Numerical Results with replacement rules [duplicate]

Question

Why are the following outputs different?

N[(x^2 + y^2) /. Thread[{x, y} -> {1., 1.}], 10]

gives 2..

N[(1^2 + 1^2), 10]

gives 2.000000000.

How can I use replacement but still retain numerical precision?

Answer 1

Once you've lost precision, you cannot get it back. Use

N[(x^2 + y^2) /. Thread[{x, y} -> {1, 1}], 10]

or

(x^2 + y^2) /. Thread[{x, y} -> {1`10, 1`10}]