Answer to Question #170517 in Real Analysis for Prathibha Rose

Question

Let f:R² to R defined by f(x)=√(x²+y²) show that f is continuous on R²

Answer 1

$f(x,y)=\sqrt{(x^2+y^2)}$

now, to show that f(x,y) is continuous everywhere on $R^2$ ,

we first fix an arbitrary $(a.b)$ and $ϵ>0$ .

here note that $f$ is indeeded the Euclidean's norm on $R^2$ ,

so we can use the reverse triangle inequality.

For any $(x,y)∈R^2$ with $|(x,y)−(a,b)|<δ=ϵ$

$d((fx,y),f(a,b))=|f(x,y)−f(a,b)|$

$=|\sqrt{x^2+y^2}-\sqrt{a^2+b^2}|\\ =|f(x,y)|−|f(a,b)||\\ ≤|(x,y)−(a,b)|\\ <ϵ$

hence, proving $f$ is continuous in our arbitrary chosen (a,b).

since, this is our arbitrary chosen ,

we conclude that $f$ is continuous everywhere.

this assumes that both domain and co-domain of $f$

are equipped with corresponding euclidean metrics.

hence,proved.