Simple question on squares of rationals

  • Thread starter xalvyn
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  • #1
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Hi all,

I think this sounds like a really simple and trivial question, but i've no clue as to where i should start:

true or false? between any two different positive rational numbers lies the square of a rational number. while i can't provide a construction of such a number, i somehow think that the statement is true.

thanks for any help
 

Answers and Replies

  • #2
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Assume there is no p, q such that p1/q1 < p^2/q^2 < p2/q2. What happens if you take the square root of all 3?
 
  • #3
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Hi,

the answer sounds obvious enough now :)

however, i was wondering if there is a rigorous way of proving this very elementary and trivial result.

for if we assume that if p1/q1 < p2/q2, then

sqrt(p1/q1) < sqrt(p2/q2), we'll have to prove this using 'dedekind section arguments', since the square root of a rational number is not necessarily another rational number.

i'll follow the definitions in my textbook closely:

i) if sqrt(p1/q1) < sqrt(p2/q2), then the lower section of sqrt(p1/q1) is contained in the lower section of qsrt(p2/q2); further

ii) there is at least one member of the upper section of sqrt(p1/q1) which is not contained in the upper section of sqrt(p2/q2), i.e. it is contained in the lower section of sqrt(p2/q2).

The first property is obvious enough; if x belongs to the lower section of sqrt(p1/q1), then x^2 < (p1/q1) < (p2/q2), and so x also belongs to the lower section of sqrt(p2/q2).

To prove the second property, we must show that there exists a rational number whose square lies between p1/q1 and p2/q2, which takes us back to the same question.

In some sense, the second property seems to follow naturally, because we 'feel' that the square roots of two different rational numbers must also be different; yet to prove this rigorously is rather frustrating.
 
  • #4
CRGreathouse
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There's always a rational between two nonequal irrationals, right? That should make this pretty easy.
 
  • #5
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Are you asking if p1/q1 < p2/q2, how to prove sqrt (p1/q1) < sqrt (p2/q2)? It works for any real positives a, b if a < b. a < b implies 1 < b/a, so sqrt (1) < sqrt (b/a) implies sqrt (a) < sqrt (b). Now, since between any two reals there exists a rational p/q, assume a and b are now positive rational numbers. It follows that there exists p/q such that sqrt (a) < p/q < sqrt (b) implying a < (p/q) ^2 < b, so between any two positive rationals there exists a square of a rational.
 
  • #6
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ok i think i understand now, thanks :)
 

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