Proof of quadratic irrationals - stuck

• aarciga
In summary, the conversation discusses the task of proving that a quadratic equation with odd and even coefficients and a non-divisible by 4 constant term has no rational solutions. The conversation explores different methods and approaches for proving this, including showing that the discriminant cannot be a perfect square and that the square root of a non-square number is irrational. The conversation also considers the possibility of using the irrationality of √2 to prove the irrationality of √2m.
aarciga

Homework Statement

I have to show a proof that if a is odd, b is even, and c is even but not divisible by 4.
a,b,c are int coefficients

ax2 + bx + c = 0

has no rational solutions

Homework Equations

all letters here are integers.

So i have,

a = 2d + 1
b = 2f
c = 2g, but c is not divisable by for so, c $$\neq$$ 4h
so, g $$\neq 2h$$, or g is not even. g is odd
c = 2(2j + 1)

The Attempt at a Solution

Im trying to show that the discriminant in the quadratic formula can't be a square, but I am having trouble showing that.

or that b2-4ac can't be a square number

$$\sqrt{4f^{2}-4(2d+1)(2(2j+1))}$$

(2d+1)(2j+1) will give some odd number 2k+1

$$\sqrt{4[f^{2}-2(2k+1)]}$$

2(2k+1) will give an even number 2m

$$\sqrt{4[f^{2}-2m]}$$

$$\sqrt{f^{2}-2m}^{2}$$ can't = some square $$\frac{S^{2}}{2^{2}}$$

$$f^{2}-2m = \frac{S^{2}}{4}$$

After I show that it isn't a square, i can show that the square root of a non square is irrational. But I feel like i might be off base on this. It seems like the discrimininant certainly could be a perfect square.

it looks like I am trying to show that a perfect square can't equal a perfect square divided by 4. which is false. ( i.e. 16/4 = 4)

so, any ideas to point me in more proper direction, or to make a point on what i already have would be greatly appreciated.

Last edited:
Another way of thinking about it??

$$\sqrt{b^{2}-4ac}$$

$$\sqrt{(2f)^{2}-4(2d+1)(2(2j+1))}$$

2f is even, so (2f)2 is even
and 8(2d+1)(2j+1) is even x (odd x odd). (odd x odd) is odd so then even x odd = even

$$\sqrt{even number -even number}$$
even number - even number = even. = 2m

so i must show
$$\sqrt{2m}$$ can't be rational, or

$$\sqrt{2m}$$ $$\neq$$ some rational number $$\frac{P}{Q}$$

and from here, I can prove $$\sqrt{2}$$ is irrational.

and assuming $$\sqrt{m}$$ is rational, an irrational time a rational can't be rational??

$$\sqrt{2}\sqrt{m} \neq \frac{P}{Q}$$

1. What is the definition of a quadratic irrational number?

A quadratic irrational number is a real number that cannot be expressed as a ratio of two integers and is the solution to a quadratic equation with rational coefficients.

2. How can we prove that a number is a quadratic irrational?

There are several methods for proving that a number is a quadratic irrational, including the continued fraction method, the quadratic surd method, and the abstract algebraic method.

3. Can all irrational numbers be classified as quadratic irrationals?

No, not all irrational numbers are quadratic irrationals. For example, pi and the square root of 2 are irrational but not quadratic irrationals.

4. Is there a specific form for expressing quadratic irrationals?

Yes, quadratic irrationals can be expressed in the form of a continued fraction, where the coefficients are integers and the denominators are positive integers.

5. What are some real-world applications of quadratic irrationals?

Quadratic irrationals have various applications in fields such as physics, engineering, and computer science. For example, they are used in the design of electronic circuits and in the calculation of trajectories in projectile motion.

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