Prove that if x and y are real numbers, then max(x,y) + min(x,y) = x+y. [Hint: Use a proof by cases, with the two cases corresponding to x≥y and x<y, respectively.]

Using hint, I supposed two cases where: x≥y or x<y.

Thus, if x≥y, then max(x,y) = x or y (since x and y might be the same), min(x,y) = y. Therefore, it is equal to 2y or x+y. Since it covers all of the cases, the equality always holds.

If x<y, then max(x,y) = y, min(x,y) = x. Thus, it is equal to x+y.

But the answer is slightly different from my solution. The answer is:

If x≤y, then max(x,y) + min(x,y) = y+x = x+y. If x≥y, then max(x,y)+min(x,y) = x+y. Because there are only two cases, the equality always holds.

You can notice that the answer supposed two different cases, x≤y or x≥y.

But I supposed x≥y or x<y.

I want to know whether these two different answers are both correct or not.

On the other hand, is it correct if I make three cases, where x > y, x = y, and x < y?

When x > y, max(x,y) = x, min(x,y) = y. Thus, x + y.

When x = y, max(x,y) = x or y, min(x,y) = x or y. Thus, 2x or 2y or x + y.

When x < y, max(x,y) = y, min(x,y) = x. Thus, x + y.

Is this answer correct? It seems clearer for me.

Thank you in advance.

  • 2
    $\begingroup$ If $x=y$ then the max equals the min so the result is true. If $x\ne y$ then one of them is the max and the other is the min so max+min is $x+y$. $\endgroup$
    – John Douma
    Jun 5, 2023 at 13:34
  • 3
    $\begingroup$ The three ways you report are equally correct. It is a matter of taste. Personally, I would write only: "wlog $x\ge y$, then $\max(x,y)+\min(x,y)=x+y$". $\endgroup$ Jun 5, 2023 at 13:35
  • 1
    $\begingroup$ They are the same. Your three way approach works, but the $x=y$ case can be incorporated into both of the others: when $x \ge y, \max(x,y) = x, \min(x,y) = y$, thus $x + y$; and when $x \le y, \max(x,y) = y, \min(x,y) = x$, thus $y + x$. It does not matter that when $x=y$ you have $\max(x,y) = x =y$ as you can use whichever is more helpful $\endgroup$
    – Henry
    Jun 5, 2023 at 13:46

1 Answer 1


Your solution is correct but i suggest to take different case for equity, because this may be confusing.
Actually the solution present in the book and your solution are almost similar. But if you ignore equality in first case and prove the statement, then i believe this is better because discussing "equality" and "greater than" at a time is confusing.
And your 3rd proof is also correct. Just write second case simply as
If $x=y$, then $max\{ x,y \} + min\{x,y\}=x+y$ as desired.
Writing 2x and 2y is unnecessary because you have to prove simply x+y.
Btw all three are correct just need little bit modification according to me.


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