Proof explanation of $``exists xinmathbb{R}$ with $x^2=2"$












3














Can someone please help me break down the proof below from $(*)$ onwards. I'm lost at what is going on and where the proceeding steps are coming from. Is this a proof by contradiction? Why are we assuming $M^2lt 2 , M^2gt 2$, and choosing $delta$ to be the minimum of $frac{2-M^2}{5}$ and $1$?



Theorem 2.13
(Square root 2 exists). There exists $xinmathbb{R}$ with $x^2=2.$



Proof.



Define $$A={yinmathbb{R}|y^2 leq{2}}.$$



As $0^2leq{2}$, we have $0in{A}$, so $A$ is non-empty. Now suppose $yinmathbb{R}$ has $ygeq{2}$, then $y^2geq{4}$, so $ynotin{A}$. Thus, $2$ is an upper bound for A and hence A is bounded above. Therefore, by the completeness axiom, $M=sup(A)$ exists. Note that $Mgeq{1}$ as $1in{A}$ and $Mleq{2}$ is an upper bound for A.



$(*)$



Suppose $M^2lt{2}$. Choose $deltagt{0}$ with $deltalt $ min$(frac{2-M^2}{5},1)$, and then define $ y=M+delta$. As $deltalt{1}$, we have $delta^2 lt{delta}$ and so



$$y^2=(M+delta)^2=M^2+2Mdelta+delta^2leq M^2+4delta+delta=M^2+5deltalt 2.$$



Thus $yin A$, but this is a contradiction as $ygt M$. Therefore $M^2geq 2$.



Suppose $M^2gt 2$. Choose $delta gt 0$ with $deltalt $min$(M,frac{M^2-2}{2M})$ so that $M^2-2Mdelta gt 2$ and $M-deltagt 0$. Then,



$$(M-delta)^2=M^2-2Mdelta+delta^2geq M^2-2Mdeltagt 2.$$



Now if $ygeq (M-delta)$, then $y^2geq (M-delta)^2 gt 2$ so $ynotin A$. Thus $M-delta$ is an upper bound for $A$, contradicting the fact that $M$ is the supremum of $A$



Since $M^2 lt 2$ and $M^2gt 2$ both lead to contradictions, we conclude that $M^2=2$, as required.










share|cite|improve this question





























    3














    Can someone please help me break down the proof below from $(*)$ onwards. I'm lost at what is going on and where the proceeding steps are coming from. Is this a proof by contradiction? Why are we assuming $M^2lt 2 , M^2gt 2$, and choosing $delta$ to be the minimum of $frac{2-M^2}{5}$ and $1$?



    Theorem 2.13
    (Square root 2 exists). There exists $xinmathbb{R}$ with $x^2=2.$



    Proof.



    Define $$A={yinmathbb{R}|y^2 leq{2}}.$$



    As $0^2leq{2}$, we have $0in{A}$, so $A$ is non-empty. Now suppose $yinmathbb{R}$ has $ygeq{2}$, then $y^2geq{4}$, so $ynotin{A}$. Thus, $2$ is an upper bound for A and hence A is bounded above. Therefore, by the completeness axiom, $M=sup(A)$ exists. Note that $Mgeq{1}$ as $1in{A}$ and $Mleq{2}$ is an upper bound for A.



    $(*)$



    Suppose $M^2lt{2}$. Choose $deltagt{0}$ with $deltalt $ min$(frac{2-M^2}{5},1)$, and then define $ y=M+delta$. As $deltalt{1}$, we have $delta^2 lt{delta}$ and so



    $$y^2=(M+delta)^2=M^2+2Mdelta+delta^2leq M^2+4delta+delta=M^2+5deltalt 2.$$



    Thus $yin A$, but this is a contradiction as $ygt M$. Therefore $M^2geq 2$.



    Suppose $M^2gt 2$. Choose $delta gt 0$ with $deltalt $min$(M,frac{M^2-2}{2M})$ so that $M^2-2Mdelta gt 2$ and $M-deltagt 0$. Then,



    $$(M-delta)^2=M^2-2Mdelta+delta^2geq M^2-2Mdeltagt 2.$$



    Now if $ygeq (M-delta)$, then $y^2geq (M-delta)^2 gt 2$ so $ynotin A$. Thus $M-delta$ is an upper bound for $A$, contradicting the fact that $M$ is the supremum of $A$



    Since $M^2 lt 2$ and $M^2gt 2$ both lead to contradictions, we conclude that $M^2=2$, as required.










    share|cite|improve this question



























      3












      3








      3


      1





      Can someone please help me break down the proof below from $(*)$ onwards. I'm lost at what is going on and where the proceeding steps are coming from. Is this a proof by contradiction? Why are we assuming $M^2lt 2 , M^2gt 2$, and choosing $delta$ to be the minimum of $frac{2-M^2}{5}$ and $1$?



      Theorem 2.13
      (Square root 2 exists). There exists $xinmathbb{R}$ with $x^2=2.$



      Proof.



      Define $$A={yinmathbb{R}|y^2 leq{2}}.$$



      As $0^2leq{2}$, we have $0in{A}$, so $A$ is non-empty. Now suppose $yinmathbb{R}$ has $ygeq{2}$, then $y^2geq{4}$, so $ynotin{A}$. Thus, $2$ is an upper bound for A and hence A is bounded above. Therefore, by the completeness axiom, $M=sup(A)$ exists. Note that $Mgeq{1}$ as $1in{A}$ and $Mleq{2}$ is an upper bound for A.



      $(*)$



      Suppose $M^2lt{2}$. Choose $deltagt{0}$ with $deltalt $ min$(frac{2-M^2}{5},1)$, and then define $ y=M+delta$. As $deltalt{1}$, we have $delta^2 lt{delta}$ and so



      $$y^2=(M+delta)^2=M^2+2Mdelta+delta^2leq M^2+4delta+delta=M^2+5deltalt 2.$$



      Thus $yin A$, but this is a contradiction as $ygt M$. Therefore $M^2geq 2$.



      Suppose $M^2gt 2$. Choose $delta gt 0$ with $deltalt $min$(M,frac{M^2-2}{2M})$ so that $M^2-2Mdelta gt 2$ and $M-deltagt 0$. Then,



      $$(M-delta)^2=M^2-2Mdelta+delta^2geq M^2-2Mdeltagt 2.$$



      Now if $ygeq (M-delta)$, then $y^2geq (M-delta)^2 gt 2$ so $ynotin A$. Thus $M-delta$ is an upper bound for $A$, contradicting the fact that $M$ is the supremum of $A$



      Since $M^2 lt 2$ and $M^2gt 2$ both lead to contradictions, we conclude that $M^2=2$, as required.










      share|cite|improve this question















      Can someone please help me break down the proof below from $(*)$ onwards. I'm lost at what is going on and where the proceeding steps are coming from. Is this a proof by contradiction? Why are we assuming $M^2lt 2 , M^2gt 2$, and choosing $delta$ to be the minimum of $frac{2-M^2}{5}$ and $1$?



      Theorem 2.13
      (Square root 2 exists). There exists $xinmathbb{R}$ with $x^2=2.$



      Proof.



      Define $$A={yinmathbb{R}|y^2 leq{2}}.$$



      As $0^2leq{2}$, we have $0in{A}$, so $A$ is non-empty. Now suppose $yinmathbb{R}$ has $ygeq{2}$, then $y^2geq{4}$, so $ynotin{A}$. Thus, $2$ is an upper bound for A and hence A is bounded above. Therefore, by the completeness axiom, $M=sup(A)$ exists. Note that $Mgeq{1}$ as $1in{A}$ and $Mleq{2}$ is an upper bound for A.



      $(*)$



      Suppose $M^2lt{2}$. Choose $deltagt{0}$ with $deltalt $ min$(frac{2-M^2}{5},1)$, and then define $ y=M+delta$. As $deltalt{1}$, we have $delta^2 lt{delta}$ and so



      $$y^2=(M+delta)^2=M^2+2Mdelta+delta^2leq M^2+4delta+delta=M^2+5deltalt 2.$$



      Thus $yin A$, but this is a contradiction as $ygt M$. Therefore $M^2geq 2$.



      Suppose $M^2gt 2$. Choose $delta gt 0$ with $deltalt $min$(M,frac{M^2-2}{2M})$ so that $M^2-2Mdelta gt 2$ and $M-deltagt 0$. Then,



      $$(M-delta)^2=M^2-2Mdelta+delta^2geq M^2-2Mdeltagt 2.$$



      Now if $ygeq (M-delta)$, then $y^2geq (M-delta)^2 gt 2$ so $ynotin A$. Thus $M-delta$ is an upper bound for $A$, contradicting the fact that $M$ is the supremum of $A$



      Since $M^2 lt 2$ and $M^2gt 2$ both lead to contradictions, we conclude that $M^2=2$, as required.







      real-analysis proof-explanation intuition real-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 3 at 23:46









      Bernard

      118k639112




      118k639112










      asked Jan 3 at 23:42









      Imran

      1815




      1815






















          4 Answers
          4






          active

          oldest

          votes


















          2














          The last part is a proof by contradiction, using the fact that $M^2$ and $2$ are related either by $M^2 < 2$ or $2 < M^2$ or $M^2 = 2$, exclusively. In both cases, we contradict the supremum property of $M$; that $M$ is greater than or equal to every number in $A$, and that it is the least such number with that property.



          First, assuming $M^2 < 2$, we construct a $y > M$ but $y in A$, which is impossible if $M$ is the supremum. It's a "little" more than $M$, so add a positive $delta$ "fudge factor." We have complete control of how little or big that $delta$ is; i.e. $y = M + delta$. So, if we want $yin A, y^2 < 2$, we want $(M + delta)^2 < 2$. We will work towards this inequality "backwards", although the engineering of the different number choices are much clearer in this direction.



          Start with $(M + delta)^2$, and expand it. Two things happen in $$M^2 + 2Mdelta + delta^2 leq M^2 + 5delta$$



          First, we use $M leq 2$ to get $2Mdelta leq 4delta$. Then, we can choose for $delta < 1$ so that $delta^2 < delta$.



          To get this inequality: $$M^2 + 5delta < 2$$



          We can actually work backwards, and choose $delta < frac{2 - M^2}{5}$. If we choose $delta < min(frac{2 - M^2}{5}, 1)$, then both the inequalities we need will be true. It's a lot more clear where they come from working backwards, though.



          I hope this helps you understand the $M^2 > 2$ part, as well, which works on a similar reasoning scheme.






          share|cite|improve this answer





























            2














            Do you know the trichotomy law of inequality? It says that for any two numbers $a,b$, either $a < b$, or $a > b$, or $a=b$.



            So now if you can prove that $a<b$ leads to a contradiction, and that $a > b$ leads to a contradiction, the only possibility left must be true, namely $a=b$. Apply this with $a=M^2$ and $b=2$.






            share|cite|improve this answer





























              2














              Yes, it is a proof by contradicion. The author proves the supremum $M$ satisfies the equality $M^2$ by deducing a contradiction in each of the other two cases, $M^2<2$ and $M^2>2$.



              In the first case, he/she obtains a contradiction by finding an element $y=M+delta$ which is both in $A$ but greater than $sup A$.



              In the second case, he/she finds an upper bound for $A$ which is smaller than the least upper bound.



              $delta<minleft(frac{2-M^2}{5},1right)$ simply means that we have both $delta< frac{2-M^2}{5}$ and $delta<1$. These conditions are technically necessary (i.e. $delta$ must be small enough) to deduce the contradiction.






              share|cite|improve this answer































                0














                Indeed, the proof is a bit confusing as it sprawls out with variable definitions and claims. Here we want to calmly prove part of the argument in theorem 2.13, highlighting some of the thinking behind the mechanics.



                Proposition: Let $r gt 0$ be a real number satisfying $r^2 gt 2$.

                Then there exists another real number $s$ with $0 lt s lt r$ satisfying $s^2 gt 2$.

                Proof

                We can write any such number $s$ as $r - delta$ with $delta gt 0$. So we are looking for any $delta$ 'that works', leading us to analyze



                $$tag 1 s^2 = (r - delta)^2= r^2 - 2delta r + delta^2$$



                Now certainly if $r^2 - 2delta r = 2$ then $s^2 = 2 + delta^2 gt 2$. Solving for $delta$,



                $$tag 2 delta = frac{r^2 - 2}{2r}$$



                So $delta$ is positive and $s = r - delta lt r$ is guaranteed. We also have



                $quad s gt 0 , text{ iff } , 2r^2 - (r^2 -2) gt 0 , text{ iff } , r^2 + 2 gt 0$



                and so defining



                $$tag 3 s = r - frac{r^2 - 2}{2r}$$



                satisfies the requirements of proposition, which completes the proof. $quad blacksquare$



                There are any number of ways of putting together the proof of the OP's theorem, one way being to incorporate the above proposition.



                Note: Using and extending the above logic, an algorithm can be implemented to calculate square roots - the Babylonian method.






                share|cite|improve this answer























                  Your Answer





                  StackExchange.ifUsing("editor", function () {
                  return StackExchange.using("mathjaxEditing", function () {
                  StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
                  StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
                  });
                  });
                  }, "mathjax-editing");

                  StackExchange.ready(function() {
                  var channelOptions = {
                  tags: "".split(" "),
                  id: "69"
                  };
                  initTagRenderer("".split(" "), "".split(" "), channelOptions);

                  StackExchange.using("externalEditor", function() {
                  // Have to fire editor after snippets, if snippets enabled
                  if (StackExchange.settings.snippets.snippetsEnabled) {
                  StackExchange.using("snippets", function() {
                  createEditor();
                  });
                  }
                  else {
                  createEditor();
                  }
                  });

                  function createEditor() {
                  StackExchange.prepareEditor({
                  heartbeatType: 'answer',
                  autoActivateHeartbeat: false,
                  convertImagesToLinks: true,
                  noModals: true,
                  showLowRepImageUploadWarning: true,
                  reputationToPostImages: 10,
                  bindNavPrevention: true,
                  postfix: "",
                  imageUploader: {
                  brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                  contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                  allowUrls: true
                  },
                  noCode: true, onDemand: true,
                  discardSelector: ".discard-answer"
                  ,immediatelyShowMarkdownHelp:true
                  });


                  }
                  });














                  draft saved

                  draft discarded


















                  StackExchange.ready(
                  function () {
                  StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061153%2fproof-explanation-of-exists-x-in-mathbbr-with-x2-2%23new-answer', 'question_page');
                  }
                  );

                  Post as a guest















                  Required, but never shown

























                  4 Answers
                  4






                  active

                  oldest

                  votes








                  4 Answers
                  4






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  2














                  The last part is a proof by contradiction, using the fact that $M^2$ and $2$ are related either by $M^2 < 2$ or $2 < M^2$ or $M^2 = 2$, exclusively. In both cases, we contradict the supremum property of $M$; that $M$ is greater than or equal to every number in $A$, and that it is the least such number with that property.



                  First, assuming $M^2 < 2$, we construct a $y > M$ but $y in A$, which is impossible if $M$ is the supremum. It's a "little" more than $M$, so add a positive $delta$ "fudge factor." We have complete control of how little or big that $delta$ is; i.e. $y = M + delta$. So, if we want $yin A, y^2 < 2$, we want $(M + delta)^2 < 2$. We will work towards this inequality "backwards", although the engineering of the different number choices are much clearer in this direction.



                  Start with $(M + delta)^2$, and expand it. Two things happen in $$M^2 + 2Mdelta + delta^2 leq M^2 + 5delta$$



                  First, we use $M leq 2$ to get $2Mdelta leq 4delta$. Then, we can choose for $delta < 1$ so that $delta^2 < delta$.



                  To get this inequality: $$M^2 + 5delta < 2$$



                  We can actually work backwards, and choose $delta < frac{2 - M^2}{5}$. If we choose $delta < min(frac{2 - M^2}{5}, 1)$, then both the inequalities we need will be true. It's a lot more clear where they come from working backwards, though.



                  I hope this helps you understand the $M^2 > 2$ part, as well, which works on a similar reasoning scheme.






                  share|cite|improve this answer


























                    2














                    The last part is a proof by contradiction, using the fact that $M^2$ and $2$ are related either by $M^2 < 2$ or $2 < M^2$ or $M^2 = 2$, exclusively. In both cases, we contradict the supremum property of $M$; that $M$ is greater than or equal to every number in $A$, and that it is the least such number with that property.



                    First, assuming $M^2 < 2$, we construct a $y > M$ but $y in A$, which is impossible if $M$ is the supremum. It's a "little" more than $M$, so add a positive $delta$ "fudge factor." We have complete control of how little or big that $delta$ is; i.e. $y = M + delta$. So, if we want $yin A, y^2 < 2$, we want $(M + delta)^2 < 2$. We will work towards this inequality "backwards", although the engineering of the different number choices are much clearer in this direction.



                    Start with $(M + delta)^2$, and expand it. Two things happen in $$M^2 + 2Mdelta + delta^2 leq M^2 + 5delta$$



                    First, we use $M leq 2$ to get $2Mdelta leq 4delta$. Then, we can choose for $delta < 1$ so that $delta^2 < delta$.



                    To get this inequality: $$M^2 + 5delta < 2$$



                    We can actually work backwards, and choose $delta < frac{2 - M^2}{5}$. If we choose $delta < min(frac{2 - M^2}{5}, 1)$, then both the inequalities we need will be true. It's a lot more clear where they come from working backwards, though.



                    I hope this helps you understand the $M^2 > 2$ part, as well, which works on a similar reasoning scheme.






                    share|cite|improve this answer
























                      2












                      2








                      2






                      The last part is a proof by contradiction, using the fact that $M^2$ and $2$ are related either by $M^2 < 2$ or $2 < M^2$ or $M^2 = 2$, exclusively. In both cases, we contradict the supremum property of $M$; that $M$ is greater than or equal to every number in $A$, and that it is the least such number with that property.



                      First, assuming $M^2 < 2$, we construct a $y > M$ but $y in A$, which is impossible if $M$ is the supremum. It's a "little" more than $M$, so add a positive $delta$ "fudge factor." We have complete control of how little or big that $delta$ is; i.e. $y = M + delta$. So, if we want $yin A, y^2 < 2$, we want $(M + delta)^2 < 2$. We will work towards this inequality "backwards", although the engineering of the different number choices are much clearer in this direction.



                      Start with $(M + delta)^2$, and expand it. Two things happen in $$M^2 + 2Mdelta + delta^2 leq M^2 + 5delta$$



                      First, we use $M leq 2$ to get $2Mdelta leq 4delta$. Then, we can choose for $delta < 1$ so that $delta^2 < delta$.



                      To get this inequality: $$M^2 + 5delta < 2$$



                      We can actually work backwards, and choose $delta < frac{2 - M^2}{5}$. If we choose $delta < min(frac{2 - M^2}{5}, 1)$, then both the inequalities we need will be true. It's a lot more clear where they come from working backwards, though.



                      I hope this helps you understand the $M^2 > 2$ part, as well, which works on a similar reasoning scheme.






                      share|cite|improve this answer












                      The last part is a proof by contradiction, using the fact that $M^2$ and $2$ are related either by $M^2 < 2$ or $2 < M^2$ or $M^2 = 2$, exclusively. In both cases, we contradict the supremum property of $M$; that $M$ is greater than or equal to every number in $A$, and that it is the least such number with that property.



                      First, assuming $M^2 < 2$, we construct a $y > M$ but $y in A$, which is impossible if $M$ is the supremum. It's a "little" more than $M$, so add a positive $delta$ "fudge factor." We have complete control of how little or big that $delta$ is; i.e. $y = M + delta$. So, if we want $yin A, y^2 < 2$, we want $(M + delta)^2 < 2$. We will work towards this inequality "backwards", although the engineering of the different number choices are much clearer in this direction.



                      Start with $(M + delta)^2$, and expand it. Two things happen in $$M^2 + 2Mdelta + delta^2 leq M^2 + 5delta$$



                      First, we use $M leq 2$ to get $2Mdelta leq 4delta$. Then, we can choose for $delta < 1$ so that $delta^2 < delta$.



                      To get this inequality: $$M^2 + 5delta < 2$$



                      We can actually work backwards, and choose $delta < frac{2 - M^2}{5}$. If we choose $delta < min(frac{2 - M^2}{5}, 1)$, then both the inequalities we need will be true. It's a lot more clear where they come from working backwards, though.



                      I hope this helps you understand the $M^2 > 2$ part, as well, which works on a similar reasoning scheme.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 4 at 0:00









                      Larry B.

                      2,776728




                      2,776728























                          2














                          Do you know the trichotomy law of inequality? It says that for any two numbers $a,b$, either $a < b$, or $a > b$, or $a=b$.



                          So now if you can prove that $a<b$ leads to a contradiction, and that $a > b$ leads to a contradiction, the only possibility left must be true, namely $a=b$. Apply this with $a=M^2$ and $b=2$.






                          share|cite|improve this answer


























                            2














                            Do you know the trichotomy law of inequality? It says that for any two numbers $a,b$, either $a < b$, or $a > b$, or $a=b$.



                            So now if you can prove that $a<b$ leads to a contradiction, and that $a > b$ leads to a contradiction, the only possibility left must be true, namely $a=b$. Apply this with $a=M^2$ and $b=2$.






                            share|cite|improve this answer
























                              2












                              2








                              2






                              Do you know the trichotomy law of inequality? It says that for any two numbers $a,b$, either $a < b$, or $a > b$, or $a=b$.



                              So now if you can prove that $a<b$ leads to a contradiction, and that $a > b$ leads to a contradiction, the only possibility left must be true, namely $a=b$. Apply this with $a=M^2$ and $b=2$.






                              share|cite|improve this answer












                              Do you know the trichotomy law of inequality? It says that for any two numbers $a,b$, either $a < b$, or $a > b$, or $a=b$.



                              So now if you can prove that $a<b$ leads to a contradiction, and that $a > b$ leads to a contradiction, the only possibility left must be true, namely $a=b$. Apply this with $a=M^2$ and $b=2$.







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 3 at 23:55









                              Lee Mosher

                              48.2k33681




                              48.2k33681























                                  2














                                  Yes, it is a proof by contradicion. The author proves the supremum $M$ satisfies the equality $M^2$ by deducing a contradiction in each of the other two cases, $M^2<2$ and $M^2>2$.



                                  In the first case, he/she obtains a contradiction by finding an element $y=M+delta$ which is both in $A$ but greater than $sup A$.



                                  In the second case, he/she finds an upper bound for $A$ which is smaller than the least upper bound.



                                  $delta<minleft(frac{2-M^2}{5},1right)$ simply means that we have both $delta< frac{2-M^2}{5}$ and $delta<1$. These conditions are technically necessary (i.e. $delta$ must be small enough) to deduce the contradiction.






                                  share|cite|improve this answer




























                                    2














                                    Yes, it is a proof by contradicion. The author proves the supremum $M$ satisfies the equality $M^2$ by deducing a contradiction in each of the other two cases, $M^2<2$ and $M^2>2$.



                                    In the first case, he/she obtains a contradiction by finding an element $y=M+delta$ which is both in $A$ but greater than $sup A$.



                                    In the second case, he/she finds an upper bound for $A$ which is smaller than the least upper bound.



                                    $delta<minleft(frac{2-M^2}{5},1right)$ simply means that we have both $delta< frac{2-M^2}{5}$ and $delta<1$. These conditions are technically necessary (i.e. $delta$ must be small enough) to deduce the contradiction.






                                    share|cite|improve this answer


























                                      2












                                      2








                                      2






                                      Yes, it is a proof by contradicion. The author proves the supremum $M$ satisfies the equality $M^2$ by deducing a contradiction in each of the other two cases, $M^2<2$ and $M^2>2$.



                                      In the first case, he/she obtains a contradiction by finding an element $y=M+delta$ which is both in $A$ but greater than $sup A$.



                                      In the second case, he/she finds an upper bound for $A$ which is smaller than the least upper bound.



                                      $delta<minleft(frac{2-M^2}{5},1right)$ simply means that we have both $delta< frac{2-M^2}{5}$ and $delta<1$. These conditions are technically necessary (i.e. $delta$ must be small enough) to deduce the contradiction.






                                      share|cite|improve this answer














                                      Yes, it is a proof by contradicion. The author proves the supremum $M$ satisfies the equality $M^2$ by deducing a contradiction in each of the other two cases, $M^2<2$ and $M^2>2$.



                                      In the first case, he/she obtains a contradiction by finding an element $y=M+delta$ which is both in $A$ but greater than $sup A$.



                                      In the second case, he/she finds an upper bound for $A$ which is smaller than the least upper bound.



                                      $delta<minleft(frac{2-M^2}{5},1right)$ simply means that we have both $delta< frac{2-M^2}{5}$ and $delta<1$. These conditions are technically necessary (i.e. $delta$ must be small enough) to deduce the contradiction.







                                      share|cite|improve this answer














                                      share|cite|improve this answer



                                      share|cite|improve this answer








                                      edited 2 days ago

























                                      answered Jan 4 at 0:03









                                      Bernard

                                      118k639112




                                      118k639112























                                          0














                                          Indeed, the proof is a bit confusing as it sprawls out with variable definitions and claims. Here we want to calmly prove part of the argument in theorem 2.13, highlighting some of the thinking behind the mechanics.



                                          Proposition: Let $r gt 0$ be a real number satisfying $r^2 gt 2$.

                                          Then there exists another real number $s$ with $0 lt s lt r$ satisfying $s^2 gt 2$.

                                          Proof

                                          We can write any such number $s$ as $r - delta$ with $delta gt 0$. So we are looking for any $delta$ 'that works', leading us to analyze



                                          $$tag 1 s^2 = (r - delta)^2= r^2 - 2delta r + delta^2$$



                                          Now certainly if $r^2 - 2delta r = 2$ then $s^2 = 2 + delta^2 gt 2$. Solving for $delta$,



                                          $$tag 2 delta = frac{r^2 - 2}{2r}$$



                                          So $delta$ is positive and $s = r - delta lt r$ is guaranteed. We also have



                                          $quad s gt 0 , text{ iff } , 2r^2 - (r^2 -2) gt 0 , text{ iff } , r^2 + 2 gt 0$



                                          and so defining



                                          $$tag 3 s = r - frac{r^2 - 2}{2r}$$



                                          satisfies the requirements of proposition, which completes the proof. $quad blacksquare$



                                          There are any number of ways of putting together the proof of the OP's theorem, one way being to incorporate the above proposition.



                                          Note: Using and extending the above logic, an algorithm can be implemented to calculate square roots - the Babylonian method.






                                          share|cite|improve this answer




























                                            0














                                            Indeed, the proof is a bit confusing as it sprawls out with variable definitions and claims. Here we want to calmly prove part of the argument in theorem 2.13, highlighting some of the thinking behind the mechanics.



                                            Proposition: Let $r gt 0$ be a real number satisfying $r^2 gt 2$.

                                            Then there exists another real number $s$ with $0 lt s lt r$ satisfying $s^2 gt 2$.

                                            Proof

                                            We can write any such number $s$ as $r - delta$ with $delta gt 0$. So we are looking for any $delta$ 'that works', leading us to analyze



                                            $$tag 1 s^2 = (r - delta)^2= r^2 - 2delta r + delta^2$$



                                            Now certainly if $r^2 - 2delta r = 2$ then $s^2 = 2 + delta^2 gt 2$. Solving for $delta$,



                                            $$tag 2 delta = frac{r^2 - 2}{2r}$$



                                            So $delta$ is positive and $s = r - delta lt r$ is guaranteed. We also have



                                            $quad s gt 0 , text{ iff } , 2r^2 - (r^2 -2) gt 0 , text{ iff } , r^2 + 2 gt 0$



                                            and so defining



                                            $$tag 3 s = r - frac{r^2 - 2}{2r}$$



                                            satisfies the requirements of proposition, which completes the proof. $quad blacksquare$



                                            There are any number of ways of putting together the proof of the OP's theorem, one way being to incorporate the above proposition.



                                            Note: Using and extending the above logic, an algorithm can be implemented to calculate square roots - the Babylonian method.






                                            share|cite|improve this answer


























                                              0












                                              0








                                              0






                                              Indeed, the proof is a bit confusing as it sprawls out with variable definitions and claims. Here we want to calmly prove part of the argument in theorem 2.13, highlighting some of the thinking behind the mechanics.



                                              Proposition: Let $r gt 0$ be a real number satisfying $r^2 gt 2$.

                                              Then there exists another real number $s$ with $0 lt s lt r$ satisfying $s^2 gt 2$.

                                              Proof

                                              We can write any such number $s$ as $r - delta$ with $delta gt 0$. So we are looking for any $delta$ 'that works', leading us to analyze



                                              $$tag 1 s^2 = (r - delta)^2= r^2 - 2delta r + delta^2$$



                                              Now certainly if $r^2 - 2delta r = 2$ then $s^2 = 2 + delta^2 gt 2$. Solving for $delta$,



                                              $$tag 2 delta = frac{r^2 - 2}{2r}$$



                                              So $delta$ is positive and $s = r - delta lt r$ is guaranteed. We also have



                                              $quad s gt 0 , text{ iff } , 2r^2 - (r^2 -2) gt 0 , text{ iff } , r^2 + 2 gt 0$



                                              and so defining



                                              $$tag 3 s = r - frac{r^2 - 2}{2r}$$



                                              satisfies the requirements of proposition, which completes the proof. $quad blacksquare$



                                              There are any number of ways of putting together the proof of the OP's theorem, one way being to incorporate the above proposition.



                                              Note: Using and extending the above logic, an algorithm can be implemented to calculate square roots - the Babylonian method.






                                              share|cite|improve this answer














                                              Indeed, the proof is a bit confusing as it sprawls out with variable definitions and claims. Here we want to calmly prove part of the argument in theorem 2.13, highlighting some of the thinking behind the mechanics.



                                              Proposition: Let $r gt 0$ be a real number satisfying $r^2 gt 2$.

                                              Then there exists another real number $s$ with $0 lt s lt r$ satisfying $s^2 gt 2$.

                                              Proof

                                              We can write any such number $s$ as $r - delta$ with $delta gt 0$. So we are looking for any $delta$ 'that works', leading us to analyze



                                              $$tag 1 s^2 = (r - delta)^2= r^2 - 2delta r + delta^2$$



                                              Now certainly if $r^2 - 2delta r = 2$ then $s^2 = 2 + delta^2 gt 2$. Solving for $delta$,



                                              $$tag 2 delta = frac{r^2 - 2}{2r}$$



                                              So $delta$ is positive and $s = r - delta lt r$ is guaranteed. We also have



                                              $quad s gt 0 , text{ iff } , 2r^2 - (r^2 -2) gt 0 , text{ iff } , r^2 + 2 gt 0$



                                              and so defining



                                              $$tag 3 s = r - frac{r^2 - 2}{2r}$$



                                              satisfies the requirements of proposition, which completes the proof. $quad blacksquare$



                                              There are any number of ways of putting together the proof of the OP's theorem, one way being to incorporate the above proposition.



                                              Note: Using and extending the above logic, an algorithm can be implemented to calculate square roots - the Babylonian method.







                                              share|cite|improve this answer














                                              share|cite|improve this answer



                                              share|cite|improve this answer








                                              edited 2 days ago

























                                              answered 2 days ago









                                              CopyPasteIt

                                              4,0651627




                                              4,0651627






























                                                  draft saved

                                                  draft discarded




















































                                                  Thanks for contributing an answer to Mathematics Stack Exchange!


                                                  • Please be sure to answer the question. Provide details and share your research!

                                                  But avoid



                                                  • Asking for help, clarification, or responding to other answers.

                                                  • Making statements based on opinion; back them up with references or personal experience.


                                                  Use MathJax to format equations. MathJax reference.


                                                  To learn more, see our tips on writing great answers.





                                                  Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                                                  Please pay close attention to the following guidance:


                                                  • Please be sure to answer the question. Provide details and share your research!

                                                  But avoid



                                                  • Asking for help, clarification, or responding to other answers.

                                                  • Making statements based on opinion; back them up with references or personal experience.


                                                  To learn more, see our tips on writing great answers.




                                                  draft saved


                                                  draft discarded














                                                  StackExchange.ready(
                                                  function () {
                                                  StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061153%2fproof-explanation-of-exists-x-in-mathbbr-with-x2-2%23new-answer', 'question_page');
                                                  }
                                                  );

                                                  Post as a guest















                                                  Required, but never shown





















































                                                  Required, but never shown














                                                  Required, but never shown












                                                  Required, but never shown







                                                  Required, but never shown

































                                                  Required, but never shown














                                                  Required, but never shown












                                                  Required, but never shown







                                                  Required, but never shown







                                                  Popular posts from this blog

                                                  1300-talet

                                                  1300-talet

                                                  Display a custom attribute below product name in the front-end Magento 1.9.3.8