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Prove xy=0 <=> (x=0 or y=0)

  1. Jan 20, 2012 #1
    Our teacher is teaching calculus 1 with a logic approach. So, he wants us to practice our math logic skills, however, I lack in providing formal methods of validity in operation and to some degree, application. So, here is my solution outline (please tell me what you think)

    if operation axiom translational invariance of order:

    for all x,y,z€ℝ
    x<y => x + z < y + z then the statement:

    xy=0 <=> (x=0 or y=0) is true

    xy=0 => for all x,y,0€ℝ:[x<y => (x + 0)< (y + 0) or y<x => (y + 0)<(x+0)] by the translation invariance of order

    y>0[x€ℝ => 0x = 0] => x>0[y€ℝ => 0y = 0] => for all x,y,0€ℝ:[x<y => (x + 0)< (y + 0) or y<x => (y + 0)<(x+0)] by the multiplication by 0 theorem

    Anything else to the proof?
     
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 20, 2012 #2

    Stephen Tashi

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    We'd have to know what definitions and assumptions your text materials make to know how to prove this using them. It's not clear what you mean by "the translational invariance of order". How does your book define that?
     
  4. Jan 20, 2012 #3
    operation axiom translational invariance of order:

    for all x,y,z€ℝ
    x<y => x + z < y + z
     
  5. Jan 20, 2012 #4
    To prove the necessary condition, use contraposition. Suppose [itex]x \neq 0 \wedge y \neq 0[/itex]. Then, you can divide by either [itex]x[/itex] or [itex]y[/itex] in [itex]x \, y = 0[/itex]. What do you get?
     
  6. Jan 20, 2012 #5
    If neither x nor y can be zero, then the statement that x or y in xy=0 is false.
     
  7. Jan 20, 2012 #6
    So, I'm sorry. I'm really trying to hard to move on with this proof. I'm just having a tough time with it.

    When it comes to proofs, I hardly know when to start and when I do, I don't know when It's complete or when it needs to be complete. I had 80 percent confidence in my answer, but apparently, I'm wrong. So, as far as this contraposition method, it's sounds like you are suggesting I use symbols to work out the logic, which I have done. Perhaps I am wrong in that interpretation. What exactly do you mean?
     
  8. Jan 20, 2012 #7

    Stephen Tashi

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    It's fine to use symbols to abbreviate thought but you if you treat them as some sort of magical substitute for thought, you only get in trouble.

    For example, in your first attempt at the proof you write down the step:
    It isn't clear what you thought you were deducing this statement from. If you were trying to deduce it from the statement
    then you were trying to deduce from the statement that you are attempting to prove, which wouldn't make sense.

    To prove a statment with an "if and ony if" condition in it, it is often simplest to break this into two implications. So "for each pair of real numbers x,y , xy = 0 if and only if x = 0 or y = 0" can be proved by proving these two statements separately:

    1) "for each pair of real numbers x,y, if (x = 0 or y = 0 ) then xy = 0 "
    2) "for each pair of real numbers x,y, if (xy = 0) then x = 0 or y = 0.

    To prove an "if.... then...." statement, we are allowed to assume the "if..." part is true.

    To prove statement 1, you need to refer to some assumption or theorem that tells you that 0 times another number is 0. You can assume x = 0 or y = 0. You can break that into two cases if you need to: Case 1) x = 0 Case 2) y = 0 . (The case of both x= 0 and y = 0 is treated by either one of those cases. The "or" in mathematics is not exclusive.)


    To prove statement 2. Dickfore suggests you use the contrapositive. The contrapostive of a statement of the form "If A then B" is the statement "If not B then not A". These two forms of a statement are logically equivalent, so if you can prove the contrapositive of a statement, it proves the original statement.

    The contrapostive of the if-then part of the statement "for each pair of real numbers, if (xy = 0) then x = 0 or y = 0" is "for each pair of real numbers if( it is not true that (x = 0 or y = 0) then it is not true that xy = 0)"

    To prove this we may assume "it is not true that (x = 0 or y = 0)". This is equivalent to assuming "x is not zero and y is not zero" by one of DeMorgans laws of logic. This law states that a statement of the form "not(A or B)" is equivalent to the statement "(not A) and (not B)".

    Dickfore did not mention this, but his suggested method of proof employs proof by contradiction. In this method of proof we assume the negation of the then-part of the if-then statement and show that this implies a statement that contradicts the if-part or some other assumption or theorem.

    Our if-part is equivalent to the statement that "x is not zero and y is not zero".
    The negation of the then-part is "not (it is not true that xy = 0)" , i.e. "xy is 0".

    In summary, we may take these facts as given:
    x is not zero
    y is not zero
    xy = 0

    You need to find the assumption or theorem in your text that states the fact that a non-zero number has a multiplicative inverse. Using that as justification, we may multiply the equation xy = 0 on both sides by the multiplicative inverse of x. This gives the new equation (1)y = 0 so y = 0. This contradicts the fact that y is not 0. Hence we have accomplished proof by contradiction.



    --------------

    Many people write proofs without stating that they are using DeMorgan's laws or contraposition or proof by contradiction or other principles of logic. Many people acquire the skill of using these rules simply by reading proofs in books. They come to think of such laws as "common sense". I don't know whether you teacher expects explain your use of logic in detail or whether you are allowed to proceed with "common sense". It might help if you glance at a book on elementary symbolic logic. I don't think you need go deeply into it in order to get the hang of things like contrapostiives and DeMorgan's laws.
     
  9. Jan 21, 2012 #8
    "then you were trying to deduce from the statement that you are attempting to prove, which wouldn't make sense.

    To prove a statment with an "if and ony if" condition in it, it is often simplest to break this into two implications. So "for each pair of real numbers x,y , xy = 0 if and only if x = 0 or y = 0" can be proved by proving these two statements separately:

    1) "for each pair of real numbers x,y, if (x = 0 or y = 0 ) then xy = 0 "
    2) "for each pair of real numbers x,y, if (xy = 0) then x = 0 or y = 0.

    To prove an "if.... then...." statement, we are allowed to assume the "if..." part is true.

    To prove statement 1, you need to refer to some assumption or theorem that tells you that 0 times another number is 0. You can assume x = 0 or y = 0. You can break that into two cases if you need to: Case 1) x = 0 Case 2) y = 0 . (The case of both x= 0 and y = 0 is treated by either one of those cases. The "or" in mathematics is not exclusive.)"

    Thanks for the response. So, based on what I just tried interpreting, I'm going to do a few things. First off, I want to get rid of this statement:'(x=0 or y=0) <=> xy=0 (What is the postulate defining the operation I just made?),' since I realize that that's pretty much what you are proving. It's like looking at something at the opposite yet equal side of a spectrum or 180 degrees. See what I'm saying. So, I'm getting rid of it.

    now, I'm going to switch the two sets I used for the last operation:

    xy=0 => for all x,y,0€ℝ:[x<y => (x + 0)< (y + 0) or y<x => (y + 0)<(x+0)] by the translation invariance of order (does this statement make sense to you? I'm reading it and it does to me.)

    Now, I will show that the theorem that proves 0 times any other number is 0 (we call it the multiplication by 0 theorem which he proved in class and I know how to prove, but is unnecessary)

    The Multiplication by 0 theorem states that x€ℝ => 0x = 0

    So, now,

    y>0[x€ℝ => 0x = 0] => x>0[y€ℝ => 0y = 0] => for all x,y,0€ℝ:[x<y => (x + 0)< (y + 0) or y<x => (y + 0)<(x+0)]

    I have a feeling my interpretation of your suggestive logic was very off, but I'm still trying to get this. So, please bare with my intermediate level of this logic stuff. I hardly knew about these symbols until a few days ago, basically. I suppose I was familiar with the symbols to some degree, but hardly. I'm surprised I'm actually getting it enough to write statements, regardless of genuine logic. Thanks!
     
  10. Jan 21, 2012 #9

    Stephen Tashi

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    I interpret you notation to mean:

    (For all real number x,y) if xy = 0 then [ for all real numbers x,y, if x < y then (x+0)< (y+0) ] or if (y < x) then (y+0) < (x+0)].

    That may be a true statement, but it makes no sense to me as a step in a proof, because you don't need to say "if (xy = 0)" to deduce the part that says "for all all real numbers x,y, if (x < y) then ..... " etc. As you indicated, that part of the statement follows from what you call "the translation invariance of order".

    What do you mean by "the theorem"? Are you saying that you are proving the "multiplication by 0 theorem? If this is already known to be a theorem, you don't have to try to prove it.

    Try writing in words what you wrote in symbols. To me, what you wrote doesn't make much sense.

    For example "y > 0{ if (x in R) then (0x = 0)". You don't need the fact that (y > 0) to conclude that "if (x in R) then (0X= 0)" if we are assuming the theorem that says 0 times any number is 0.

    If you are trying to prove that theorem, the theorem isn' proved by the condition "if (y > 0)".

    Unless you are studying a very advanced form of symbolic logic, you cannot write this proof completely in symbols. My advice is to think about the proof in words and then translate some of your verbal statements to symbols. Don't try to translate all of them to symbols.

    You better take my advice about breaking the proof of a statement with a "<=>" condition into two statements, one with "<=" and one with "=>".

    The proof I indicated makes no use of "the translation invariance of order".

    It looks you are approaching this work as person would "prove" a trigonometric identity. In that type of problem, you write down a symbolic expression and then manipulate individual parts of it. This is not a good way to write most proofs and the so-called "proofs" people create for trigonometric identities are not real proofs.
     
  11. Jan 21, 2012 #10
    Yes! You are probably correct on that idea. I am so much more used to proving trigonometric identities, since I practiced my pre-calculus a lot and this is a calculus 1 class, that I have approached this form of mathematics the same way I approach trigonometric proofs. Most of the time, It's about substitution from identities and also algebraic operation. What type of approach should one go for trigonometric proofs over other proofs? It sounds like I'm truly not grasping the firm idea of proving statements to be correct.
     
  12. Jan 21, 2012 #11
    Starting from the axioms of addition and multiplication, can you prove the following:
    [tex]
    \left( \forall x \in \mathbb{R} \right) 0 \cdot x = 0
    [/tex]
     
  13. Jan 21, 2012 #12
    0x=0
    <=> 0x - 0x same as 0
    <=> (0+0)x - 0x substitution
    <=> 0x + (0x - 0x) associative law
    <=> 0x transitive property of equality
     
  14. Jan 22, 2012 #13

    Stephen Tashi

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    If you are trying to prove (as Dickfore suggested) that "for each real number x, (x)(0) = 0", you can't begin the proof by assuming "0x = 0".

    As you said, you are copying the method used on trigonometric identities, where the tradition is to begin by writing down the identity to be proven and then manipulate it. But trigonometric identities are proven by a rote process that is not an actual proof. Forget about using that approach.

    Also, don't bother to prove "for each real number x, (x)(0) = 0" if you your book has already established that as a theorem. Just use the theorem as justification if you need that fact.

    Think about the problem in ordinary language. If you were having a political argument and you said "If Grod Fressington is elected, he'll hire his brother-in-law as the city attorney", you couldn't convincing argue you case by beginning with "Let's assume if Grod Fresston is elected then he will hire his brother-in-law as city attorney."


    The pattern for proving the statement "If A then B" directly goes like this:

    Assume A
    Show B is true
    ----------------------
    Conclude "if A then B".

    If you want to prove "if x is a real number then (x)(0) = 0" then you are only allowed:

    Assume x is a real number.

    You must show x(0) = 0 using that assumption plus all the theorems and assumptions already established

    We can't tell you exactly how to do this proof in a manner that is consistent with your textbook unless we know exactly how your textbook states the assumptons and theorems that are relevant. (There are many different ways of formulating the the ideas needed to develop the properties of real number.)

    As I said, before, don't worry about proving this if your textbook as already presented this as a theorem.
     
  15. Jan 22, 2012 #14
    He didn't assume [itex]0 \cdot x = 0[/itex]. He just wrote it in the first line. All the other steps are, in fact, a correct way of proving it. So, you are done with the
    [tex]
    x = 0 \vee y = 0 \Rightarrow x \cdot y = 0
    [/tex]
    direction of your equivalence.

    Now, you are left with proving the direction:
    [tex]
    x \cdot y = 0 \Rightarrow x = 0 \vee y = 0
    [/tex]
    I suggest you use the method "reductio ad absurdum".

    Your implication is equivalent to the following statement:
    [tex]
    \neg (x \cdot y = 0) \vee \left( x = 0 \vee y = 0 \right)
    [/tex]
    and its negation, by De Morgan's rules, is:
    [tex]
    \neg \left[ \neg (x \cdot y = 0) \vee \left( x = 0 \vee y = 0 \right) \right]
    [/tex]
    [tex]
    x \cdot y = 0 \wedge \neg (x = 0) \wedge \neg (y = 0)
    [/tex]
    [tex]
    x \cdot y = 0 \wedge x \neq 0 \wedge y \neq 0
    [/tex]
    Can you prove this statement incorrect?
     
    Last edited: Jan 22, 2012
  16. Jan 22, 2012 #15

    Stephen Tashi

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    I disagree. For example, by that interpretation, the first step is then:

    "<=> 0x - 0x same as 0"

    Which in words (considering the previous line), says:

    "( for each real number x, 0x = 0) if and only if ( 0x -0x = 0)"

    There isn't any reason given for this step. We'd have to know what Rayquesto's textbook has defined and proven up to this point in order to advise him what reason to give.

    It isn't clear that Rayquesto needs to prove (0)(x) = 0. Prior to assigning this proof, his textbook may have defined a zero to be a number such at for any real number x, (x)(0) = 0 and assumed or proved that zero is unique ( thus establishing the fact that the symbol "0" can be used unambiguously). If that work has already been done, then it would not be appropriate to re-do it in writing the proof for the theorem in the original post.

    I agree with the content of the rest of your post, but I think Rayquesto would do better to think in verbal terms unless his class is employing that level of symbolism already. (We are all being sloppy about the quantifiers that could be included in symbolic expressions to make them correct.)
     
  17. Jan 22, 2012 #16
    And, if you went through the rest of the argument, you would see that he never used that "assumption". I agree that he made a sloppy notation, but his logic deduction is solid.

    I am interpreting what has been written. It was asked to prove an equivalence. This means you need to prove both directions of the implication.

    I don't think verbal terms are any easier to understand than shorthand notation. For example, no one uses "necessary and sufficient condition" everyday talk without being a student of logic. Also, De Morgan's laws as well as the identity [itex]p \Rightarrow q \equiv \neg p \vee q[/itex] are rules of (formal, mathematical) logic, and not language and I don't see any "common-sense way" to deduce their general validity. They are crucial in obtaining a negation of the expression he is trying to verify. Now, the expression is one step away from demonstrating it is a contradiction (I already have given a hint how to do it in a previous post).
     
  18. Jan 22, 2012 #17
    I just received a very large amount of text from you tutors and I thank you deeply for this. I will read it and try to understand it, but I feel compelled to state something. My calculus 1 book does not teach us the language of logic or go over it at all. My teacher, however, gave us a 50 page packet that includes information pertaining to this language of mathematics. He did not cover De Morgan's law or any of that. He covers general concepts and general applications so we can take calculus 1 into a different approach. Technically, I don't need to know this. I just have a strong curious nature to me that leads me to feel compelled to learn certain material even if it isn't necessary. So, I'm going to read what you guys wrote and I'll get back to you soon. Thanks for the help!
     
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