of $\displaystyle c$. Since $\displaystyle c$ is an integer, this would make it so that $\displaystyle r$ has to be an integer. Thus, we have shown it is a rational.

The roots $\displaystyle r_1$ and $\displaystyle r_2$ are given by: $\displaystyle r_{1,2} = \frac {-b \pm \sqrt{b^2 - 4c}}2$For this to be rational, we need $\displaystyle -b \pm \sqrt{b^2 - 4c} = k$ where $\displaystyle k$ is an integer

So now I need to show $\displaystyle k$ is even which will then mean dividing $\displaystyle k$ by $\displaystyle 2$ (hence getting $\displaystyle r_{1,2}$) will give an integer which will conclude the proof. How do I show it's even though?

Thanks for the help daon.

If

x^2 + bx + c = 0

....- b +- sqrt(b^2 - 4c)
x = ----------------------
..............2

If the sqrt(...) is an integer, the top must be even. Why?

b^2 has the same parity as b.
b^2 - 4c has the same parity as b.
sqrt(b^2 - 4c) has the same parity as b.
- b has the same parity as b.

So the top is a sum-difference of two integers of like parity. Take it from there.

 

[pka]

Here is totally different approach to this question.
In order to have a rational root at all, $\displaystyle b^2-4c=k^2$ for some positive integer $\displaystyle k$.
Otherwise, $\displaystyle \sqrt{b^2-4c}$ is irrational.
Moreover that means $\displaystyle 4c=b^2-k^2=(b-k)(b+k)$ so the factors are both even.
Note that the solution set is $\displaystyle \left\{ {\frac{{ - b + k}}{2},\frac{{ - b - k}}{2}} \right\}$ both integers.