Taking complex solution on $frac{1}{x^2 + 1} = 0$












1












$begingroup$


Is $i$ a complex solution to $$ frac{1}{x^2 + 1} = 0 ;?$$

Knowing that it has no real solutions, substituting $i$ leaves $frac{1}{0}$ which for me is equal to zero. Is this right?






algebra-precalculus complex-numbers







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Is $i$ a complex solution to $$ frac{1}{x^2 + 1} = 0 ;?$$

    Knowing that it has no real solutions, substituting $i$ leaves $frac{1}{0}$ which for me is equal to zero. Is this right?






    algebra-precalculus complex-numbers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Is $i$ a complex solution to $$ frac{1}{x^2 + 1} = 0 ;?$$

      Knowing that it has no real solutions, substituting $i$ leaves $frac{1}{0}$ which for me is equal to zero. Is this right?






      algebra-precalculus complex-numbers







      share|cite|improve this question











      $endgroup$




      Is $i$ a complex solution to $$ frac{1}{x^2 + 1} = 0 ;?$$

      Knowing that it has no real solutions, substituting $i$ leaves $frac{1}{0}$ which for me is equal to zero. Is this right?






      algebra-precalculus complex-numbers




      algebra-precalculus complex-numbers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 19 '18 at 20:53









      Eric Wofsey

      184k13211338




      184k13211338










      asked Nov 10 '18 at 10:34









      MMJMMMJM

      3351111




      3351111






















          4 Answers
          4






          active

          oldest

          votes


















          7












          $begingroup$

          No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



          $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



          The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



          The equation



          $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



          is false, as it has no solutions at all.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
            $endgroup$
            – MMJM
            Nov 10 '18 at 11:13












          • $begingroup$
            @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
            $endgroup$
            – Rebellos
            Nov 10 '18 at 11:15












          • $begingroup$
            I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
            $endgroup$
            – MMJM
            Nov 10 '18 at 11:20



















          2












          $begingroup$

          We have that



          $$frac{1}{x^2 + 1} = 0$$



          is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



          Indeed for any $zin mathbb{C}$ we have



          $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



          but



          $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$






          share|cite|improve this answer











          $endgroup$





















            1












            $begingroup$

            No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



            That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



              $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



              Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.






              share|cite|improve this answer









              $endgroup$













                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%2f2992467%2ftaking-complex-solution-on-frac1x2-1-0%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









                7












                $begingroup$

                No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



                $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



                The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



                The equation



                $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



                is false, as it has no solutions at all.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:13












                • $begingroup$
                  @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                  $endgroup$
                  – Rebellos
                  Nov 10 '18 at 11:15












                • $begingroup$
                  I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:20
















                7












                $begingroup$

                No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



                $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



                The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



                The equation



                $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



                is false, as it has no solutions at all.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:13












                • $begingroup$
                  @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                  $endgroup$
                  – Rebellos
                  Nov 10 '18 at 11:15












                • $begingroup$
                  I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:20














                7












                7








                7





                $begingroup$

                No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



                $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



                The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



                The equation



                $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



                is false, as it has no solutions at all.






                share|cite|improve this answer









                $endgroup$



                No, this is wrong. $1/0$ is definitely not equal to zero, it is rather an undefined expression. Note that in complex analysis functions are defined in similar ways. If you wanted to consider that expression over $mathbb C$, it would be :



                $$f(z) = frac{1}{z^2+1}, quad z in mathbb Csetminus{pm i}$$



                The denominator can never be zero no matter if you're studying strictly real numbers or complex numbers.



                The equation



                $$f(z) = 0 implies frac{1}{z^2+1} = 0$$



                is false, as it has no solutions at all.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 10 '18 at 10:38









                RebellosRebellos

                14.6k31247




                14.6k31247












                • $begingroup$
                  If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:13












                • $begingroup$
                  @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                  $endgroup$
                  – Rebellos
                  Nov 10 '18 at 11:15












                • $begingroup$
                  I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:20


















                • $begingroup$
                  If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:13












                • $begingroup$
                  @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                  $endgroup$
                  – Rebellos
                  Nov 10 '18 at 11:15












                • $begingroup$
                  I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                  $endgroup$
                  – MMJM
                  Nov 10 '18 at 11:20
















                $begingroup$
                If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                $endgroup$
                – MMJM
                Nov 10 '18 at 11:13






                $begingroup$
                If we solve for $$frac{1}{x^2 + 1} = frac{1}{0}$$ will it have $i$ be a solution or not ?(assuming that $frac{1}{0}$ would not be simplified) since begin{align} frac{1}{x^2 + 1} &= frac{1}{0}\ frac{1}{(i)^2 + 1} &= frac{1}{0}\ frac{1}{0} &= frac{1}{0} end{align}
                $endgroup$
                – MMJM
                Nov 10 '18 at 11:13














                $begingroup$
                @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                $endgroup$
                – Rebellos
                Nov 10 '18 at 11:15






                $begingroup$
                @MMJM You cannot use $1/0$ as an expression in whatever you do, it is not a properly defined mathematical substance. It is an undefined expression and that's why you yield contradictions when you use it. Recall that a fraction $frac{a}{b}$ is defined iff $b neq 0$.
                $endgroup$
                – Rebellos
                Nov 10 '18 at 11:15














                $begingroup$
                I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                $endgroup$
                – MMJM
                Nov 10 '18 at 11:20




                $begingroup$
                I misunderstood $frac{1}{0}$ at first when I posted the question. Now I get the point.
                $endgroup$
                – MMJM
                Nov 10 '18 at 11:20











                2












                $begingroup$

                We have that



                $$frac{1}{x^2 + 1} = 0$$



                is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



                Indeed for any $zin mathbb{C}$ we have



                $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



                but



                $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$






                share|cite|improve this answer











                $endgroup$


















                  2












                  $begingroup$

                  We have that



                  $$frac{1}{x^2 + 1} = 0$$



                  is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



                  Indeed for any $zin mathbb{C}$ we have



                  $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



                  but



                  $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$






                  share|cite|improve this answer











                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    We have that



                    $$frac{1}{x^2 + 1} = 0$$



                    is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



                    Indeed for any $zin mathbb{C}$ we have



                    $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



                    but



                    $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$






                    share|cite|improve this answer











                    $endgroup$



                    We have that



                    $$frac{1}{x^2 + 1} = 0$$



                    is an expression defined for $x^2+1 neq 0$ and it has not (real nor complex) solutions.



                    Indeed for any $zin mathbb{C}$ we have



                    $$frac{1}{z^2 + 1} = 0 implies left|frac{1}{z^2 + 1}right| = |0| $$



                    but



                    $$left|frac{1}{z^2 + 1}right|=frac{1}{|z^2 + 1|}neq 0quad forall z$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 10 '18 at 10:52

























                    answered Nov 10 '18 at 10:38









                    gimusigimusi

                    92.8k84494




                    92.8k84494























                        1












                        $begingroup$

                        No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



                        That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



                          That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



                            That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.






                            share|cite|improve this answer









                            $endgroup$



                            No, the complex field does not provide solution to $$frac {1}{x^2+1}=0$$



                            That is equivalent to $$1=0times (x^2+1)$$ and as you know the product of $0$ with any number is $0$ not $1$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 10 '18 at 10:42









                            Mohammad Riazi-KermaniMohammad Riazi-Kermani

                            41.5k42061




                            41.5k42061























                                0












                                $begingroup$

                                Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



                                $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



                                Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



                                  $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



                                  Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



                                    $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



                                    Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.






                                    share|cite|improve this answer









                                    $endgroup$



                                    Note that for $xinmathbb{C}$ and polynomials $P$ and $Q$,



                                    $$ frac{P(x)}{Q(x)} = 0 Leftrightarrow P(x) = 0 text{ and } Q(x)neq 0 $$



                                    Since in this case $P(x) equiv 1 neq 0$, there are no $x$ such that $frac{P(x)}{Q(x)} = 0$.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Nov 10 '18 at 10:43









                                    AlexanderJ93AlexanderJ93

                                    6,158823




                                    6,158823






























                                        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.




                                        draft saved


                                        draft discarded














                                        StackExchange.ready(
                                        function () {
                                        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2992467%2ftaking-complex-solution-on-frac1x2-1-0%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

                                        Bressuire

                                        Image
                                        .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                                        Read more

                                        Cabo Verde

                                        Image
                                          Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
                                        Read more

                                        Gyllenstierna

                                        Image
                                        .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid
                                        Read more