Proof, quadratic equation and coefficients

In summary, the problem states that given that a, b, and c are odd integers, the equation ax2 + bx + c = 0 cannot have a rational root. This is proven using a "proof by contradiction" approach, showing that if the conclusion is false, then it leads to a contradiction. Therefore, the conclusion must be correct.
  • #1
SpartanG345
70
1
http://www.cs.odu.edu/~toida/nerzic/content/problem_solving/problem_solving.html
example 3

========================================= Quote
Problem: Given that a, b, and c are odd integers, prove that equation ax2 + bx + c = 0 can not have a rational root.

Understanding the Problem: This is a "prove" type problem.
The hypothesis is that a, b, and c are odd integers, and the conclusion is that equation ax2 + bx + c = 0 can not have a rational root.
The hypothesis is straightforward. In the conclusion, "rational root" means a root, that is, the value of x that satisfies the equation, and that can be expressed as m/n, where m and n are integers. So the conclusion means that there is no number of the form m/n that satisfies the equation under the hypothesis.

Devising a Solution Plan: For this problem, let us try "proof by contradiction". When you are asked to prove the impossibility of an event or non-existence of certain things, this approach often is quite helpful.

Following the "proof by contradiction", let us assume that the conclusion is false, that is the equation ax2 + bx + c = 0 has a rational root m/n, where m and n are integers, when a, b, and c are odd integers. We can assume without loss of generality that m and n do not have any factors in common. Then
a(m/n)2 + b(m/n) + c = 0 . ------------------------ (1)

Let us try to derive a contradiction from this.
First let us make this equation simpler, that is, let us get rid of fractions.
Since n is not equal to 0, multiplying the both sides of (1) by n2, we get
am2 + bmn + cn2 = 0 . ------------------------ (2)

Since m is an integer, it is either even or odd. We are going to consider those cases one by one. That is "divide into cases".
Let us first consider the case when m is even.
Then n is odd, since otherwise m and n have a common factor 2. Now am2 + bmn is even, and cn2 is odd. Hence am2 + bmn + cn2 can not be 0.

Next let us consider the case when m is odd.
By an argument similar to the previous case, we can see that n is also odd.

If m and n are odd, then am2, bmn, and cn2 are all odd, since a, b, and c are odd integers. However, the sum of three odd numbers can not be equal to 0.

Thus by assuming that the conclusion is false, we have arrived at a contradiction, that is m/n does not satisfy the equation. Hence our assumption must be wrong, and therefore the conclusion is correct.

========================================================== End quote
I found this problem online, however i do not understand the last part, the one in bold. Would the last statement result in loss of generality.

Eg how would you get 3/2 this fits in the second case where m is odd. However in either case n is odd. so it is impossible to get 3/2...? If n is any even number then there can be common factors ie 9/6.

Any ideas?
 
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  • #2
Easy way: m even -> n odd, can't be. n even -> m odd, can't be. Both odd, can't be. Both even, not allowed.
 

1. What is a proof in mathematics?

A proof is a logical argument that shows the validity of a mathematical statement or theorem. It involves using previously established facts and rules to demonstrate that a statement is true.

2. How do you solve a quadratic equation?

A quadratic equation is an equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. To solve a quadratic equation, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. You can also solve it by factoring the equation or completing the square.

3. What are coefficients in a quadratic equation?

Coefficients are the numerical values that are multiplied by the variable in a quadratic equation. In the standard form, ax^2 + bx + c = 0, a is the coefficient of the squared term, b is the coefficient of the linear term, and c is the constant term.

4. How can you tell if a quadratic equation has real solutions?

A quadratic equation will have real solutions if the discriminant, b^2 - 4ac, is greater than or equal to 0. If the discriminant is equal to 0, the equation will have one real solution, and if it is greater than 0, the equation will have two distinct real solutions.

5. Can a quadratic equation have imaginary solutions?

Yes, a quadratic equation can have imaginary solutions if the discriminant, b^2 - 4ac, is negative. In this case, the solutions will be complex numbers of the form a + bi, where i is the imaginary unit (√-1).

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