Use proof by contradiction to show there is no rational number such that....

[r0bHadz]

Homework Statement

Use proof by contradiction to show there is no rational number r for which r^3+r+1 = 0

Homework Equations

The Attempt at a Solution

Assume there is a rational number r for which r^3+r+1=0. Then r = (a/b) with a,b ∈ℤ and b ≠ 0, and a/b is in lowest terms

Then a/b is a root, => (a/b)^3 + (a/b) + 1 = 0

=> $\frac {a^3}{b^3} + (a/b) + 1 = 0 => a^3 + ab^2 + b^3 = 0$

since a/b is in lowest terms, one must be odd, the other even. Let's examine a odd b even

let any letter variable you see from here on be ∈ ℤ

$(2p+1)^3 + (2p+1)(2f)^2 + (2f)^3 = 0 => [8p^3 + 12p^2 + 6p + 1] + [8f^2p + 4f^2] + 8f^3 = 0 => 8p^3+8f^3 + 8f^2p + 12p^2 + 4f^2 + 6p + 1 = 0 => 2[4p^3 + 3f^3 + 4f^2p + 6p^2 + 2f^2 + 3p] + 1 = 0$

here we have the contradiction that 0 is an odd number.

now a even b odd:

$a^3 + b^3 + b^2a = 0 => (2y)^3 + (2x+1)^3 + (2x+1)^2(2y) = 0 => 8y^3 + [8x^3 + 12x^2 + 6x + 1] + [8x^2y + 8xy + 2y] = 0 => 8y^3 + 8x^3 + 8x^2y + 12x^2 + 8xy + 6x + 2y + 1 = 0 2[4y^3 + 4x^3 + 4x^2y + 6x^2 + 4xy + 3x + y] + 1 = 0$

as we can see, again zero is not an odd number.

So there is a contradiction. Since a/b is in lowest form, a/b cannot be a root.
I've been having problems with proof by contradictions guys. Every other proof method that I've learned so far is really easy for me but this one always kicks my butt. Does my proof seem okay to y'all?

 

[SammyS]

Homework Statement

Use proof by contradiction to show there is no rational number r for which r^3+r+1 = 0

Homework Equations

The Attempt at a Solution

Assume there is a rational number r for which r^3+r+1=0. Then r = (a/b) with a,b ∈ℤ and b ≠ 0, and a/b is in lowest terms

Then a/b is a root, => (a/b)^3 + (a/b) + 1 = 0

=> $\frac {a^3}{b^3} + (a/b) + 1 = 0 => a^3 + ab^2 + b^3 = 0$

since a/b is in lowest terms, one must be odd, the other even. Let's examine a odd b even

This is a false assumption. All that is necessary for $\dfrac{a}{b}$ to be in lowest terms is that $a$ and $b$ have no prime factors in common, neither is required to be even.

For example: $\dfrac 3 5$ and $\dfrac{35}{33}$ are in lowest terms

I've been having problems with proof by contradictions guys. Every other proof method that I've learned so far is really easy for me but this one always kicks my butt. Does my proof seem okay to y'all?

As to the issue of a proper proof by contradiction, yes, you have correct idea for this proof in this regard.

Note:

In a previous thread of yours, https://www.physicsforums.com/threads/is-my-proof-by-contradiction-valid.964666/, you were to prove:

If 3n+2 is odd, then n is odd.​

Restating that in the form of the proof you are attempting in the current thread you could say:

Prove: There is no even number, $n$ , for which $3n+2$ is odd.​

.
.

 

[Ray Vickson]

Homework Statement

Use proof by contradiction to show there is no rational number r for which r^3+r+1 = 0

Homework Equations

The Attempt at a Solution

Assume there is a rational number r for which r^3+r+1=0. Then r = (a/b) with a,b ∈ℤ and b ≠ 0, and a/b is in lowest terms

Then a/b is a root, => (a/b)^3 + (a/b) + 1 = 0

=> $\frac {a^3}{b^3} + (a/b) + 1 = 0 => a^3 + ab^2 + b^3 = 0$

since a/b is in lowest terms, one must be odd, the other even. Let's examine a odd b even

******************************************

I've been having problems with proof by contradictions guys. Every other proof method that I've learned so far is really easy for me but this one always kicks my butt. Does my proof seem okay to y'all?

As others have pointed out, you cannot assume that one of $p,q$ must be even and the other odd.

Anyway, look at your equation $a^3 + a b^2 + b^3 = 0,$ which can be re-written as
$$ a^3 = -b^3 - a b^2 \hspace{3ex} (1) $$ The right-hand-side of (1) is divisible by $b$, so $a^3$ is divisible by $b$. If you can show that this implies $a$ is divisible by $b$, you will have the contradiction you want.

 

[r0bHadz]

Whoops sorry guys good catch. If I change it to: for p/q, p and q can be either odd and odd, or even and odd, but not even and even,and then show the case when they're both odd:

$a^3+ ab^2 + b^3 = 0 => [8x^3 + 12x^2 + 6x + 1] + [8p^2x + 4p^2 + 8px + 4p + 2x + 1] + [8p^3 + 12p^2 +6p +1] = 0 => 8x^3 + 8p^3 + 8p^2x + 12x^2 + 16p^2 + 8px + 6x + 10p + 2x + 2 + 1 = 0 => 2[4x^3 + 4p^3 + 4p^2x + 6x^2 + 8p^2 + 4px + 3x + 5p + x + 1] + 1 = 0$and we see since 0 is of form 2u and the form of the left side is odd form, there is a contradiction.Does this complete my proof? (taking into account that p/q is assumed every combination besides even and even)

Ray I appreciate the input but I am already too deep into this problem to change what I am doing now xD thank you tho

 

[Mark44]

If I change it to: for p/q, p and q can be either odd and odd, or even and odd, but not even and even,

Are p and q the numerator and denominator, respectively, of the rational number, in reduced form? You started out with a/b, so I'm not sure where p and q came from.

If p and q are both odd, there's no guarantee that the fraction p/q will be in reduced form -- for example, 3/9 meets the condition that both numbers are odd, but it is not in lowest terms.

 

[SammyS]

@r0bHadz : As Mark44 indicated, it's confusing to suddenly switch from a/b to p/q .

Whoops sorry guys good catch. If I change it to: for p/q, p and q can be either odd and odd, or even and odd, but not even and even,
...

Does this complete my proof? (taking into account that p/q is assumed every combination besides even and even)

Ray, I appreciate the input but I'm already too deep into this problem to change what I'm doing now x'D thank you tho

Looks like now you have sketched out a proof in some detail. Normally at this point you would review what you have and see if you can possibly make it more compact and more cohesive.

Your proof relies on the even-ness and/or odd-ness of a & b (or p & q), so consider the following.

A couple of lemmas addressing the combining of EVEN:ODD integers could aid your proof a lot.

Adding Two Integers:

Case 1: Sum of EVEN and ODD gives ODD.
Case2: Otherwise, sum of two integers id EVEN​

Multiplying Two Integers:

Case 1: One of them is EVEN (doesn't matter about the other): The product is EVEN.
Case 2: Both are ODD: The product is ODD.​

Basic proof for each of these, then quite simple to work with $a^3 +a ~ b^2 + b^3$.

 

[r0bHadz]

Are p and q the numerator and denominator, respectively, of the rational number, in reduced form? You started out with a/b, so I'm not sure where p and q came from.

If p and q are both odd, there's no guarantee that the fraction p/q will be in reduced form -- for example, 3/9 meets the condition that both numbers are odd, but it is not in lowest terms.

Yes sorry for changing it. I meant a and b for p and q. Man I'm all over the place, I don't know how you guys have enough time in the day to actually learn anything -_-

Anyways, I started the proof off with "assume" a/b is in lowest form, so for a odd, b odd, a/b is in lowest form, according to the hypothesis.

 

[PeroK]

Yes sorry for changing it. I meant a and b for p and q. Man I'm all over the place, I don't know how you guys have enough time in the day to actually learn anything -_-

Anyways, I started the proof off with "assume" a/b is in lowest form, so for a odd, b odd, a/b is in lowest form, according to the hypothesis.

The concept you are not using here is that of prime factors. You have dug yourself into a deep hole by focusing on oddness and evenness.

 

[r0bHadz]

The concept you are not using here is that of prime factors. You have dug yourself into a deep hole by focusing on oddness and evenness.

I will be learning more about primes very soon then. It seems like primes are pretty damn important in math so I guess I have no choice. But as my proof is now, after the extra posts, is it valid? I understand its not the best but I'm just worried about the validity right now.

 

[Ray Vickson]

The concept you are not using here is that of prime factors. You have dug yourself into a deep hole by focusing on oddness and evenness.

I think what he did was perfectly OK: if $a$ and $b$ are relatively prime then it is a fact that either one is even and the other is odd, or else both are odd. His post #4 covered the latter case, and his post #1 covered the first two.

 

[WWGD]

You may also consider general conditions in https://en.wikipedia.org/wiki/Rational_root_theorem

 

[SammyS]

@r0bHadz
A couple of lemmas addressing the combining of EVEN:ODD integers could aid your proof a lot.
$\dots$

a Basic proof for each of these, then quite simple to work with $a^3 +a ~ b^2 + b^3$.

Rather than mess around with that suggestion of mine:
A similar result can be easily accomplished using modular arithmetic .

Examine the above equation mod 2, for (a , b): (ODD, EVEN) : (EVEN,ODD) : (ODD, ODD)

 

[r0bHadz]

you guys are the best <3