Simple way of explaining the empty product

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I understand that 3 raised to the power of, say, 2 is 9 because 3 times 3 equals 9.



But is it possible to explain in same simple terms how 3 raised to the power of 0 is 1?






abstract-algebra







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    up vote
    2
    down vote

    favorite












    I understand that 3 raised to the power of, say, 2 is 9 because 3 times 3 equals 9.



    But is it possible to explain in same simple terms how 3 raised to the power of 0 is 1?






    abstract-algebra







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      I understand that 3 raised to the power of, say, 2 is 9 because 3 times 3 equals 9.



      But is it possible to explain in same simple terms how 3 raised to the power of 0 is 1?






      abstract-algebra







      share|cite|improve this question













      I understand that 3 raised to the power of, say, 2 is 9 because 3 times 3 equals 9.



      But is it possible to explain in same simple terms how 3 raised to the power of 0 is 1?






      abstract-algebra




      abstract-algebra






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      share|cite|improve this question











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      asked 29 mins ago









      brilliant

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      1797




















          2 Answers
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          $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



          We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



          Using this definition, we can notice the following interesting property:




          For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




          Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.






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            $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.






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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              4
              down vote



              accepted










              $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



              We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



              Using this definition, we can notice the following interesting property:




              For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




              Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.






              share|cite|improve this answer
























                up vote
                4
                down vote



                accepted










                $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



                We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



                Using this definition, we can notice the following interesting property:




                For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




                Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.






                share|cite|improve this answer






















                  up vote
                  4
                  down vote



                  accepted







                  up vote
                  4
                  down vote



                  accepted






                  $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



                  We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



                  Using this definition, we can notice the following interesting property:




                  For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




                  Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.






                  share|cite|improve this answer












                  $x^0$ doesn't mean anything before we define what it means. We can define it to mean anything, but we usually want to define things so that they make sense.



                  We start by defining $x^k$ as $underbracexcdot x cdots x_ktext times$ because it's a useful way of writing the product.



                  Using this definition, we can notice the following interesting property:




                  For any pair of integers $k_1, k_2$ such that $k_1 > k_2$ and any positive real number $xin mathbb R$ we have $$fracx^k_1x^k_2 = x^k_1-k_2$$




                  Now, we like this property. We want this property be true for other pairs of $k_1, k_2$. In order for this rule to also be true if $k_1=k_2$, we have to define $x^0=1$, so that's what we define it as. We do it because it's useful.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 24 mins ago









                  5xum

                  84.3k386151




                  84.3k386151




















                      up vote
                      5
                      down vote













                      $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.






                      share|cite|improve this answer
























                        up vote
                        5
                        down vote













                        $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.






                        share|cite|improve this answer






















                          up vote
                          5
                          down vote










                          up vote
                          5
                          down vote









                          $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.






                          share|cite|improve this answer












                          $$3=3^1=3^1+0=3^1times3^0=3times3^0$$implying that $3^0=1$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 24 mins ago









                          Vera

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