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prime numbers and expressing non-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

edited 2 hours ago

Mr. Brooks

22211338

asked 4 hours ago

Ahmed M. ElsonbatyAhmed M. Elsonbaty

624

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited 2 hours ago

Mr. Brooks

22211338

asked 4 hours ago

Ahmed M. ElsonbatyAhmed M. Elsonbaty

624

$endgroup$

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

$begingroup$

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

edited 2 hours ago

Mr. Brooks

22211338

asked 4 hours ago

Ahmed M. ElsonbatyAhmed M. Elsonbaty

624

$endgroup$

share|cite|improve this question

edited 2 hours ago

Mr. Brooks

22211338

asked 4 hours ago

Ahmed M. ElsonbatyAhmed M. Elsonbaty

624

share|cite|improve this question

edited 2 hours ago

Mr. Brooks

22211338

asked 4 hours ago

Ahmed M. ElsonbatyAhmed M. Elsonbaty

624

edited 2 hours ago

Mr. Brooks

22211338

edited 2 hours ago

Mr. Brooks

22211338

asked 4 hours ago

Ahmed M. ElsonbatyAhmed M. Elsonbaty

624

asked 4 hours ago

Ahmed M. ElsonbatyAhmed M. Elsonbaty

624

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

answered 4 hours ago

Peter ForemanPeter Foreman

7,8751320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 3 hours ago

Barry CipraBarry Cipra

60.7k655129

$endgroup$

add a comment | 

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

answered 3 hours ago

Mr. BrooksMr. Brooks

22211338

$endgroup$

add a comment | 

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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

answered 4 hours ago

Peter ForemanPeter Foreman

7,8751320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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

answered 4 hours ago

Peter ForemanPeter Foreman

7,8751320

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

$begingroup$

"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

answered 4 hours ago

Peter ForemanPeter Foreman

7,8751320

$endgroup$

share|cite|improve this answer

answered 4 hours ago

Peter ForemanPeter Foreman

7,8751320

answered 4 hours ago

Peter ForemanPeter Foreman

7,8751320

answered 4 hours ago

Peter ForemanPeter Foreman

7,8751320

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

answered 3 hours ago

Barry CipraBarry Cipra

60.7k655129

$endgroup$

add a comment | 

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

answered 3 hours ago

Barry CipraBarry Cipra

60.7k655129

$endgroup$

add a comment | 

$begingroup$

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

answered 3 hours ago

Barry CipraBarry Cipra

60.7k655129

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Barry CipraBarry Cipra

60.7k655129

answered 3 hours ago

Barry CipraBarry Cipra

60.7k655129

answered 3 hours ago

Barry CipraBarry Cipra

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

answered 3 hours ago

Mr. BrooksMr. Brooks

22211338

$endgroup$

add a comment | 

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

answered 3 hours ago

Mr. BrooksMr. Brooks

22211338

$endgroup$

add a comment | 

$begingroup$

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

answered 3 hours ago

Mr. BrooksMr. Brooks

22211338

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Mr. BrooksMr. Brooks

22211338

answered 3 hours ago

Mr. BrooksMr. Brooks

22211338

answered 3 hours ago

Mr. BrooksMr. Brooks

22211338