Irrationality of the square root of a prime

  • Thread starter LeBrad
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I saw a proof saying the root of a prime is always irrational, and it went something like this:

sqrt(r) = p/q where p/q is reduced
r = p^2/q^2
r*q^2 = p^2 therefore, r divides p
so define p = c*r
r*q^2 = (c*r)^2 = c^2*r^2
q^2 = c^2*r therefore, r divides q also
since r divides p and q, p/q is not reduced and we have a contradition so sqrt(r) is irrational.

Now my question is, assuming the above is correct, why does this work for prime r and not, say, r = 4.
 

Answers and Replies

  • #2
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LeBrad said:
I saw a proof saying the root of a prime is always irrational, and it went something like this:

sqrt(r) = p/q where p/q is reduced
r = p^2/q^2
r*q^2 = p^2 therefore, r divides p
so define p = c*r
r*q^2 = (c*r)^2 = c^2*r^2
q^2 = c^2*r therefore, r divides q also
since r divides p and q, p/q is not reduced and we have a contradition so sqrt(r) is irrational.

Now my question is, assuming the above is correct, why does this work for prime r and not, say, r = 4.
This proof is based on the fact that a power has the same factors than it's root. Since 2 is prime, then it must be a factor of p. But with the number 4 this is no longer revelant since 4=2x2 and it would be fallascious to say 4 is a factor of p.
 
  • #3
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Werg22 said:
This proof is based on the fact that a power has the same factors than it's root. Since 2 is prime, then it must be a factor of p. But with the number 4 this is no longer revelant since 4=2x2 and it would be fallascious to say 4 is a factor of p.
I figured that's where the difference was but I couldn't quite see why. Thanks.
 
  • #4
matt grime
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another point to bear in mind is the "real" definition of primeness:

p is prime if whenever p diveds ab p divides one onf a or b. 4 fails this test.

here is another way of showing the same result:

suppose p=a^2/b^2, then rearranging pa^2=b^2. By the uniqueness of prime decomposition since the RHS has an even number of prime factors and the left an odd number we have a contradiction.
 
  • #5
Tide
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The argument is more general than that. If p is an natural number that is not a perfect square then its square root is irrational.
 
  • #6
SGT
LeBrad said:
I saw a proof saying the root of a prime is always irrational, and it went something like this:

sqrt(r) = p/q where p/q is reduced
r = p^2/q^2
r*q^2 = p^2 therefore, r divides p
so define p = c*r
r*q^2 = (c*r)^2 = c^2*r^2
q^2 = c^2*r therefore, r divides q also
since r divides p and q, p/q is not reduced and we have a contradition so sqrt(r) is irrational.

Now my question is, assuming the above is correct, why does this work for prime r and not, say, r = 4.
There is no way to express 4 as p/q, where p/q is reduced, except for the trivial case q = 1.
 
  • #7
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SGT said:
There is no way to express 4 as p/q, where p/q is reduced, except for the trivial case q = 1.
If every rational number can be expressed as the ratio of two integers, I don't see why 4 = 4/1 is trivial.
 
  • #8
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sqrt(r) = p/q where p/q is reduced
r = p^2/q^2
r*q^2 = p^2 therefore, r divides p

Hi, can you tell me why r divides p? Why not q^2=P^2/r ? Please teach me. Thank you very much!
 
  • #9
shmoe
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logic2b1 said:
sqrt(r) = p/q where p/q is reduced
r = p^2/q^2
r*q^2 = p^2 therefore, r divides p

Hi, can you tell me why r divides p? Why not q^2=P^2/r ? Please teach me. Thank you very much!
r*q^2=p^2 means r divides p^2. LeBrad was assuming r was a prime, so r divides p^2 means r divides p (see matt's post for the definition of prime).
 
  • #10
Gokul43201
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matt grime said:
suppose p=a^2/b^2, then rearranging pa^2=b^2. By the uniqueness of prime decomposition since the RHS has an even number of prime factors and the left an odd number we have a contradiction.
How about this ?

In the above, the RHS has an odd number of divisors while the LHS has an even number, unless p is a perfect square.
 
  • #11
mathwonk
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here is another similar proof:

if sqrt(p) = a/b, is in lowest tems, then p/1 = a^2/b^2 is also in lowest terms.

but lowest terms is unique so a^2 = p and b^2 = 1, i.e. p must be a perfect square.
 
  • #12
mathwonk
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here is another one: if X^2 - p =0 has a rational solution then by the rational roots theorem, then if p is prime, the solution is p, -p, 1, or -1, none of which work.
 
  • #13
matt grime's method is elegant! I have never seen it done that way...
 

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