Proving a statement about real numbers












3












$begingroup$



If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.




Okay so I went about this by proving the contrapositive.



Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.



Is this a valid proof or not?






real-analysis proof-verification real-numbers







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    The conclusion should be $xleq y$.
    $endgroup$
    – gamma
    6 hours ago










  • $begingroup$
    If $$epsilon=x-y$$ then your epsilon would be negative!
    $endgroup$
    – Dr. Sonnhard Graubner
    6 hours ago










  • $begingroup$
    Why would it be negative if $x ge y$?
    $endgroup$
    – Ash
    6 hours ago






  • 1




    $begingroup$
    1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
    $endgroup$
    – fleablood
    6 hours ago












  • $begingroup$
    I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
    $endgroup$
    – fleablood
    5 hours ago
















3












$begingroup$



If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.




Okay so I went about this by proving the contrapositive.



Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.



Is this a valid proof or not?






real-analysis proof-verification real-numbers







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    The conclusion should be $xleq y$.
    $endgroup$
    – gamma
    6 hours ago










  • $begingroup$
    If $$epsilon=x-y$$ then your epsilon would be negative!
    $endgroup$
    – Dr. Sonnhard Graubner
    6 hours ago










  • $begingroup$
    Why would it be negative if $x ge y$?
    $endgroup$
    – Ash
    6 hours ago






  • 1




    $begingroup$
    1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
    $endgroup$
    – fleablood
    6 hours ago












  • $begingroup$
    I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
    $endgroup$
    – fleablood
    5 hours ago














3












3








3





$begingroup$



If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.




Okay so I went about this by proving the contrapositive.



Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.



Is this a valid proof or not?






real-analysis proof-verification real-numbers







share|cite|improve this question











$endgroup$





If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.




Okay so I went about this by proving the contrapositive.



Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.



Is this a valid proof or not?






real-analysis proof-verification real-numbers




real-analysis proof-verification real-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









J.G.

29.4k22846




29.4k22846










asked 6 hours ago









AshAsh

384




384








  • 4




    $begingroup$
    The conclusion should be $xleq y$.
    $endgroup$
    – gamma
    6 hours ago










  • $begingroup$
    If $$epsilon=x-y$$ then your epsilon would be negative!
    $endgroup$
    – Dr. Sonnhard Graubner
    6 hours ago










  • $begingroup$
    Why would it be negative if $x ge y$?
    $endgroup$
    – Ash
    6 hours ago






  • 1




    $begingroup$
    1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
    $endgroup$
    – fleablood
    6 hours ago












  • $begingroup$
    I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
    $endgroup$
    – fleablood
    5 hours ago














  • 4




    $begingroup$
    The conclusion should be $xleq y$.
    $endgroup$
    – gamma
    6 hours ago










  • $begingroup$
    If $$epsilon=x-y$$ then your epsilon would be negative!
    $endgroup$
    – Dr. Sonnhard Graubner
    6 hours ago










  • $begingroup$
    Why would it be negative if $x ge y$?
    $endgroup$
    – Ash
    6 hours ago






  • 1




    $begingroup$
    1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
    $endgroup$
    – fleablood
    6 hours ago












  • $begingroup$
    I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
    $endgroup$
    – fleablood
    5 hours ago








4




4




$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
6 hours ago




$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
6 hours ago












$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago




$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
6 hours ago












$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
6 hours ago




$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
6 hours ago




1




1




$begingroup$
1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
$endgroup$
– fleablood
6 hours ago






$begingroup$
1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
$endgroup$
– fleablood
6 hours ago














$begingroup$
I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
$endgroup$
– fleablood
5 hours ago




$begingroup$
I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
$endgroup$
– fleablood
5 hours ago










3 Answers
3






active

oldest

votes


















5












$begingroup$

The statement you want to prove is not true.



Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



    The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



    Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      1) The statement is false.



      Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



      2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



      Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



      3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



      4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        5












        $begingroup$

        The statement you want to prove is not true.



        Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.






        share|cite|improve this answer









        $endgroup$


















          5












          $begingroup$

          The statement you want to prove is not true.



          Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.






          share|cite|improve this answer









          $endgroup$
















            5












            5








            5





            $begingroup$

            The statement you want to prove is not true.



            Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.






            share|cite|improve this answer









            $endgroup$



            The statement you want to prove is not true.



            Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 6 hours ago









            TheSilverDoeTheSilverDoe

            3,448112




            3,448112























                3












                $begingroup$

                The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



                The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



                Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



                  The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



                  Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



                    The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



                    Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.






                    share|cite|improve this answer









                    $endgroup$



                    The proof is invalid for the simple fact that the statement is false, so you cannot prove it.



                    The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.



                    Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 6 hours ago









                    egregegreg

                    184k1486205




                    184k1486205























                        0












                        $begingroup$

                        1) The statement is false.



                        Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



                        2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



                        Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



                        3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



                        4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          1) The statement is false.



                          Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



                          2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



                          Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



                          3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



                          4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            1) The statement is false.



                            Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



                            2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



                            Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



                            3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



                            4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.






                            share|cite|improve this answer









                            $endgroup$



                            1) The statement is false.



                            Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.



                            2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.



                            Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.



                            3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.



                            4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 6 hours ago









                            fleabloodfleablood

                            72.1k22687




                            72.1k22687






























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