Does this inequality make sense? 1 = |1|?











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-1
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Okay so suppose x = -9.



Then we have x < 1 .



But 1 = |1|
Hence x < |1|
Implies -1 < x < 1



But this clearly is not true.
Just wondering what the limitations are when using inequalities or is there a limitation with the absolute value. Or am I just missing something?



Also I wondered. If x< 4 and x > 4, can we say x = 4?










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  • 3




    "x < |1| Implies -1 < x < 1" No it doesn't. $|x|<1$ implies $-1<x<1$.
    – Rahul
    Nov 18 at 10:39










  • Ahhhhhhh i am so silly! xD
    – Sashin Chetty
    Nov 19 at 13:34















up vote
-1
down vote

favorite












Okay so suppose x = -9.



Then we have x < 1 .



But 1 = |1|
Hence x < |1|
Implies -1 < x < 1



But this clearly is not true.
Just wondering what the limitations are when using inequalities or is there a limitation with the absolute value. Or am I just missing something?



Also I wondered. If x< 4 and x > 4, can we say x = 4?










share|cite|improve this question


















  • 3




    "x < |1| Implies -1 < x < 1" No it doesn't. $|x|<1$ implies $-1<x<1$.
    – Rahul
    Nov 18 at 10:39










  • Ahhhhhhh i am so silly! xD
    – Sashin Chetty
    Nov 19 at 13:34













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











Okay so suppose x = -9.



Then we have x < 1 .



But 1 = |1|
Hence x < |1|
Implies -1 < x < 1



But this clearly is not true.
Just wondering what the limitations are when using inequalities or is there a limitation with the absolute value. Or am I just missing something?



Also I wondered. If x< 4 and x > 4, can we say x = 4?










share|cite|improve this question













Okay so suppose x = -9.



Then we have x < 1 .



But 1 = |1|
Hence x < |1|
Implies -1 < x < 1



But this clearly is not true.
Just wondering what the limitations are when using inequalities or is there a limitation with the absolute value. Or am I just missing something?



Also I wondered. If x< 4 and x > 4, can we say x = 4?







real-analysis algebra-precalculus inequality






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 18 at 10:33









Sashin Chetty

222




222








  • 3




    "x < |1| Implies -1 < x < 1" No it doesn't. $|x|<1$ implies $-1<x<1$.
    – Rahul
    Nov 18 at 10:39










  • Ahhhhhhh i am so silly! xD
    – Sashin Chetty
    Nov 19 at 13:34














  • 3




    "x < |1| Implies -1 < x < 1" No it doesn't. $|x|<1$ implies $-1<x<1$.
    – Rahul
    Nov 18 at 10:39










  • Ahhhhhhh i am so silly! xD
    – Sashin Chetty
    Nov 19 at 13:34








3




3




"x < |1| Implies -1 < x < 1" No it doesn't. $|x|<1$ implies $-1<x<1$.
– Rahul
Nov 18 at 10:39




"x < |1| Implies -1 < x < 1" No it doesn't. $|x|<1$ implies $-1<x<1$.
– Rahul
Nov 18 at 10:39












Ahhhhhhh i am so silly! xD
– Sashin Chetty
Nov 19 at 13:34




Ahhhhhhh i am so silly! xD
– Sashin Chetty
Nov 19 at 13:34










3 Answers
3






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up vote
1
down vote













You are confusing $|x| < 1$ with $x < |1|$.



when $x=-9$, the first inequality that I have written above is not true.



Also, there is no number that satisfies $x<4$ and $x > 4$.



However, if $x le 4$ and $x ge 4$, then we can say that $x=4$.






share|cite|improve this answer




























    up vote
    1
    down vote













    There is some confusion here. It is not true that $x<lvert1rvertimplies-1<x<1$. Perhaps that you are thinking about $lvert xrvert<1implies-1<x<1$.



    And if $x<4$ and $x>4$, then there is no such $x$. In particular, you can't say that $x=4$.






    share|cite|improve this answer




























      up vote
      0
      down vote













      You seem to be assuming that if $a<b$ then $|a|<|b|$, but that is not true. More generally, if $a<b$ that does not necessarily imply $f(a)<f(b)$ for any function $f$. However, it is true if $f$ is strictly increasing (which the absolute value function isn't).



      If $x<|1|$ that is the same as saying $x<1$ because the absolute value of 1 is simply 1. However, If $|x|<1$ then it is true that $-1<x<1$.



      If $x<4$ and $x>4$ then $x$ does not exist, because no real number satisfies both of those conditions.






      share|cite|improve this answer





















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        3 Answers
        3






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        3 Answers
        3






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        active

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        up vote
        1
        down vote













        You are confusing $|x| < 1$ with $x < |1|$.



        when $x=-9$, the first inequality that I have written above is not true.



        Also, there is no number that satisfies $x<4$ and $x > 4$.



        However, if $x le 4$ and $x ge 4$, then we can say that $x=4$.






        share|cite|improve this answer

























          up vote
          1
          down vote













          You are confusing $|x| < 1$ with $x < |1|$.



          when $x=-9$, the first inequality that I have written above is not true.



          Also, there is no number that satisfies $x<4$ and $x > 4$.



          However, if $x le 4$ and $x ge 4$, then we can say that $x=4$.






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            You are confusing $|x| < 1$ with $x < |1|$.



            when $x=-9$, the first inequality that I have written above is not true.



            Also, there is no number that satisfies $x<4$ and $x > 4$.



            However, if $x le 4$ and $x ge 4$, then we can say that $x=4$.






            share|cite|improve this answer












            You are confusing $|x| < 1$ with $x < |1|$.



            when $x=-9$, the first inequality that I have written above is not true.



            Also, there is no number that satisfies $x<4$ and $x > 4$.



            However, if $x le 4$ and $x ge 4$, then we can say that $x=4$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 18 at 10:38









            Siong Thye Goh

            95.9k1462116




            95.9k1462116






















                up vote
                1
                down vote













                There is some confusion here. It is not true that $x<lvert1rvertimplies-1<x<1$. Perhaps that you are thinking about $lvert xrvert<1implies-1<x<1$.



                And if $x<4$ and $x>4$, then there is no such $x$. In particular, you can't say that $x=4$.






                share|cite|improve this answer

























                  up vote
                  1
                  down vote













                  There is some confusion here. It is not true that $x<lvert1rvertimplies-1<x<1$. Perhaps that you are thinking about $lvert xrvert<1implies-1<x<1$.



                  And if $x<4$ and $x>4$, then there is no such $x$. In particular, you can't say that $x=4$.






                  share|cite|improve this answer























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    There is some confusion here. It is not true that $x<lvert1rvertimplies-1<x<1$. Perhaps that you are thinking about $lvert xrvert<1implies-1<x<1$.



                    And if $x<4$ and $x>4$, then there is no such $x$. In particular, you can't say that $x=4$.






                    share|cite|improve this answer












                    There is some confusion here. It is not true that $x<lvert1rvertimplies-1<x<1$. Perhaps that you are thinking about $lvert xrvert<1implies-1<x<1$.



                    And if $x<4$ and $x>4$, then there is no such $x$. In particular, you can't say that $x=4$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 18 at 10:39









                    José Carlos Santos

                    143k20112212




                    143k20112212






















                        up vote
                        0
                        down vote













                        You seem to be assuming that if $a<b$ then $|a|<|b|$, but that is not true. More generally, if $a<b$ that does not necessarily imply $f(a)<f(b)$ for any function $f$. However, it is true if $f$ is strictly increasing (which the absolute value function isn't).



                        If $x<|1|$ that is the same as saying $x<1$ because the absolute value of 1 is simply 1. However, If $|x|<1$ then it is true that $-1<x<1$.



                        If $x<4$ and $x>4$ then $x$ does not exist, because no real number satisfies both of those conditions.






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          You seem to be assuming that if $a<b$ then $|a|<|b|$, but that is not true. More generally, if $a<b$ that does not necessarily imply $f(a)<f(b)$ for any function $f$. However, it is true if $f$ is strictly increasing (which the absolute value function isn't).



                          If $x<|1|$ that is the same as saying $x<1$ because the absolute value of 1 is simply 1. However, If $|x|<1$ then it is true that $-1<x<1$.



                          If $x<4$ and $x>4$ then $x$ does not exist, because no real number satisfies both of those conditions.






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            You seem to be assuming that if $a<b$ then $|a|<|b|$, but that is not true. More generally, if $a<b$ that does not necessarily imply $f(a)<f(b)$ for any function $f$. However, it is true if $f$ is strictly increasing (which the absolute value function isn't).



                            If $x<|1|$ that is the same as saying $x<1$ because the absolute value of 1 is simply 1. However, If $|x|<1$ then it is true that $-1<x<1$.



                            If $x<4$ and $x>4$ then $x$ does not exist, because no real number satisfies both of those conditions.






                            share|cite|improve this answer












                            You seem to be assuming that if $a<b$ then $|a|<|b|$, but that is not true. More generally, if $a<b$ that does not necessarily imply $f(a)<f(b)$ for any function $f$. However, it is true if $f$ is strictly increasing (which the absolute value function isn't).



                            If $x<|1|$ that is the same as saying $x<1$ because the absolute value of 1 is simply 1. However, If $|x|<1$ then it is true that $-1<x<1$.



                            If $x<4$ and $x>4$ then $x$ does not exist, because no real number satisfies both of those conditions.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 18 at 10:39









                            Eff

                            11.5k21638




                            11.5k21638






























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