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Problem: Show that the operation of addition is a continuous function from $\mathbf{R}\times\mathbf{R}$ to $\mathbf{R.}$
Solution: The function is defined by $f(x,y)=x+y.$ We prove that it is continuous at $(a,b)\in\mathbf{R}\times\mathbf{R.}$ Let $c=f(a,b)=a+b.$ Given $\epsilon>0.$ Let $\delta=\epsilon/2.$ Suppose $(x,y)$ lies within $\delta$ from $(a,b).$ Then $$\sqrt{(x-a)^{2}+(y-b)^{2}}<\delta.$$ Hence $$(x-a)^{2}+(y-b)^{2}<\delta^{2}.$$ Therefore $$(x-a)^{2}<\delta^{2}$$ and $$(y-b)^{2}<\delta^{2}.$$ It follows that $|x-a|<\delta$ and $|y-b|<\delta.$ Consider now $z=f(x,y)=x+y.$ From the above inequalities and the triangle inequality it follows that $$|z-c|=|(x+y)-(a+b)|=|(x-a)+(y-b)|\leq |x-a|+|y-b|<\delta+\delta=2\delta=\epsilon.$$ We have proven that if $(x,y)\in D((a,b),\delta)$ then $z=f(x,y)\in D(c,\epsilon).$ Since $f(a,b)=c,$ this means that $f(D((a,b),\delta))\subset D(f(a,b),\epsilon).$ Therefore $f$ is continuous at $(a,b)$ and, since $(a,b)$ was chosen arbitrarily, it is continuous on $\mathbf{R}\times\mathbf{R.}$