Thread: 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.

Originally Posted by jstarks44444

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

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.

Originally Posted by tonio

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.

Originally Posted by Deveno

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