Multiplication and division by addition and subtraction

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite
2












I know that this sounds like the stupidest question possible, but I had to ask, is it possible to express any multiplication and division solely by addition and subtraction? Such as... 7*0.3 or 7/0.3?



You can express sine and cosine by a Taylor series so I wondered this purely out of curiosity. My searching didn't seem to come up with an answer anywhere though hence I'm asking.










share|cite|improve this question

























    up vote
    1
    down vote

    favorite
    2












    I know that this sounds like the stupidest question possible, but I had to ask, is it possible to express any multiplication and division solely by addition and subtraction? Such as... 7*0.3 or 7/0.3?



    You can express sine and cosine by a Taylor series so I wondered this purely out of curiosity. My searching didn't seem to come up with an answer anywhere though hence I'm asking.










    share|cite|improve this question























      up vote
      1
      down vote

      favorite
      2









      up vote
      1
      down vote

      favorite
      2






      2





      I know that this sounds like the stupidest question possible, but I had to ask, is it possible to express any multiplication and division solely by addition and subtraction? Such as... 7*0.3 or 7/0.3?



      You can express sine and cosine by a Taylor series so I wondered this purely out of curiosity. My searching didn't seem to come up with an answer anywhere though hence I'm asking.










      share|cite|improve this question













      I know that this sounds like the stupidest question possible, but I had to ask, is it possible to express any multiplication and division solely by addition and subtraction? Such as... 7*0.3 or 7/0.3?



      You can express sine and cosine by a Taylor series so I wondered this purely out of curiosity. My searching didn't seem to come up with an answer anywhere though hence I'm asking.







      functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Aug 30 at 4:26









      John Ernest

      83




      83




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          0
          down vote



          accepted










          Well,first let's look at the rules. For multiplication,



          if, $x,yin mathbbQ$
          $$x.y=underbrace(x+x+x+x+...+x)_texty times$$



          For division, if
          $$ x/y=n$$
          $$Then,x=y.n$$
          $$means,x-underbrace(y+y+y+...+y)_textn times=0$$



          Now,lets work on your given numbers.
          Here, $$7*0.3=7*dfrac310$$
          That means we have to multiply 7, 3 times.than yiels $implies$
          $$7*3=7+7+7=21$$
          Now, we have $$dfrac2110$$
          Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
          $$21-(10+10)=1$$
          1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$






          share|cite|improve this answer





























            up vote
            0
            down vote













            Yes it is possible



            $$xcdot y=
            underbracex+x+cdots+x _text$y$ times$$



            if $$fracxy=n$$
            $$0=x-underbrace(y+y+cdots+y)_text $n$ times$$



            $fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0



            But here is one limitation $x,y$ belongs to Rational Numbers






            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',
              convertImagesToLinks: true,
              noModals: false,
              showLowRepImageUploadWarning: true,
              reputationToPostImages: 10,
              bindNavPrevention: true,
              postfix: "",
              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%2f2899123%2fmultiplication-and-division-by-addition-and-subtraction%23new-answer', 'question_page');

              );

              Post as a guest






























              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              0
              down vote



              accepted










              Well,first let's look at the rules. For multiplication,



              if, $x,yin mathbbQ$
              $$x.y=underbrace(x+x+x+x+...+x)_texty times$$



              For division, if
              $$ x/y=n$$
              $$Then,x=y.n$$
              $$means,x-underbrace(y+y+y+...+y)_textn times=0$$



              Now,lets work on your given numbers.
              Here, $$7*0.3=7*dfrac310$$
              That means we have to multiply 7, 3 times.than yiels $implies$
              $$7*3=7+7+7=21$$
              Now, we have $$dfrac2110$$
              Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
              $$21-(10+10)=1$$
              1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$






              share|cite|improve this answer


























                up vote
                0
                down vote



                accepted










                Well,first let's look at the rules. For multiplication,



                if, $x,yin mathbbQ$
                $$x.y=underbrace(x+x+x+x+...+x)_texty times$$



                For division, if
                $$ x/y=n$$
                $$Then,x=y.n$$
                $$means,x-underbrace(y+y+y+...+y)_textn times=0$$



                Now,lets work on your given numbers.
                Here, $$7*0.3=7*dfrac310$$
                That means we have to multiply 7, 3 times.than yiels $implies$
                $$7*3=7+7+7=21$$
                Now, we have $$dfrac2110$$
                Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
                $$21-(10+10)=1$$
                1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$






                share|cite|improve this answer
























                  up vote
                  0
                  down vote



                  accepted







                  up vote
                  0
                  down vote



                  accepted






                  Well,first let's look at the rules. For multiplication,



                  if, $x,yin mathbbQ$
                  $$x.y=underbrace(x+x+x+x+...+x)_texty times$$



                  For division, if
                  $$ x/y=n$$
                  $$Then,x=y.n$$
                  $$means,x-underbrace(y+y+y+...+y)_textn times=0$$



                  Now,lets work on your given numbers.
                  Here, $$7*0.3=7*dfrac310$$
                  That means we have to multiply 7, 3 times.than yiels $implies$
                  $$7*3=7+7+7=21$$
                  Now, we have $$dfrac2110$$
                  Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
                  $$21-(10+10)=1$$
                  1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$






                  share|cite|improve this answer














                  Well,first let's look at the rules. For multiplication,



                  if, $x,yin mathbbQ$
                  $$x.y=underbrace(x+x+x+x+...+x)_texty times$$



                  For division, if
                  $$ x/y=n$$
                  $$Then,x=y.n$$
                  $$means,x-underbrace(y+y+y+...+y)_textn times=0$$



                  Now,lets work on your given numbers.
                  Here, $$7*0.3=7*dfrac310$$
                  That means we have to multiply 7, 3 times.than yiels $implies$
                  $$7*3=7+7+7=21$$
                  Now, we have $$dfrac2110$$
                  Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
                  $$21-(10+10)=1$$
                  1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Aug 30 at 5:11

























                  answered Aug 30 at 5:05









                  Rakibul Islam Prince

                  3398




                  3398




















                      up vote
                      0
                      down vote













                      Yes it is possible



                      $$xcdot y=
                      underbracex+x+cdots+x _text$y$ times$$



                      if $$fracxy=n$$
                      $$0=x-underbrace(y+y+cdots+y)_text $n$ times$$



                      $fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0



                      But here is one limitation $x,y$ belongs to Rational Numbers






                      share|cite|improve this answer
























                        up vote
                        0
                        down vote













                        Yes it is possible



                        $$xcdot y=
                        underbracex+x+cdots+x _text$y$ times$$



                        if $$fracxy=n$$
                        $$0=x-underbrace(y+y+cdots+y)_text $n$ times$$



                        $fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0



                        But here is one limitation $x,y$ belongs to Rational Numbers






                        share|cite|improve this answer






















                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Yes it is possible



                          $$xcdot y=
                          underbracex+x+cdots+x _text$y$ times$$



                          if $$fracxy=n$$
                          $$0=x-underbrace(y+y+cdots+y)_text $n$ times$$



                          $fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0



                          But here is one limitation $x,y$ belongs to Rational Numbers






                          share|cite|improve this answer












                          Yes it is possible



                          $$xcdot y=
                          underbracex+x+cdots+x _text$y$ times$$



                          if $$fracxy=n$$
                          $$0=x-underbrace(y+y+cdots+y)_text $n$ times$$



                          $fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0



                          But here is one limitation $x,y$ belongs to Rational Numbers







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Aug 30 at 4:42









                          Deepesh Meena

                          3,2492824




                          3,2492824



























                               

                              draft saved


                              draft discarded















































                               


                              draft saved


                              draft discarded














                              StackExchange.ready(
                              function ()
                              StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2899123%2fmultiplication-and-division-by-addition-and-subtraction%23new-answer', 'question_page');

                              );

                              Post as a guest













































































                              這個網誌中的熱門文章

                              How to combine Bézier curves to a surface?

                              Mutual Information Always Non-negative

                              Why am i infinitely getting the same tweet with the Twitter Search API?