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How can I prove that the square root of a prime is irrational?
Preferably through contradiction. So far, I have:
Assume the square root of some prime p is rational
Therefore, square root of p can be represented by the division of two non-canceling integers, a and b.
If we square both sides, we get p is equal to a squared over b squared.
....
Where do I go from here?
Assume the square root of some prime p is rational
Therefore, square root of p can be represented by the division of two non-canceling integers, a and b.
If we square both sides, we get p is equal to a squared over b squared.
....
Where do I go from here?
4 Answers
First, assume that there exists some prime number 'p' whose square root is a rational numer 'r'.
√p = r
Now square both sides:
p = r²
Since 'r' is a rational number, it can be represented as a fraction 'a/b', in which 'a' and 'b' are both positive, prime integers:
p = (a/b)² = (a/b)*(a/b)
*** You can clearly see that (a/b)² = (a/b)*(a/b) = p. But this means that 'p' isn't prime, because it has 'a/b' (an integer) as a divisor, so we have a contradiction of the given fact that 'p' is prime.
The girl below me literally copied her proof from this website:
http://www.ever…
√p = r
Now square both sides:
p = r²
Since 'r' is a rational number, it can be represented as a fraction 'a/b', in which 'a' and 'b' are both positive, prime integers:
p = (a/b)² = (a/b)*(a/b)
*** You can clearly see that (a/b)² = (a/b)*(a/b) = p. But this means that 'p' isn't prime, because it has 'a/b' (an integer) as a divisor, so we have a contradiction of the given fact that 'p' is prime.
The girl below me literally copied her proof from this website:
http://www.ever…
Let p be any prime number.
Assume √p is a rational number. √p can therefore be written as a fraction, a/b where a and b are coprime integers. (That it can be written as a fraction comes from the definition of rational but to choose a and b to be coprime we require the fundamental theorem of arithmetic. If you don't know a proof of this I suggest you read the node for the sake of thoroughness.)
√p= a/b
b×√p=a
Ok next stage (this isn't related to the first bit.) Take the highest integer lower than √p and call this number c.(e.g. √5 is approximately 2.236 so c would be 2.)
b×(√p-c)=b×√p-b×c
Now b×√p is an integer and b×c is an integer so the result, let's call it d, must also be an integer. The next step is to multiply this result by √p
b×√p-b×c=d
d×√p=b×p- b×c×√p
Now b×p is an integer and c×(b×√p) is also an integer. Therefore d×√p is an integer.
d is less than b. {d=b(√p-c} but in choosing and a and b to be comprime we ensured that b was the smallest interger which when multiplied by √p gave an interger.
Voila the contradiction!
This proof not only covers primes but extendeds to all intergers with nonintergal square roots. (If p is a perfect square then (√p - c) is zero and the rest of the proof goes down the tube, unsurprisingly.)
Assume √p is a rational number. √p can therefore be written as a fraction, a/b where a and b are coprime integers. (That it can be written as a fraction comes from the definition of rational but to choose a and b to be coprime we require the fundamental theorem of arithmetic. If you don't know a proof of this I suggest you read the node for the sake of thoroughness.)
√p= a/b
b×√p=a
Ok next stage (this isn't related to the first bit.) Take the highest integer lower than √p and call this number c.(e.g. √5 is approximately 2.236 so c would be 2.)
b×(√p-c)=b×√p-b×c
Now b×√p is an integer and b×c is an integer so the result, let's call it d, must also be an integer. The next step is to multiply this result by √p
b×√p-b×c=d
d×√p=b×p- b×c×√p
Now b×p is an integer and c×(b×√p) is also an integer. Therefore d×√p is an integer.
d is less than b. {d=b(√p-c} but in choosing and a and b to be comprime we ensured that b was the smallest interger which when multiplied by √p gave an interger.
Voila the contradiction!
This proof not only covers primes but extendeds to all intergers with nonintergal square roots. (If p is a perfect square then (√p - c) is zero and the rest of the proof goes down the tube, unsurprisingly.)
Same idea but a little different nonetheless
Assum sqrt(p) is rational for some prime p. Then we can write sqrt(p) = a/b for relatively prime positive integers a and b. Squaring both sides yields
p = (a^2)/(b^2)
p(b^2) = a^2.
Clearly the right hand side is a perfect square. So the LHS should be a perfect square as well. For this to happen p must be of the form p = c^2 for positive integral c. This contradicts the fact that p is prime. As a result, our assumption was false, and the square root of a prime is irrational.
Assum sqrt(p) is rational for some prime p. Then we can write sqrt(p) = a/b for relatively prime positive integers a and b. Squaring both sides yields
p = (a^2)/(b^2)
p(b^2) = a^2.
Clearly the right hand side is a perfect square. So the LHS should be a perfect square as well. For this to happen p must be of the form p = c^2 for positive integral c. This contradicts the fact that p is prime. As a result, our assumption was false, and the square root of a prime is irrational.
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