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Prove that sqrt of a prime is irrational

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Homework Statement



If a is a prime number, prove that √a is not a rational number. (You may assume the uniqueness of prime factorization.)

Homework Equations



Per the text: A positive integer a is said to be prime if a > 1 and whenever a is written as the product of two positive integers, a=n*m, one of the factors is a.

[I assume what this means is that if a=n*m, then n=a or m=a.]

The Attempt at a Solution



Proof by contradiction. Assume √a is rational. Then √a = (b/c), where b and c are positive integers with no common factors. So a=(b/c)^2=(b/c)*(b/c). Using the definition of a prime from above, a=(b/c) and a*c=b. Using substitution, a=((a*c)/c)*((a*c)/c), which reduces to a=a*a. Multiply both sides by (1/a) to get 1=a, which contradicts the definition of a prime that a > 1. So √a is not rational.


I have seen other proofs that the square root of a prime is irrational, but they usually end up with a*(c^2)=(b^2) and use the language "we know a divides b^2, and therefore a must divide b," or something similar. I'm trying to stay close to the examples and materials presented so far in my text, and nothing about assuming an integer divides another integer has come up, so I'm not sure if that's something I can assume yet. Is the above proof, where I say a=n*m means a=n or a=m, correct? I don't think I used the uniqueness of prime factorization as mentioned in the question, so I'm worried I'm missing something.

Thank you for reading, and for any insight you may provide.
 

Answers and Replies

  • #2
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Homework Statement



If a is a prime number, prove that √a is not a rational number. (You may assume the uniqueness of prime factorization.)

Homework Equations



Per the text: A positive integer a is said to be prime if a > 1 and whenever a is written as the product of two positive integers, a=n*m, one of the factors is a.

[I assume what this means is that if a=n*m, then n=a or m=a.]

The Attempt at a Solution



Proof by contradiction. Assume √a is rational. Then √a = (b/c), where b and c are positive integers with no common factors. So a=(b/c)^2=(b/c)*(b/c). Using the definition of a prime from above, a=(b/c)
No, this is incorrect. The definition in the text says that for integers n and m it holds that if a=n*m, then a=n or n=m. But when you say that [itex]a=(b/c)(b/c)[/itex], then b/c is not an integer, so you can't apply this.
 
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Thank you for the response micromass. I knew I must have been overlooking something like that; it seemed just a little too easy. I will go back to the drawing board and look through some more examples in my text, and see what I can come up with. I appreciate the quick feedback!
 

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