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prime numbers and expressing non-prime numbers

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prime numbers and expressing non-prime numbers



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Expressing a Non Negative Integer as Sums of Two SquaresAre all prime numbers finite?How can can you write a prime number as a product of prime numbers?Who generates the prime numbers for encryption?In Fermat's little theorem, if mod is not prime?Prime numbers like 113Non-unique prime factorisationWhy are all non-prime numbers divisible by a prime number?Generating Prime Numbers From Composite NumbersAlternate definition of prime numbers










2












$begingroup$


My textbook says if $b$ is a non-prime number then it can be expressed as a product of prime numbers. But if $1$ isn't prime how it can be expressed as a product of prime numbers?






number-theory elementary-number-theory prime-numbers







share|cite|improve this question











$endgroup$







  • 5




    $begingroup$
    An empty product is still a product.
    $endgroup$
    – lulu
    4 hours ago






  • 1




    $begingroup$
    The standard modern day view is that the number 1 is neither prime nor composite.
    $endgroup$
    – Martin Hansen
    4 hours ago















2












$begingroup$


My textbook says if $b$ is a non-prime number then it can be expressed as a product of prime numbers. But if $1$ isn't prime how it can be expressed as a product of prime numbers?






number-theory elementary-number-theory prime-numbers







share|cite|improve this question











$endgroup$







  • 5




    $begingroup$
    An empty product is still a product.
    $endgroup$
    – lulu
    4 hours ago






  • 1




    $begingroup$
    The standard modern day view is that the number 1 is neither prime nor composite.
    $endgroup$
    – Martin Hansen
    4 hours ago













2












2








2





$begingroup$


My textbook says if $b$ is a non-prime number then it can be expressed as a product of prime numbers. But if $1$ isn't prime how it can be expressed as a product of prime numbers?






number-theory elementary-number-theory prime-numbers







share|cite|improve this question











$endgroup$




My textbook says if $b$ is a non-prime number then it can be expressed as a product of prime numbers. But if $1$ isn't prime how it can be expressed as a product of prime numbers?






number-theory elementary-number-theory prime-numbers




number-theory elementary-number-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









Mr. Brooks

22211338




22211338










asked 4 hours ago









Ahmed M. ElsonbatyAhmed M. Elsonbaty

624




624







  • 5




    $begingroup$
    An empty product is still a product.
    $endgroup$
    – lulu
    4 hours ago






  • 1




    $begingroup$
    The standard modern day view is that the number 1 is neither prime nor composite.
    $endgroup$
    – Martin Hansen
    4 hours ago












  • 5




    $begingroup$
    An empty product is still a product.
    $endgroup$
    – lulu
    4 hours ago






  • 1




    $begingroup$
    The standard modern day view is that the number 1 is neither prime nor composite.
    $endgroup$
    – Martin Hansen
    4 hours ago







5




5




$begingroup$
An empty product is still a product.
$endgroup$
– lulu
4 hours ago




$begingroup$
An empty product is still a product.
$endgroup$
– lulu
4 hours ago




1




1




$begingroup$
The standard modern day view is that the number 1 is neither prime nor composite.
$endgroup$
– Martin Hansen
4 hours ago




$begingroup$
The standard modern day view is that the number 1 is neither prime nor composite.
$endgroup$
– Martin Hansen
4 hours ago










3 Answers
3






active

oldest

votes


















1












$begingroup$

"In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to (except for) the order of the factors."



Your error lies in stating the fundamental theorem of arithmetic incorrectly.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    What about non-UFDs? 1 is still not prime in those, so FTA does not apply, and the asker didn't even bring up the FTA.
    $endgroup$
    – Mr. Brooks
    3 hours ago










  • $begingroup$
    I assumed that their textbook meant the FTA and they accepted my answer, what is the problem?
    $endgroup$
    – Peter Foreman
    3 hours ago






  • 1




    $begingroup$
    There is no problem at all if you treat this website as a competition for points.
    $endgroup$
    – Mr. Brooks
    3 hours ago


















2












$begingroup$

This is mainly just an extended comment on Peter Foreman's answer. The (relatively difficult) uniqueness aspect of the Fundamental Theorem of Arithmetic is not needed for the OP's question, just the (easier) existence aspect.



What's missing from the OP's textbook is the qualifier in the correct assertion that every non-prime number greater than $1$ can be expressed as a product of primes. This is the existence aspect of FTA, and it can be proved by strong induction: If $ngt1$ is not a prime, then $n=ab$ for some pair of integers with $1lt a,b$. Both $a$ and $b$ must be less than $n$ (otherwise their product would be more than $n$), so we can assume, by strong induction, that each of them can be written as a product of primes, hence so can their product, which is $n$.



Remark: "Strong" induction means that you don't just assume an assertion is true for $n-1$ and then prove it for $n$, you assume it's true for all positive integers $klt n$. In this case the assertion is "if $kgt1$ and $k$ is non-prime, then $k$ can be written as a product of primes." Note that the base case, $k=1$, is vacuously true, because $1$ is not greater than $1$.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$

    What is the sum of no numbers at all? Zero, of course, since it is the additive identity: $x + 0 = 0$, where $x neq 0$, or even if it is.



    Now, what is the product of no numbers at all? It can't be zero, since, maintaining the stipulation that $x neq 0$, we have $x times 0 = 0$, and we said $x neq 0$. The multiplicative identity is $1$, since $x times 1 = 1$.



    Hence, the product of no primes at all is $1$. The fundamental theorem of arithmetic is a subtlety that's unnecessary for answering your question.






    share|cite|improve this answer









    $endgroup$













      Your Answer








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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      "In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to (except for) the order of the factors."



      Your error lies in stating the fundamental theorem of arithmetic incorrectly.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        What about non-UFDs? 1 is still not prime in those, so FTA does not apply, and the asker didn't even bring up the FTA.
        $endgroup$
        – Mr. Brooks
        3 hours ago










      • $begingroup$
        I assumed that their textbook meant the FTA and they accepted my answer, what is the problem?
        $endgroup$
        – Peter Foreman
        3 hours ago






      • 1




        $begingroup$
        There is no problem at all if you treat this website as a competition for points.
        $endgroup$
        – Mr. Brooks
        3 hours ago















      1












      $begingroup$

      "In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to (except for) the order of the factors."



      Your error lies in stating the fundamental theorem of arithmetic incorrectly.






      share|cite|improve this answer









      $endgroup$












      • $begingroup$
        What about non-UFDs? 1 is still not prime in those, so FTA does not apply, and the asker didn't even bring up the FTA.
        $endgroup$
        – Mr. Brooks
        3 hours ago










      • $begingroup$
        I assumed that their textbook meant the FTA and they accepted my answer, what is the problem?
        $endgroup$
        – Peter Foreman
        3 hours ago






      • 1




        $begingroup$
        There is no problem at all if you treat this website as a competition for points.
        $endgroup$
        – Mr. Brooks
        3 hours ago













      1












      1








      1





      $begingroup$

      "In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to (except for) the order of the factors."



      Your error lies in stating the fundamental theorem of arithmetic incorrectly.






      share|cite|improve this answer









      $endgroup$



      "In number theory, the fundamental theorem of arithmetic, also called the unique factorization theorem or the unique-prime-factorization theorem, states that every integer greater than 1 either is a prime number itself or can be represented as the product of prime numbers and that, moreover, this representation is unique, up to (except for) the order of the factors."



      Your error lies in stating the fundamental theorem of arithmetic incorrectly.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered 4 hours ago









      Peter ForemanPeter Foreman

      7,8931320




      7,8931320











      • $begingroup$
        What about non-UFDs? 1 is still not prime in those, so FTA does not apply, and the asker didn't even bring up the FTA.
        $endgroup$
        – Mr. Brooks
        3 hours ago










      • $begingroup$
        I assumed that their textbook meant the FTA and they accepted my answer, what is the problem?
        $endgroup$
        – Peter Foreman
        3 hours ago






      • 1




        $begingroup$
        There is no problem at all if you treat this website as a competition for points.
        $endgroup$
        – Mr. Brooks
        3 hours ago
















      • $begingroup$
        What about non-UFDs? 1 is still not prime in those, so FTA does not apply, and the asker didn't even bring up the FTA.
        $endgroup$
        – Mr. Brooks
        3 hours ago










      • $begingroup$
        I assumed that their textbook meant the FTA and they accepted my answer, what is the problem?
        $endgroup$
        – Peter Foreman
        3 hours ago






      • 1




        $begingroup$
        There is no problem at all if you treat this website as a competition for points.
        $endgroup$
        – Mr. Brooks
        3 hours ago















      $begingroup$
      What about non-UFDs? 1 is still not prime in those, so FTA does not apply, and the asker didn't even bring up the FTA.
      $endgroup$
      – Mr. Brooks
      3 hours ago




      $begingroup$
      What about non-UFDs? 1 is still not prime in those, so FTA does not apply, and the asker didn't even bring up the FTA.
      $endgroup$
      – Mr. Brooks
      3 hours ago












      $begingroup$
      I assumed that their textbook meant the FTA and they accepted my answer, what is the problem?
      $endgroup$
      – Peter Foreman
      3 hours ago




      $begingroup$
      I assumed that their textbook meant the FTA and they accepted my answer, what is the problem?
      $endgroup$
      – Peter Foreman
      3 hours ago




      1




      1




      $begingroup$
      There is no problem at all if you treat this website as a competition for points.
      $endgroup$
      – Mr. Brooks
      3 hours ago




      $begingroup$
      There is no problem at all if you treat this website as a competition for points.
      $endgroup$
      – Mr. Brooks
      3 hours ago











      2












      $begingroup$

      This is mainly just an extended comment on Peter Foreman's answer. The (relatively difficult) uniqueness aspect of the Fundamental Theorem of Arithmetic is not needed for the OP's question, just the (easier) existence aspect.



      What's missing from the OP's textbook is the qualifier in the correct assertion that every non-prime number greater than $1$ can be expressed as a product of primes. This is the existence aspect of FTA, and it can be proved by strong induction: If $ngt1$ is not a prime, then $n=ab$ for some pair of integers with $1lt a,b$. Both $a$ and $b$ must be less than $n$ (otherwise their product would be more than $n$), so we can assume, by strong induction, that each of them can be written as a product of primes, hence so can their product, which is $n$.



      Remark: "Strong" induction means that you don't just assume an assertion is true for $n-1$ and then prove it for $n$, you assume it's true for all positive integers $klt n$. In this case the assertion is "if $kgt1$ and $k$ is non-prime, then $k$ can be written as a product of primes." Note that the base case, $k=1$, is vacuously true, because $1$ is not greater than $1$.






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        This is mainly just an extended comment on Peter Foreman's answer. The (relatively difficult) uniqueness aspect of the Fundamental Theorem of Arithmetic is not needed for the OP's question, just the (easier) existence aspect.



        What's missing from the OP's textbook is the qualifier in the correct assertion that every non-prime number greater than $1$ can be expressed as a product of primes. This is the existence aspect of FTA, and it can be proved by strong induction: If $ngt1$ is not a prime, then $n=ab$ for some pair of integers with $1lt a,b$. Both $a$ and $b$ must be less than $n$ (otherwise their product would be more than $n$), so we can assume, by strong induction, that each of them can be written as a product of primes, hence so can their product, which is $n$.



        Remark: "Strong" induction means that you don't just assume an assertion is true for $n-1$ and then prove it for $n$, you assume it's true for all positive integers $klt n$. In this case the assertion is "if $kgt1$ and $k$ is non-prime, then $k$ can be written as a product of primes." Note that the base case, $k=1$, is vacuously true, because $1$ is not greater than $1$.






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          This is mainly just an extended comment on Peter Foreman's answer. The (relatively difficult) uniqueness aspect of the Fundamental Theorem of Arithmetic is not needed for the OP's question, just the (easier) existence aspect.



          What's missing from the OP's textbook is the qualifier in the correct assertion that every non-prime number greater than $1$ can be expressed as a product of primes. This is the existence aspect of FTA, and it can be proved by strong induction: If $ngt1$ is not a prime, then $n=ab$ for some pair of integers with $1lt a,b$. Both $a$ and $b$ must be less than $n$ (otherwise their product would be more than $n$), so we can assume, by strong induction, that each of them can be written as a product of primes, hence so can their product, which is $n$.



          Remark: "Strong" induction means that you don't just assume an assertion is true for $n-1$ and then prove it for $n$, you assume it's true for all positive integers $klt n$. In this case the assertion is "if $kgt1$ and $k$ is non-prime, then $k$ can be written as a product of primes." Note that the base case, $k=1$, is vacuously true, because $1$ is not greater than $1$.






          share|cite|improve this answer









          $endgroup$



          This is mainly just an extended comment on Peter Foreman's answer. The (relatively difficult) uniqueness aspect of the Fundamental Theorem of Arithmetic is not needed for the OP's question, just the (easier) existence aspect.



          What's missing from the OP's textbook is the qualifier in the correct assertion that every non-prime number greater than $1$ can be expressed as a product of primes. This is the existence aspect of FTA, and it can be proved by strong induction: If $ngt1$ is not a prime, then $n=ab$ for some pair of integers with $1lt a,b$. Both $a$ and $b$ must be less than $n$ (otherwise their product would be more than $n$), so we can assume, by strong induction, that each of them can be written as a product of primes, hence so can their product, which is $n$.



          Remark: "Strong" induction means that you don't just assume an assertion is true for $n-1$ and then prove it for $n$, you assume it's true for all positive integers $klt n$. In this case the assertion is "if $kgt1$ and $k$ is non-prime, then $k$ can be written as a product of primes." Note that the base case, $k=1$, is vacuously true, because $1$ is not greater than $1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Barry CipraBarry Cipra

          60.7k655129




          60.7k655129





















              1












              $begingroup$

              What is the sum of no numbers at all? Zero, of course, since it is the additive identity: $x + 0 = 0$, where $x neq 0$, or even if it is.



              Now, what is the product of no numbers at all? It can't be zero, since, maintaining the stipulation that $x neq 0$, we have $x times 0 = 0$, and we said $x neq 0$. The multiplicative identity is $1$, since $x times 1 = 1$.



              Hence, the product of no primes at all is $1$. The fundamental theorem of arithmetic is a subtlety that's unnecessary for answering your question.






              share|cite|improve this answer









              $endgroup$

















                1












                $begingroup$

                What is the sum of no numbers at all? Zero, of course, since it is the additive identity: $x + 0 = 0$, where $x neq 0$, or even if it is.



                Now, what is the product of no numbers at all? It can't be zero, since, maintaining the stipulation that $x neq 0$, we have $x times 0 = 0$, and we said $x neq 0$. The multiplicative identity is $1$, since $x times 1 = 1$.



                Hence, the product of no primes at all is $1$. The fundamental theorem of arithmetic is a subtlety that's unnecessary for answering your question.






                share|cite|improve this answer









                $endgroup$















                  1












                  1








                  1





                  $begingroup$

                  What is the sum of no numbers at all? Zero, of course, since it is the additive identity: $x + 0 = 0$, where $x neq 0$, or even if it is.



                  Now, what is the product of no numbers at all? It can't be zero, since, maintaining the stipulation that $x neq 0$, we have $x times 0 = 0$, and we said $x neq 0$. The multiplicative identity is $1$, since $x times 1 = 1$.



                  Hence, the product of no primes at all is $1$. The fundamental theorem of arithmetic is a subtlety that's unnecessary for answering your question.






                  share|cite|improve this answer









                  $endgroup$



                  What is the sum of no numbers at all? Zero, of course, since it is the additive identity: $x + 0 = 0$, where $x neq 0$, or even if it is.



                  Now, what is the product of no numbers at all? It can't be zero, since, maintaining the stipulation that $x neq 0$, we have $x times 0 = 0$, and we said $x neq 0$. The multiplicative identity is $1$, since $x times 1 = 1$.



                  Hence, the product of no primes at all is $1$. The fundamental theorem of arithmetic is a subtlety that's unnecessary for answering your question.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 3 hours ago









                  Mr. BrooksMr. Brooks

                  22211338




                  22211338



























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                      東螺溪清水溪加走寮溪阿里山溪東埔蚋溪獅尾堀清水溪清水溝溪水里溪陳有蘭溪郡坑溪十八重溪那馬嘎班溪松柏坑溪卓棍溪瓠瓢坑溪益則坑溪玉崙溪丹大溪郡大溪巒大溪卡社溪栗栖溪萬大溪萬大北溪萬大南溪霧社溪塔羅灣溪馬赫坡溪萬大發電廠大觀發電廠日月潭第一發電所明潭發電廠鉅工分廠水里機組濁水機組名間水力發電廠西濱大橋自強大橋高鐵濁水溪橋西螺大橋溪州大橋中沙大橋彰雲大橋台鐵濁水溪橋國道三號濁水溪橋名竹大橋集集橋集鹿大橋玉峰大橋永興橋龍神橋靜和橋寶石橋雙龍橋雙龍吊橋孫海橋思源橋松林橋親愛橋濁水溪橋德魯灣橋雲龍橋 名間鄉臺灣水力發電廠 臺灣水力發電廠南投縣名間鄉集集攔河堰集集攔河堰濁水溪台灣BOT水力發電廠經濟部經濟部水利署集集攔河堰名間鄉名間水力公司日本關西電力公頃台灣電力公司 body.skin-minerva .mw-parser-output table.infobox captiontext-align:center 名間水力發電廠 名間水力發電廠 名間水力發電廠在臺灣的位置 國家   中華民國 位置 南投縣名間鄉濁水村 坐標 23°50′10″N 120°42′41″E  /  23.83611°N 120.71139°E  / 23.83611; 120.71139 坐标： 23°50′10″N 120°42′41″E  /  23.83611°N 120.71139°E  / 23.83611; 120.71139 現況 運轉中 始建日期 2004年 啟用日期 2007年 建造費用 新台幣832,858,000元 持有單位 經濟部水利署 运营单位 名間水力公司 建造者 名間水力公司 設計監造：中興工程顧問公司 機電工程：大將作公司 土建工程：順達營造工程公司 慣常式水力發電 水力發電形式 川流式水力發電 取水來源 濁水溪 發電機型號 額定容量9280kV，同步發電機兩部 水輪機型號 豎軸卡布蘭式水輪機兩部 廠房類型 半地下式 壓力鋼管 兩條，長113公尺、直徑4公尺 水頭高度 有效落差31.0公尺 用水量 單部：16.704CMS 兩部：33CMS 發電概況 额定容量 16.704MW 年发电量 每年約7,594萬度 名間水力發電廠 ，是臺灣一座小型水力發電廠，位於南投縣名間鄉濁水村集集攔河堰下游五公里，於2007年啟用，係利用集集攔河堰截取濁水溪之水源發電
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