Proof by contradiction - polynomials and infinite primes

In summary, the conversation discusses two questions that involve proving by contradiction. The first question involves showing that -1 is not a root of a quadratic function with integer coefficients, while the second question involves proving the infinitude of prime numbers by finding a contradiction in the assumption that there are only a finite number of primes. The conversation also includes a discussion about the definition of a root and the meaning of "dividing" in relation to prime numbers.
  • #1
bloynoys
25
0

Homework Statement



Two Questions:

1. Prove, by contradiction, that if a and b are integers and b is odd,, then -1 is not a root of f(x)= ax^2+bx+a.

2. Prove, by contradiction, that there are infinitely many primes as follows. Assume that there only finite primes. Let P be the largest prime. Explain why there is a prime dividing P!+1 and find the the contradiction.

Homework Equations



For both, assume the contradiction work towards finding it is impossible.

The Attempt at a Solution



1. (x-1)(x-a)=ax^2+bx+a
Not sure where to go from there

2. This is not a normal infinite prime solutions as we have gone over a few of the solutions in class. I am not sure what she means by "there is a prime dividing P!+1" as that isn't really a clear sentence. Any Ideas?
 
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  • #2
bloynoys said:

Homework Statement



Two Questions:

1. Prove, by contradiction, that if a and b are integers and b is odd,, then -1 is not a root of f(x)= ax^2+bx+a.

2. Prove, by contradiction, that there are infinitely many primes as follows. Assume that there only finite primes. Let P be the largest prime. Explain why there is a prime dividing P!+1 and find the the contradiction.

Homework Equations



For both, assume the contradiction work towards finding it is impossible.

The Attempt at a Solution



1. (x-1)(x-a)=ax^2+bx+a
Not sure where to go from there

What does it mean for -1 to be a root of f?? What is the definition??

2. This is not a normal infinite prime solutions as we have gone over a few of the solutions in class. I am not sure what she means by "there is a prime dividing P!+1" as that isn't really a clear sentence. Any Ideas?

What's wrong with that sentence?? "There is a prime number that divides P!+1" is a perfectly good sentence...
 
  • #3
Maybe I am just confused to how to relate the P!+1 to P. Is it P/(P!+1)? Or (P!+1)/P and that just eliminates the "largest prime?" So it is the second largest prime plus one?

And for the first one, a root means that it is one of the solutions to an equation. I realize this looks dumb as this is an easyish problem, but we are working on this in a high level undergrad math course. I am just drawing a blank on the intermediate steps to set these problems up. Oh and I believe I made a mistake. It should read:

(x+1)(x+a)=ax^2+bx+a correct?
 
  • #4
bloynoys said:
Maybe I am just confused to how to relate the P!+1 to P. Is it P/(P!+1)? Or (P!+1)/P and that just eliminates the "largest prime?" So it is the second largest prime plus one?

You're making this far too difficult. For example, let P=3, then P!+1=7. So a prime that divides P!+1 is 7.

In general, there is always a prime that divides P!+1 for any P.

And for the first one, a root means that it is one of the solutions to an equation. I realize this looks dumb as this is an easyish problem, but we are working on this in a high level undergrad math course. I am just drawing a blank on the intermediate steps to set these problems up. Oh and I believe I made a mistake. It should read:

(x+1)(x+a)=ax^2+bx+a correct?

-1 being a root means that f(-1)=0 right? Now write out the definition of f and see what it gives you.
 
  • #5
Wow, holy cow, that was a whole lot easier than I thought for the first one. I just had my head focused in on a multiple root function and didn't think about the f(x)=0. Thank you for that!

And wow on the second one as well. She basically spelled out the definition of prime numbers, and then added one. It was the word dividing that had me confused. I knew they weren't as hard as I was making them but wow, I really shouldn't overlook stuff like that.

Thank you for all your help!
 

1. What is a proof by contradiction?

A proof by contradiction is a mathematical proof technique in which one assumes the opposite of what is being proven and then shows that this assumption leads to a contradiction. This contradiction then proves that the original statement must be true.

2. How is proof by contradiction used in polynomials?

In polynomials, proof by contradiction is often used to show that a certain polynomial does not have a certain root. For example, if we assume that a polynomial has a root at a certain value, we can then manipulate the polynomial algebraically to show that this assumption leads to a contradiction, thus proving that the polynomial does not have that root.

3. Can proof by contradiction be used to prove that there are infinite primes?

Yes, proof by contradiction is commonly used to prove that there are infinite primes. This proof, known as Euclid's proof, assumes that there are only a finite number of primes and then shows that this leads to a contradiction, thus proving that there must be infinite primes.

4. What is the relationship between proof by contradiction and infinity?

Proof by contradiction is often used to prove statements about infinity, such as the infinite number of primes. By assuming the opposite of what is being proven and showing that this leads to a contradiction, we are able to prove that the original statement is true, even if it involves infinity.

5. Are there any limitations to using proof by contradiction?

While proof by contradiction is a powerful proof technique, it is not always applicable. In some cases, the contradiction may not be obvious or easy to reach, making it difficult to use this method. Additionally, some mathematical systems may not allow for the use of proof by contradiction. In these cases, alternative proof techniques must be used.

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