Is the converse of this implication true?












2












$begingroup$


If we have,



begin{align}
|x-2|leq {1} iff & -1leq{x-2}leq1\ iff & 4leq {x+3}leq6 \ iff & 5leq {x+4}leq7.end{align}



Then In particular,



$ |x-2|leq {1} implies 4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7.$



Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?



$4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7 implies |x-2|leq {1}$.



This is a question in my textbook. I feel like the implication can be reversed.






inequality logic proof-explanation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you mean $|x-2|leq 1$?
    $endgroup$
    – Lucas Corrêa
    Dec 29 '18 at 2:51








  • 2




    $begingroup$
    Yes, that's what I meant.
    $endgroup$
    – user503154
    Dec 29 '18 at 2:53
















2












$begingroup$


If we have,



begin{align}
|x-2|leq {1} iff & -1leq{x-2}leq1\ iff & 4leq {x+3}leq6 \ iff & 5leq {x+4}leq7.end{align}



Then In particular,



$ |x-2|leq {1} implies 4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7.$



Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?



$4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7 implies |x-2|leq {1}$.



This is a question in my textbook. I feel like the implication can be reversed.






inequality logic proof-explanation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Did you mean $|x-2|leq 1$?
    $endgroup$
    – Lucas Corrêa
    Dec 29 '18 at 2:51








  • 2




    $begingroup$
    Yes, that's what I meant.
    $endgroup$
    – user503154
    Dec 29 '18 at 2:53














2












2








2





$begingroup$


If we have,



begin{align}
|x-2|leq {1} iff & -1leq{x-2}leq1\ iff & 4leq {x+3}leq6 \ iff & 5leq {x+4}leq7.end{align}



Then In particular,



$ |x-2|leq {1} implies 4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7.$



Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?



$4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7 implies |x-2|leq {1}$.



This is a question in my textbook. I feel like the implication can be reversed.






inequality logic proof-explanation







share|cite|improve this question











$endgroup$




If we have,



begin{align}
|x-2|leq {1} iff & -1leq{x-2}leq1\ iff & 4leq {x+3}leq6 \ iff & 5leq {x+4}leq7.end{align}



Then In particular,



$ |x-2|leq {1} implies 4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7.$



Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?



$4leq {|x+3|}leq6$ and $5leq {|x+4|}leq7 implies |x-2|leq {1}$.



This is a question in my textbook. I feel like the implication can be reversed.






inequality logic proof-explanation




inequality logic proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 3:13

























asked Dec 29 '18 at 2:34







user503154















  • 1




    $begingroup$
    Did you mean $|x-2|leq 1$?
    $endgroup$
    – Lucas Corrêa
    Dec 29 '18 at 2:51








  • 2




    $begingroup$
    Yes, that's what I meant.
    $endgroup$
    – user503154
    Dec 29 '18 at 2:53














  • 1




    $begingroup$
    Did you mean $|x-2|leq 1$?
    $endgroup$
    – Lucas Corrêa
    Dec 29 '18 at 2:51








  • 2




    $begingroup$
    Yes, that's what I meant.
    $endgroup$
    – user503154
    Dec 29 '18 at 2:53








1




1




$begingroup$
Did you mean $|x-2|leq 1$?
$endgroup$
– Lucas Corrêa
Dec 29 '18 at 2:51






$begingroup$
Did you mean $|x-2|leq 1$?
$endgroup$
– Lucas Corrêa
Dec 29 '18 at 2:51






2




2




$begingroup$
Yes, that's what I meant.
$endgroup$
– user503154
Dec 29 '18 at 2:53




$begingroup$
Yes, that's what I meant.
$endgroup$
– user503154
Dec 29 '18 at 2:53










2 Answers
2






active

oldest

votes


















1












$begingroup$

For the first question: suppose that $|x-2|leq 1$. So,
$$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$



For the second:
$$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
$$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
$$|-9-2| = |-11| = 11 > 1$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$


    Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?




    Note that:
    $$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
    4le |x+3|le 6 &notRightarrow 4le x+3le 6, end{align}$$
    because $4le x+3le 6$ has a solution $xin [1,3]$:
    enter image description here



    whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
    enter image description here






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      1












      $begingroup$

      For the first question: suppose that $|x-2|leq 1$. So,
      $$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$



      For the second:
      $$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
      $$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
      then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
      $$|-9-2| = |-11| = 11 > 1$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        For the first question: suppose that $|x-2|leq 1$. So,
        $$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$



        For the second:
        $$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
        $$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
        then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
        $$|-9-2| = |-11| = 11 > 1$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          For the first question: suppose that $|x-2|leq 1$. So,
          $$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$



          For the second:
          $$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
          $$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
          then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
          $$|-9-2| = |-11| = 11 > 1$$






          share|cite|improve this answer









          $endgroup$



          For the first question: suppose that $|x-2|leq 1$. So,
          $$|x+k| = |(x-2) + (k+2)| leq |x-2| + |k+2| leq 1 + |k+2|.$$



          For the second:
          $$4 leq |x+3| leq 6 Longrightarrow x in [1,3]cup[-9,-7]$$
          $$5 leq |x+4| leq 7 Longrightarrow x in [1,3]cup[-11,-9]$$
          then if $x$ satisfies both, $x in [1,3]cup{-9}$. Take $x=-9$,
          $$|-9-2| = |-11| = 11 > 1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 29 '18 at 3:13









          Lucas CorrêaLucas Corrêa

          1,5751321




          1,5751321























              0












              $begingroup$


              Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?




              Note that:
              $$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
              4le |x+3|le 6 &notRightarrow 4le x+3le 6, end{align}$$
              because $4le x+3le 6$ has a solution $xin [1,3]$:
              enter image description here



              whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
              enter image description here






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$


                Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?




                Note that:
                $$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
                4le |x+3|le 6 &notRightarrow 4le x+3le 6, end{align}$$
                because $4le x+3le 6$ has a solution $xin [1,3]$:
                enter image description here



                whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
                enter image description here






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$


                  Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?




                  Note that:
                  $$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
                  4le |x+3|le 6 &notRightarrow 4le x+3le 6, end{align}$$
                  because $4le x+3le 6$ has a solution $xin [1,3]$:
                  enter image description here



                  whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
                  enter image description here






                  share|cite|improve this answer









                  $endgroup$




                  Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?




                  Note that:
                  $$begin{align}4le x+3le 6 &Rightarrow 4le |x+3|le 6, text{but} \
                  4le |x+3|le 6 &notRightarrow 4le x+3le 6, end{align}$$
                  because $4le x+3le 6$ has a solution $xin [1,3]$:
                  enter image description here



                  whereas $4le|x+3|le 6$ has a solution $xin [-9,-7]cup [1,3]$:
                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 29 '18 at 8:35









                  farruhotafarruhota

                  20.4k2739




                  20.4k2739






























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