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I found the answer to this question where it asks if this function is onto or not, but I don't understand why they made it s = sqrt(3). Can someone help explain that to me as well as explain how the derived answer means that it's not onto? Thank you!
Define J : Q x Q -> R by the rule J (r, s) = r sqrt(2) * s
Observe that, J is not onto.
Counter example:
Observe that, sqrt(6) ∈ R and
sqrt(6) = sqrt(2) * sqrt(3)
= 0 sqrt(2) * sqrt(3)
= r sqrt(2) * s
Here, r = 0 and s = sqrt(3)
r = 0 ∈ Q, s = sqrt(3) ∉ Q
= (0, sqrt(3)) ∉ Q x Q
This means that, sqrt(6) has no pre-image in Q x Q under the mapping J (r, s) = r sqrt(2) * s
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OH OKAY THAT MAKES SO MUCH MORE SENSE, THANK YOU FOR ANSWERING!