Question
Prove that if $x$ and $y$ are real numbers, then max $(x, y)+$ $\min (x, y)=x+y .[\text { Hint: Use a proof by cases, with the }$ two cases corresponding to $x \geq y$ and $x<y,$ respectively. $]$
Step 1
In this case, $\max(x, y)$ would be $x$ and $\min(x, y)$ would be $y$. Therefore, $\max(x, y) + \min(x, y) = x + y$. Show more…
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Use proof by cases to prove that $\max \{x, y \mid+\min (x, y)=$ $x+y$ for all real numbers $x$ and $y$.
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