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Math Help - Square Root Uniqueness

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    Square Root Uniqueness

    Given any r in positive R (reals), the number sqroot(r) is unique in the sense that, if x is a positive real number such that x^2 = r, then x = sqroot(r)

    Any help with this proof would be appreciated.
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    Quote Originally Posted by jstarks44444 View Post
    Given any r in positive R (reals), the number sqroot(r) is unique in the sense that, if x is a positive real number such that x^2 = r, then x = sqroot(r)

    Any help with this proof would be appreciated.


    Suppose x^2 = y^2 ==> (x-y)(x+y) = 0 , and since we're in a field then x = y as we're talking of positive square roots .

    Tonio
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  3. #3
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    It's also pretty straightforward to prove that for positive reals, a>b ==> a^2 > b^2. So you get the contrapositive: if x does not equal sqrt(r), then x^2 does not equal r.
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    Quote Originally Posted by tonio View Post
    Suppose x^2 = y^2 ==> (x-y)(x+y) = 0 , and since we're in a field then x = y as we're talking of positive square roots .

    Tonio
    i think this is only true for ordered fields (and thus fields of characteristic 0, R qualifies on both counts).

    you need the order properties of R to show that x+y is non-zero.
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    Quote Originally Posted by Deveno View Post
    i think this is only true for ordered fields (and thus fields of characteristic 0, R qualifies on both counts).

    you need the order properties of R to show that x+y is non-zero.


    First, the question was about reals, so ordered fields and stuff is way too much for this EXCEPT for the part that we talk about POSITIVE numbers.

    Second, the equalities x^2 = y^2 <==> (x-y)(x+y) = 0 are true in any commutative ring, and if we add some other condition, not necessarily of ordered

    fields, we can decide whether x = y or x = -y (for example, if we take the usual representatives of the residue classes

    modulo a prime p, we can decide that the solution to x^2 = has to be between

    0 and (p-1)/2 (mod p), say...

    Tonio
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