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Consider the problem of proving that, “If $x$ and $y$ are non-negative real numbers that satisfy $x + y = 0$ then $x = 0$ and $y = 0$.”

For the following condensed proof, write an analysis indicating the forward and backward steps and the key questions and answers.

Proof. First, it will be shown that $x \le 0$, for then, because $x \ge 0$, it must be that $x = 0$. To see that $x \le 0$, by the hypothesis, $x + y = 0$, so $x = −y$. Also, because $y \ge 0$, it follows that $−y \le 0$ and hence $x = −y \le 0$. Finally, because $x = 0$ and $x + y = 0$, it follows that $0 = x + y = 0 + y = y$.

I am of the opinion that the textbook's solution/reasoning for this problem is so unstructured as to be logically incorrect.

Textbook Solution:

A represents the hypothesis, and B represents the conclusion.

My comments to the textbook solution have been added in italics.

Analysis of Proof. A key question associated with the conclusion is, "How can I show that a real number (namely, $x$) is $0$?". To show that $x = 0$, it will be established that

B1: $x \le 0$ and $x \ge 0$

Working forward from the hypothesis immediately establishes that

A1: $x \ge 0$

To see that $x \le 0$, it will be shown that

B2: $x = -y$ and $-y \le 0$. I think this is an error. One is supposed to derive statements from the conclusion by using the backwards method, which first necessitates asking a key question. One then derives the next statement by answering this key question. However, in this case, B2 is not an answer to a key question. Furthermore, the mathematical expression $x = -y$ and $-y \le 0$ is not logically derivable from the information contained in B1 alone; it must be derived from the information in A (the hypothesis) and A1. This requires the use of the forward method, rather than the backward method. Indeed, the next section of the solution seems to clarify this, but it does not change the fact that this part is incorrect.

Both of these statements follow by working forward from the hypotheses that

A2: $x + y = 0$ (so $x = -y$) and

A3: $y \ge 0$ (so $-y \le 0$)

It remains only to show that

B3: $y = 0$

which follows by working forward from the fact that

A5: $x + y = 0$

so,

A6: $0 = x + y = 0 + y = y$

End of Analysis of Proof.

I think my proof is correct and structured in a superior way.

My solution:

B: $x = 0 \land y = 0$

How can I show that a real number is equal to $0$?

Show that the number is greater than or equal to $0$ and less than or equal to $0$.

B1: $0 \ge x \ge 0$

A: $x + y = 0$ for all $x$ and $y$ that are elements of the real numbers that are greater than or equal to $0$.

A1: $x = -y \le 0$ Since the hypothesis states that $x$ and $y$ are elements of the real numbers greater than or equal to $0$

We now assume that the last statement in the forward process (A1) is true and combine it with the last statement in the backwards process (B1):

A2: $0 = -y \le 0$

$$\therefore y = 0 \land x = 0$$ Q.E.D

I would greatly appreciate it if any of the more experienced mathematicians on MSE could take the time to provide feedback.

Is my reasoning correct? If not please explain why and what the correct reasoning should be.

Thank you.

EDIT:

I have used user21820's excellent answer in redoing my proof.

My second attempt at a solution:

B: $x = 0$

How can I show that a number is equal to $0$?

Show that the number is greater than or equal to $0$ and less than or equal to $0$.

B1: $0 \ge x \ge 0$ If B1 is true then B is also true

A: $x + y = 0$

  1. For all $y$ that are elements of the real numbers that are greater than or equal to $0$.
  2. For all $x$ that are elements of the real numbers that are greater than or equal to $0$.

A1: $x + y = 0$

$\Rightarrow x = -y$

A2: $y \ge 0$ ($y$ is an element of the real numbers that are greater than or equal to $0$) (Shown in A)

$\Rightarrow -y \le 0$

$\therefore x = -y \le 0$ For all $x$ that are elements of the real numbers that are greater than or equal to $0$. (Show in A)

A3: $0 \ge -y = x \ge 0$

$\therefore x = 0$ Q.E.D We have now linked up to the last statement in the backward process (B1).

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Yours is the one that makes no sense at all. You say that you need to show $x \ge 0$, but you did not do so. Instead, you used it as if it were true, without proof!

The given solution in the textbook is correct, and you analyzed it wrongly. In mathematics there is no such thing as "key questions" that are "necessitated". Furthermore, it is false that you must derive statements from the conclusion, because to do so would be circular or useless. Let us see exactly how the textbook solution should be understood.


First, it will be shown that $x \le 0$, for then, because $x \ge 0$, it must be that $x = 0$. To see that $x \le 0$, by the hypothesis, $x + y = 0$, so $x = −y$. Also, because $y \ge 0$, it follows that $−y \le 0$ and hence $x = −y \le 0$. Finally, because $x = 0$ and $x + y = 0$, it follows that $0 = x + y = 0 + y = y$.

First, note that ( if $x \le 0$ then $x = 0$ because $x \ge 0$ ).

Next, since $x+y = 0$, we have $x = -y$.

Also, since $y \ge 0$, we have $-y \le 0$.

Thus $x \le 0$ by substitution.

Hence, by the first note, we have $x = 0$.

Thus $y = -x = -0 = 0$.


Observe that every single statement in my rephrasing is true (in the context of the question). This is the hallmark of a proper proof. Whenever one does so-called backward reasoning, one does not derive anything from the desired conclusion. Rather, if one wishes to prove a statement "$B$", then backward reasoning simply means that we try to find a statement of the form "$A \to B$" (meaning "$A$ implies $B$") that we can prove, and then try to prove "$A$". If we succeed, then from "$A$" and "$A \to B$" we of course can prove "$B$". How to come up with "$A$" relies on your ingenuity or your mathematical experience (or both).

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  • $\begingroup$ I would like to thank you for taking the time to post such an elaborate and clear answer. You are absolutely correct: I incorrectly applied the backward process. After reading your illuminating response, I have redone my proof and edited the OP to include it. I hope the new proof reflects the clear reasoning you conveyed in your answer. $\endgroup$ – The Pointer Dec 20 '16 at 9:30
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    $\begingroup$ @ThePointer: You're welcome. I don't see an edited question, but you shouldn't change your question too much otherwise it would invalidate existing answers. Anyway if you're satisfied with my answer you can accept it. If you have further questions about logic feel free to come to chat.stackexchange.com/rooms/44058/logic and ask me there. $\endgroup$ – user21820 Dec 20 '16 at 9:38
  • $\begingroup$ I just finished the edit. Thank you for the generous assistance! $\endgroup$ – The Pointer Dec 20 '16 at 9:45
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    $\begingroup$ @ThePointer: Apart from a missing minus-sign, the basic idea is correct, but your side-explanations are not correct. The problem asks for a proof of a statement about every non-negative real numbers $x,y$. Hence in a proof you are not free to choose what $x,y$ are; they are given to you and you must somehow show that if $x+y = 0$ then $x = y = 0$. Thus your bullet-points under (A) should not read "for all ..." but "and ...". $\endgroup$ – user21820 Dec 20 '16 at 9:57

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