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prime numbers and expressing non-prime numbers
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prime numbers and expressing non-prime numbers
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$begingroup$
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
$endgroup$
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
$endgroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
"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.
$endgroup$
$begingroup$
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 |
$begingroup$
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 $1$.
$endgroup$
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.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
"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.
$endgroup$
$begingroup$
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.
$endgroup$
$begingroup$
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.
$endgroup$
"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.
answered 4 hours ago
Peter ForemanPeter Foreman
7,8921320
7,8921320
$begingroup$
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 |
$begingroup$
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
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 $1$.
$endgroup$
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 $1$.
$endgroup$
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 $1$.
$endgroup$
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 $1$.
answered 3 hours ago
Barry CipraBarry Cipra
60.7k655129
60.7k655129
add a comment |
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
answered 3 hours ago
Mr. BrooksMr. Brooks
22211338
22211338
add a comment |
add a comment |
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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