Sixth grade math (number related) problem












1












$begingroup$


We have this statement (about rational numbers, btw):



If $m-n+p = p$ and $ m neq n neq 0$ then $ m = -n$



Is this true?

a) always

b) never

c) sometimes



The given answer is b) but:

1) this textbook often has some wrong answers and

2) this question got me really thinking...



According to the truth table of the implication if the premise is false, then the whole implication is true, no matter what the conclusion is.
In this case the premise is a conjunction of several facts, two of which are contradictory ($m = n $ and $m neq n$).
So the premise is false. And so I conclude the whole implication is true.

Then... what does that mean? I guess it means the correct answer is a).






logic education rational-numbers







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    We have this statement (about rational numbers, btw):



    If $m-n+p = p$ and $ m neq n neq 0$ then $ m = -n$



    Is this true?

    a) always

    b) never

    c) sometimes



    The given answer is b) but:

    1) this textbook often has some wrong answers and

    2) this question got me really thinking...



    According to the truth table of the implication if the premise is false, then the whole implication is true, no matter what the conclusion is.
    In this case the premise is a conjunction of several facts, two of which are contradictory ($m = n $ and $m neq n$).
    So the premise is false. And so I conclude the whole implication is true.

    Then... what does that mean? I guess it means the correct answer is a).






    logic education rational-numbers







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      We have this statement (about rational numbers, btw):



      If $m-n+p = p$ and $ m neq n neq 0$ then $ m = -n$



      Is this true?

      a) always

      b) never

      c) sometimes



      The given answer is b) but:

      1) this textbook often has some wrong answers and

      2) this question got me really thinking...



      According to the truth table of the implication if the premise is false, then the whole implication is true, no matter what the conclusion is.
      In this case the premise is a conjunction of several facts, two of which are contradictory ($m = n $ and $m neq n$).
      So the premise is false. And so I conclude the whole implication is true.

      Then... what does that mean? I guess it means the correct answer is a).






      logic education rational-numbers







      share|cite|improve this question









      $endgroup$




      We have this statement (about rational numbers, btw):



      If $m-n+p = p$ and $ m neq n neq 0$ then $ m = -n$



      Is this true?

      a) always

      b) never

      c) sometimes



      The given answer is b) but:

      1) this textbook often has some wrong answers and

      2) this question got me really thinking...



      According to the truth table of the implication if the premise is false, then the whole implication is true, no matter what the conclusion is.
      In this case the premise is a conjunction of several facts, two of which are contradictory ($m = n $ and $m neq n$).
      So the premise is false. And so I conclude the whole implication is true.

      Then... what does that mean? I guess it means the correct answer is a).






      logic education rational-numbers




      logic education rational-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 4 hours ago









      peter.petrovpeter.petrov

      5,422821




      5,422821






















          2 Answers
          2






          active

          oldest

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          2












          $begingroup$

          I believe this is a poorly worded math problem, sixth grade or otherwise. The statement



          $$m neq n neq 0$$



          is possibly meant to be something like



          $$m, n neq 0$$



          instead. This would be to ensure that $m = n$ and $m = -n$ can't both be true simultaneously as it only occurs when $m = n = 0$.



          This is only a guess, but it would mean the question would then be consistent with the textbook's provided answer.






          share|cite|improve this answer











          $endgroup$





















            2












            $begingroup$

            It's unclear what $m neq n neq 0$ means. But, suppose that instead the statement were:



            "If $m - n + p = p$ and $m neq n$ then $m = -n$."



            In this case, the statement would be true (vacuously true). I wouldn't describe the statement as "always true" ; the statement is simply "true".






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              "always" because it is never true that $(m-n+p=p) land(mneq n)$.
              $endgroup$
              – jordan_glen
              3 hours ago












            • $begingroup$
              The phrase "always true" seems redundant. There is no need for the word "always". The statement I mentioned is simply "true".
              $endgroup$
              – littleO
              3 hours ago










            • $begingroup$
              Could be "true for all values of $p$".
              $endgroup$
              – dbx
              3 hours ago










            • $begingroup$
              It is always true in the sense that $forall m, forall n, forall pBig((mneq n land m-n+p=p)to (m= -n)Big)$ is true. Don't be so patronizing in your tone, please. Be kind.
              $endgroup$
              – jordan_glen
              3 hours ago












            • $begingroup$
              I didn't intend for my tone to sound patronizing, sorry if it did! However, I stand by my point that the word "always" is unnecessary. For example, one wouldn't describe Fermat's Last Theorem as being "always true", it's just a statement that is true.
              $endgroup$
              – littleO
              3 hours ago











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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            I believe this is a poorly worded math problem, sixth grade or otherwise. The statement



            $$m neq n neq 0$$



            is possibly meant to be something like



            $$m, n neq 0$$



            instead. This would be to ensure that $m = n$ and $m = -n$ can't both be true simultaneously as it only occurs when $m = n = 0$.



            This is only a guess, but it would mean the question would then be consistent with the textbook's provided answer.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              I believe this is a poorly worded math problem, sixth grade or otherwise. The statement



              $$m neq n neq 0$$



              is possibly meant to be something like



              $$m, n neq 0$$



              instead. This would be to ensure that $m = n$ and $m = -n$ can't both be true simultaneously as it only occurs when $m = n = 0$.



              This is only a guess, but it would mean the question would then be consistent with the textbook's provided answer.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                I believe this is a poorly worded math problem, sixth grade or otherwise. The statement



                $$m neq n neq 0$$



                is possibly meant to be something like



                $$m, n neq 0$$



                instead. This would be to ensure that $m = n$ and $m = -n$ can't both be true simultaneously as it only occurs when $m = n = 0$.



                This is only a guess, but it would mean the question would then be consistent with the textbook's provided answer.






                share|cite|improve this answer











                $endgroup$



                I believe this is a poorly worded math problem, sixth grade or otherwise. The statement



                $$m neq n neq 0$$



                is possibly meant to be something like



                $$m, n neq 0$$



                instead. This would be to ensure that $m = n$ and $m = -n$ can't both be true simultaneously as it only occurs when $m = n = 0$.



                This is only a guess, but it would mean the question would then be consistent with the textbook's provided answer.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 3 hours ago

























                answered 4 hours ago









                John OmielanJohn Omielan

                1,47629




                1,47629























                    2












                    $begingroup$

                    It's unclear what $m neq n neq 0$ means. But, suppose that instead the statement were:



                    "If $m - n + p = p$ and $m neq n$ then $m = -n$."



                    In this case, the statement would be true (vacuously true). I wouldn't describe the statement as "always true" ; the statement is simply "true".






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      "always" because it is never true that $(m-n+p=p) land(mneq n)$.
                      $endgroup$
                      – jordan_glen
                      3 hours ago












                    • $begingroup$
                      The phrase "always true" seems redundant. There is no need for the word "always". The statement I mentioned is simply "true".
                      $endgroup$
                      – littleO
                      3 hours ago










                    • $begingroup$
                      Could be "true for all values of $p$".
                      $endgroup$
                      – dbx
                      3 hours ago










                    • $begingroup$
                      It is always true in the sense that $forall m, forall n, forall pBig((mneq n land m-n+p=p)to (m= -n)Big)$ is true. Don't be so patronizing in your tone, please. Be kind.
                      $endgroup$
                      – jordan_glen
                      3 hours ago












                    • $begingroup$
                      I didn't intend for my tone to sound patronizing, sorry if it did! However, I stand by my point that the word "always" is unnecessary. For example, one wouldn't describe Fermat's Last Theorem as being "always true", it's just a statement that is true.
                      $endgroup$
                      – littleO
                      3 hours ago
















                    2












                    $begingroup$

                    It's unclear what $m neq n neq 0$ means. But, suppose that instead the statement were:



                    "If $m - n + p = p$ and $m neq n$ then $m = -n$."



                    In this case, the statement would be true (vacuously true). I wouldn't describe the statement as "always true" ; the statement is simply "true".






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      "always" because it is never true that $(m-n+p=p) land(mneq n)$.
                      $endgroup$
                      – jordan_glen
                      3 hours ago












                    • $begingroup$
                      The phrase "always true" seems redundant. There is no need for the word "always". The statement I mentioned is simply "true".
                      $endgroup$
                      – littleO
                      3 hours ago










                    • $begingroup$
                      Could be "true for all values of $p$".
                      $endgroup$
                      – dbx
                      3 hours ago










                    • $begingroup$
                      It is always true in the sense that $forall m, forall n, forall pBig((mneq n land m-n+p=p)to (m= -n)Big)$ is true. Don't be so patronizing in your tone, please. Be kind.
                      $endgroup$
                      – jordan_glen
                      3 hours ago












                    • $begingroup$
                      I didn't intend for my tone to sound patronizing, sorry if it did! However, I stand by my point that the word "always" is unnecessary. For example, one wouldn't describe Fermat's Last Theorem as being "always true", it's just a statement that is true.
                      $endgroup$
                      – littleO
                      3 hours ago














                    2












                    2








                    2





                    $begingroup$

                    It's unclear what $m neq n neq 0$ means. But, suppose that instead the statement were:



                    "If $m - n + p = p$ and $m neq n$ then $m = -n$."



                    In this case, the statement would be true (vacuously true). I wouldn't describe the statement as "always true" ; the statement is simply "true".






                    share|cite|improve this answer











                    $endgroup$



                    It's unclear what $m neq n neq 0$ means. But, suppose that instead the statement were:



                    "If $m - n + p = p$ and $m neq n$ then $m = -n$."



                    In this case, the statement would be true (vacuously true). I wouldn't describe the statement as "always true" ; the statement is simply "true".







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 3 hours ago

























                    answered 3 hours ago









                    littleOlittleO

                    29.5k645109




                    29.5k645109












                    • $begingroup$
                      "always" because it is never true that $(m-n+p=p) land(mneq n)$.
                      $endgroup$
                      – jordan_glen
                      3 hours ago












                    • $begingroup$
                      The phrase "always true" seems redundant. There is no need for the word "always". The statement I mentioned is simply "true".
                      $endgroup$
                      – littleO
                      3 hours ago










                    • $begingroup$
                      Could be "true for all values of $p$".
                      $endgroup$
                      – dbx
                      3 hours ago










                    • $begingroup$
                      It is always true in the sense that $forall m, forall n, forall pBig((mneq n land m-n+p=p)to (m= -n)Big)$ is true. Don't be so patronizing in your tone, please. Be kind.
                      $endgroup$
                      – jordan_glen
                      3 hours ago












                    • $begingroup$
                      I didn't intend for my tone to sound patronizing, sorry if it did! However, I stand by my point that the word "always" is unnecessary. For example, one wouldn't describe Fermat's Last Theorem as being "always true", it's just a statement that is true.
                      $endgroup$
                      – littleO
                      3 hours ago


















                    • $begingroup$
                      "always" because it is never true that $(m-n+p=p) land(mneq n)$.
                      $endgroup$
                      – jordan_glen
                      3 hours ago












                    • $begingroup$
                      The phrase "always true" seems redundant. There is no need for the word "always". The statement I mentioned is simply "true".
                      $endgroup$
                      – littleO
                      3 hours ago










                    • $begingroup$
                      Could be "true for all values of $p$".
                      $endgroup$
                      – dbx
                      3 hours ago










                    • $begingroup$
                      It is always true in the sense that $forall m, forall n, forall pBig((mneq n land m-n+p=p)to (m= -n)Big)$ is true. Don't be so patronizing in your tone, please. Be kind.
                      $endgroup$
                      – jordan_glen
                      3 hours ago












                    • $begingroup$
                      I didn't intend for my tone to sound patronizing, sorry if it did! However, I stand by my point that the word "always" is unnecessary. For example, one wouldn't describe Fermat's Last Theorem as being "always true", it's just a statement that is true.
                      $endgroup$
                      – littleO
                      3 hours ago
















                    $begingroup$
                    "always" because it is never true that $(m-n+p=p) land(mneq n)$.
                    $endgroup$
                    – jordan_glen
                    3 hours ago






                    $begingroup$
                    "always" because it is never true that $(m-n+p=p) land(mneq n)$.
                    $endgroup$
                    – jordan_glen
                    3 hours ago














                    $begingroup$
                    The phrase "always true" seems redundant. There is no need for the word "always". The statement I mentioned is simply "true".
                    $endgroup$
                    – littleO
                    3 hours ago




                    $begingroup$
                    The phrase "always true" seems redundant. There is no need for the word "always". The statement I mentioned is simply "true".
                    $endgroup$
                    – littleO
                    3 hours ago












                    $begingroup$
                    Could be "true for all values of $p$".
                    $endgroup$
                    – dbx
                    3 hours ago




                    $begingroup$
                    Could be "true for all values of $p$".
                    $endgroup$
                    – dbx
                    3 hours ago












                    $begingroup$
                    It is always true in the sense that $forall m, forall n, forall pBig((mneq n land m-n+p=p)to (m= -n)Big)$ is true. Don't be so patronizing in your tone, please. Be kind.
                    $endgroup$
                    – jordan_glen
                    3 hours ago






                    $begingroup$
                    It is always true in the sense that $forall m, forall n, forall pBig((mneq n land m-n+p=p)to (m= -n)Big)$ is true. Don't be so patronizing in your tone, please. Be kind.
                    $endgroup$
                    – jordan_glen
                    3 hours ago














                    $begingroup$
                    I didn't intend for my tone to sound patronizing, sorry if it did! However, I stand by my point that the word "always" is unnecessary. For example, one wouldn't describe Fermat's Last Theorem as being "always true", it's just a statement that is true.
                    $endgroup$
                    – littleO
                    3 hours ago




                    $begingroup$
                    I didn't intend for my tone to sound patronizing, sorry if it did! However, I stand by my point that the word "always" is unnecessary. For example, one wouldn't describe Fermat's Last Theorem as being "always true", it's just a statement that is true.
                    $endgroup$
                    – littleO
                    3 hours ago


















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