Proof of Lemma: Every nonzero integer can be written as a product of primes
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I'm new to number theory. This might be kind of a silly question, so I'm sorry if it is.
I encountered the classic lemma about every nonzero integer being the product of primes in a textbook about number theory. In this textbook there is also a proof for it provided, and I'd like to understand why it is that the proof actually works.
The proof is as follows:
Assume, for contradiction, that there is an integer $N$ that cannot be written as a product of primes. Let $N$ be the smallest positive integer with this property. Since $N$ cannot itself be prime we must have $N = mn$, where $1 < m$, $n < N$. However, since $m$, $n$ are positive and smaller than $N$ they must each be a product of primes. But then so is $N = mn$. This is a contradiction.
I feel like this proof kind of presupposes the lemma. I think this line of reasoning could be strengthened using induction, and I've seen other proofs of this lemma that use induction. Can someone help me out? What am I missing and why do I think that this proof of the lemma is circular?
elementary-number-theory prime-numbers proof-explanation integers
edited 3 hours ago
Robert Soupe
11.4k21950
asked 3 hours ago
Alena GusakovAlena Gusakov
112
New contributor
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I'm new to number theory. This might be kind of a silly question, so I'm sorry if it is.
I encountered the classic lemma about every nonzero integer being the product of primes in a textbook about number theory. In this textbook there is also a proof for it provided, and I'd like to understand why it is that the proof actually works.
The proof is as follows:
Assume, for contradiction, that there is an integer $N$ that cannot be written as a product of primes. Let $N$ be the smallest positive integer with this property. Since $N$ cannot itself be prime we must have $N = mn$, where $1 < m$, $n < N$. However, since $m$, $n$ are positive and smaller than $N$ they must each be a product of primes. But then so is $N = mn$. This is a contradiction.
I feel like this proof kind of presupposes the lemma. I think this line of reasoning could be strengthened using induction, and I've seen other proofs of this lemma that use induction. Can someone help me out? What am I missing and why do I think that this proof of the lemma is circular?
elementary-number-theory prime-numbers proof-explanation integers
edited 3 hours ago
Robert Soupe
11.4k21950
asked 3 hours ago
Alena GusakovAlena Gusakov
112
New contributor
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
2
$begingroup$
That argument is by induction. the result is easy to check for small numbers, so assume it holds up to $N-1$. Then $N$ is either prime, in which case we are done, or it factors as $atimes b$ with $1<a≤b<N-1$ and you can apply the inductive hypothesis to $a,b$. Same argument.
$endgroup$
– lulu
3 hours ago
1
$begingroup$
There is nothing missing in this proof. It is just fine. And why “two primes”?
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
@JoséCarlosSantos Typo. Fixed.
$endgroup$
– Alena Gusakov
3 hours ago
$begingroup$
It's not circular, but it could be a lot clearer. It's not strictly necessary to say $n > 1$, since $m$ is positive and $mn$ is also positive.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
It's a valid proof by infinite descent (a.k.a. minimal criminal), the contrapositive of induction - see the Remark here. You should master both this (negative) and the normal (positive) form of induction.
$endgroup$
– Bill Dubuque
46 mins ago
add a comment |
$begingroup$
I'm new to number theory. This might be kind of a silly question, so I'm sorry if it is.
I encountered the classic lemma about every nonzero integer being the product of primes in a textbook about number theory. In this textbook there is also a proof for it provided, and I'd like to understand why it is that the proof actually works.
The proof is as follows:
Assume, for contradiction, that there is an integer $N$ that cannot be written as a product of primes. Let $N$ be the smallest positive integer with this property. Since $N$ cannot itself be prime we must have $N = mn$, where $1 < m$, $n < N$. However, since $m$, $n$ are positive and smaller than $N$ they must each be a product of primes. But then so is $N = mn$. This is a contradiction.
I feel like this proof kind of presupposes the lemma. I think this line of reasoning could be strengthened using induction, and I've seen other proofs of this lemma that use induction. Can someone help me out? What am I missing and why do I think that this proof of the lemma is circular?
elementary-number-theory prime-numbers proof-explanation integers
edited 3 hours ago
Robert Soupe
11.4k21950
asked 3 hours ago
Alena GusakovAlena Gusakov
112
New contributor
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I'm new to number theory. This might be kind of a silly question, so I'm sorry if it is.
I encountered the classic lemma about every nonzero integer being the product of primes in a textbook about number theory. In this textbook there is also a proof for it provided, and I'd like to understand why it is that the proof actually works.
The proof is as follows:
Assume, for contradiction, that there is an integer $N$ that cannot be written as a product of primes. Let $N$ be the smallest positive integer with this property. Since $N$ cannot itself be prime we must have $N = mn$, where $1 < m$, $n < N$. However, since $m$, $n$ are positive and smaller than $N$ they must each be a product of primes. But then so is $N = mn$. This is a contradiction.
I feel like this proof kind of presupposes the lemma. I think this line of reasoning could be strengthened using induction, and I've seen other proofs of this lemma that use induction. Can someone help me out? What am I missing and why do I think that this proof of the lemma is circular?
elementary-number-theory prime-numbers proof-explanation integers
elementary-number-theory prime-numbers proof-explanation integers
edited 3 hours ago
Robert Soupe
11.4k21950
asked 3 hours ago
Alena GusakovAlena Gusakov
112
New contributor
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Robert Soupe
11.4k21950
asked 3 hours ago
Alena GusakovAlena Gusakov
112
New contributor
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Robert Soupe
11.4k21950
edited 3 hours ago
Robert Soupe
11.4k21950
edited 3 hours ago
Robert Soupe
11.4k21950
11.4k21950
asked 3 hours ago
Alena GusakovAlena Gusakov
112
New contributor
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Alena GusakovAlena Gusakov
112
asked 3 hours ago
Alena GusakovAlena Gusakov
112
112
New contributor
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alena Gusakov is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
That argument is by induction. the result is easy to check for small numbers, so assume it holds up to $N-1$. Then $N$ is either prime, in which case we are done, or it factors as $atimes b$ with $1<a≤b<N-1$ and you can apply the inductive hypothesis to $a,b$. Same argument.
$endgroup$
– lulu
3 hours ago
1
$begingroup$
There is nothing missing in this proof. It is just fine. And why “two primes”?
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
@JoséCarlosSantos Typo. Fixed.
$endgroup$
– Alena Gusakov
3 hours ago
$begingroup$
It's not circular, but it could be a lot clearer. It's not strictly necessary to say $n > 1$, since $m$ is positive and $mn$ is also positive.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
It's a valid proof by infinite descent (a.k.a. minimal criminal), the contrapositive of induction - see the Remark here. You should master both this (negative) and the normal (positive) form of induction.
$endgroup$
– Bill Dubuque
46 mins ago
add a comment |
2
$begingroup$
That argument is by induction. the result is easy to check for small numbers, so assume it holds up to $N-1$. Then $N$ is either prime, in which case we are done, or it factors as $atimes b$ with $1<a≤b<N-1$ and you can apply the inductive hypothesis to $a,b$. Same argument.
$endgroup$
– lulu
3 hours ago
1
$begingroup$
There is nothing missing in this proof. It is just fine. And why “two primes”?
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
@JoséCarlosSantos Typo. Fixed.
$endgroup$
– Alena Gusakov
3 hours ago
$begingroup$
It's not circular, but it could be a lot clearer. It's not strictly necessary to say $n > 1$, since $m$ is positive and $mn$ is also positive.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
It's a valid proof by infinite descent (a.k.a. minimal criminal), the contrapositive of induction - see the Remark here. You should master both this (negative) and the normal (positive) form of induction.
$endgroup$
– Bill Dubuque
46 mins ago
2
2
$begingroup$
That argument is by induction. the result is easy to check for small numbers, so assume it holds up to $N-1$. Then $N$ is either prime, in which case we are done, or it factors as $atimes b$ with $1<a≤b<N-1$ and you can apply the inductive hypothesis to $a,b$. Same argument.
$endgroup$
– lulu
3 hours ago
$begingroup$
That argument is by induction. the result is easy to check for small numbers, so assume it holds up to $N-1$. Then $N$ is either prime, in which case we are done, or it factors as $atimes b$ with $1<a≤b<N-1$ and you can apply the inductive hypothesis to $a,b$. Same argument.
$endgroup$
– lulu
3 hours ago
1
1
$begingroup$
There is nothing missing in this proof. It is just fine. And why “two primes”?
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
There is nothing missing in this proof. It is just fine. And why “two primes”?
$endgroup$
– José Carlos Santos
3 hours ago
$begingroup$
@JoséCarlosSantos Typo. Fixed.
$endgroup$
– Alena Gusakov
3 hours ago
$begingroup$
@JoséCarlosSantos Typo. Fixed.
$endgroup$
– Alena Gusakov
3 hours ago
$begingroup$
It's not circular, but it could be a lot clearer. It's not strictly necessary to say $n > 1$, since $m$ is positive and $mn$ is also positive.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
It's not circular, but it could be a lot clearer. It's not strictly necessary to say $n > 1$, since $m$ is positive and $mn$ is also positive.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
It's a valid proof by infinite descent (a.k.a. minimal criminal), the contrapositive of induction - see the Remark here. You should master both this (negative) and the normal (positive) form of induction.
$endgroup$
– Bill Dubuque
46 mins ago
$begingroup$
It's a valid proof by infinite descent (a.k.a. minimal criminal), the contrapositive of induction - see the Remark here. You should master both this (negative) and the normal (positive) form of induction.
$endgroup$
– Bill Dubuque
46 mins ago
add a comment |
3 Answers
3
active
oldest
votes
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The proof is not circular, the key is in the second sentence: Let N be the smallest positive integer with this property.
We are allowed to say a least $N$ exists because of the well-ordering principle.
answered 3 hours ago
Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez
1065
$endgroup$
$begingroup$
I don't know if it's because of the well-ordering principle ... that's like using a hammer to slice through butter. One does not need the full strength of the AOC to prove such a simple statement.
$endgroup$
– Don Thousand
3 hours ago
$begingroup$
@Don What's AOC? I presume you're not talking about Alexandria Ocasio-Cortez.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
@RobertSoupe: Axiom of choice. The more usual abbreviation is AC.
$endgroup$
– Nate Eldredge
1 hour ago
1
$begingroup$
@DonThousand: I think "well-ordering principle" here refers to the statement "the usual ordering on the natural numbers is a well order". The Axiom of Choice equivalent is "every set admits an ordering which is a well order" - that wouldn't really even help with this proof, since it would only tell us that there is some ordering of the natural numbers which is a well order - it doesn't tell us that the usual ordering is one.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
@NateEldredge: Indeed, the well-ordering principle (not the similarly-named well-ordering theorem) is equivalent to induction (and probably also to infinite descent, but I haven't worked through that one yet), so if you disallow WOP, then you are going to have a hard time proving a lot of things.
$endgroup$
– Kevin
25 mins ago
add a comment |
$begingroup$
Although the proof by contradiction is correct, your feeling of unease is fine, because the direct proof by induction is so much clearer:
Take an integer $N$. If $N$ is prime, there is nothing to prove. Otherwise, we must have $N = mn$, where $1 < m, n < N$. By induction, since $m, n$ are smaller than $N$, they must each be a product of primes. Then so is $N = mn$. Done.
answered 3 hours ago
lhflhf
166k11172402
$endgroup$
add a comment |
$begingroup$
A proof by induction has base case(s), Let m and n be said base cases. if it's true for m and true for n (not necessarily distinct), then because it's a product, it follows for mn. All the proof supposes, is N=mn for some base case ( primes or prime powers to start, in these cases) with m and n proved. Then it follows for N, which by saying N which
originally was consider the least element of a set of counterexamples, has one, it eliminates all possible least elements for the set we originally supposed existed.
answered 1 hour ago
Roddy MacPheeRoddy MacPhee
537118
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The proof is not circular, the key is in the second sentence: Let N be the smallest positive integer with this property.
We are allowed to say a least $N$ exists because of the well-ordering principle.
answered 3 hours ago
Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez
1065
$endgroup$
$begingroup$
I don't know if it's because of the well-ordering principle ... that's like using a hammer to slice through butter. One does not need the full strength of the AOC to prove such a simple statement.
$endgroup$
– Don Thousand
3 hours ago
$begingroup$
@Don What's AOC? I presume you're not talking about Alexandria Ocasio-Cortez.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
@RobertSoupe: Axiom of choice. The more usual abbreviation is AC.
$endgroup$
– Nate Eldredge
1 hour ago
1
$begingroup$
@DonThousand: I think "well-ordering principle" here refers to the statement "the usual ordering on the natural numbers is a well order". The Axiom of Choice equivalent is "every set admits an ordering which is a well order" - that wouldn't really even help with this proof, since it would only tell us that there is some ordering of the natural numbers which is a well order - it doesn't tell us that the usual ordering is one.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
@NateEldredge: Indeed, the well-ordering principle (not the similarly-named well-ordering theorem) is equivalent to induction (and probably also to infinite descent, but I haven't worked through that one yet), so if you disallow WOP, then you are going to have a hard time proving a lot of things.
$endgroup$
– Kevin
25 mins ago
add a comment |
$begingroup$
The proof is not circular, the key is in the second sentence: Let N be the smallest positive integer with this property.
We are allowed to say a least $N$ exists because of the well-ordering principle.
answered 3 hours ago
Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez
1065
$endgroup$
$begingroup$
I don't know if it's because of the well-ordering principle ... that's like using a hammer to slice through butter. One does not need the full strength of the AOC to prove such a simple statement.
$endgroup$
– Don Thousand
3 hours ago
$begingroup$
@Don What's AOC? I presume you're not talking about Alexandria Ocasio-Cortez.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
@RobertSoupe: Axiom of choice. The more usual abbreviation is AC.
$endgroup$
– Nate Eldredge
1 hour ago
1
$begingroup$
@DonThousand: I think "well-ordering principle" here refers to the statement "the usual ordering on the natural numbers is a well order". The Axiom of Choice equivalent is "every set admits an ordering which is a well order" - that wouldn't really even help with this proof, since it would only tell us that there is some ordering of the natural numbers which is a well order - it doesn't tell us that the usual ordering is one.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
@NateEldredge: Indeed, the well-ordering principle (not the similarly-named well-ordering theorem) is equivalent to induction (and probably also to infinite descent, but I haven't worked through that one yet), so if you disallow WOP, then you are going to have a hard time proving a lot of things.
$endgroup$
– Kevin
25 mins ago
add a comment |
$begingroup$
The proof is not circular, the key is in the second sentence: Let N be the smallest positive integer with this property.
We are allowed to say a least $N$ exists because of the well-ordering principle.
answered 3 hours ago
Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez
1065
$endgroup$
The proof is not circular, the key is in the second sentence: Let N be the smallest positive integer with this property.
We are allowed to say a least $N$ exists because of the well-ordering principle.
answered 3 hours ago
Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez
1065
answered 3 hours ago
Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez
1065
answered 3 hours ago
Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez
1065
answered 3 hours ago
Edgar Jaramillo RodriguezEdgar Jaramillo Rodriguez
1065
1065
$begingroup$
I don't know if it's because of the well-ordering principle ... that's like using a hammer to slice through butter. One does not need the full strength of the AOC to prove such a simple statement.
$endgroup$
– Don Thousand
3 hours ago
$begingroup$
@Don What's AOC? I presume you're not talking about Alexandria Ocasio-Cortez.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
@RobertSoupe: Axiom of choice. The more usual abbreviation is AC.
$endgroup$
– Nate Eldredge
1 hour ago
1
$begingroup$
@DonThousand: I think "well-ordering principle" here refers to the statement "the usual ordering on the natural numbers is a well order". The Axiom of Choice equivalent is "every set admits an ordering which is a well order" - that wouldn't really even help with this proof, since it would only tell us that there is some ordering of the natural numbers which is a well order - it doesn't tell us that the usual ordering is one.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
@NateEldredge: Indeed, the well-ordering principle (not the similarly-named well-ordering theorem) is equivalent to induction (and probably also to infinite descent, but I haven't worked through that one yet), so if you disallow WOP, then you are going to have a hard time proving a lot of things.
$endgroup$
– Kevin
25 mins ago
add a comment |
$begingroup$
I don't know if it's because of the well-ordering principle ... that's like using a hammer to slice through butter. One does not need the full strength of the AOC to prove such a simple statement.
$endgroup$
– Don Thousand
3 hours ago
$begingroup$
@Don What's AOC? I presume you're not talking about Alexandria Ocasio-Cortez.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
@RobertSoupe: Axiom of choice. The more usual abbreviation is AC.
$endgroup$
– Nate Eldredge
1 hour ago
1
$begingroup$
@DonThousand: I think "well-ordering principle" here refers to the statement "the usual ordering on the natural numbers is a well order". The Axiom of Choice equivalent is "every set admits an ordering which is a well order" - that wouldn't really even help with this proof, since it would only tell us that there is some ordering of the natural numbers which is a well order - it doesn't tell us that the usual ordering is one.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
@NateEldredge: Indeed, the well-ordering principle (not the similarly-named well-ordering theorem) is equivalent to induction (and probably also to infinite descent, but I haven't worked through that one yet), so if you disallow WOP, then you are going to have a hard time proving a lot of things.
$endgroup$
– Kevin
25 mins ago
$begingroup$
I don't know if it's because of the well-ordering principle ... that's like using a hammer to slice through butter. One does not need the full strength of the AOC to prove such a simple statement.
$endgroup$
– Don Thousand
3 hours ago
$begingroup$
I don't know if it's because of the well-ordering principle ... that's like using a hammer to slice through butter. One does not need the full strength of the AOC to prove such a simple statement.
$endgroup$
– Don Thousand
3 hours ago
$begingroup$
@Don What's AOC? I presume you're not talking about Alexandria Ocasio-Cortez.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
@Don What's AOC? I presume you're not talking about Alexandria Ocasio-Cortez.
$endgroup$
– Robert Soupe
2 hours ago
$begingroup$
@RobertSoupe: Axiom of choice. The more usual abbreviation is AC.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
@RobertSoupe: Axiom of choice. The more usual abbreviation is AC.
$endgroup$
– Nate Eldredge
1 hour ago
1
1
$begingroup$
@DonThousand: I think "well-ordering principle" here refers to the statement "the usual ordering on the natural numbers is a well order". The Axiom of Choice equivalent is "every set admits an ordering which is a well order" - that wouldn't really even help with this proof, since it would only tell us that there is some ordering of the natural numbers which is a well order - it doesn't tell us that the usual ordering is one.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
@DonThousand: I think "well-ordering principle" here refers to the statement "the usual ordering on the natural numbers is a well order". The Axiom of Choice equivalent is "every set admits an ordering which is a well order" - that wouldn't really even help with this proof, since it would only tell us that there is some ordering of the natural numbers which is a well order - it doesn't tell us that the usual ordering is one.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
@NateEldredge: Indeed, the well-ordering principle (not the similarly-named well-ordering theorem) is equivalent to induction (and probably also to infinite descent, but I haven't worked through that one yet), so if you disallow WOP, then you are going to have a hard time proving a lot of things.
$endgroup$
– Kevin
25 mins ago
$begingroup$
@NateEldredge: Indeed, the well-ordering principle (not the similarly-named well-ordering theorem) is equivalent to induction (and probably also to infinite descent, but I haven't worked through that one yet), so if you disallow WOP, then you are going to have a hard time proving a lot of things.
$endgroup$
– Kevin
25 mins ago
add a comment |
$begingroup$
Although the proof by contradiction is correct, your feeling of unease is fine, because the direct proof by induction is so much clearer:
Take an integer $N$. If $N$ is prime, there is nothing to prove. Otherwise, we must have $N = mn$, where $1 < m, n < N$. By induction, since $m, n$ are smaller than $N$, they must each be a product of primes. Then so is $N = mn$. Done.
answered 3 hours ago
lhflhf
166k11172402
$endgroup$
add a comment |
$begingroup$
Although the proof by contradiction is correct, your feeling of unease is fine, because the direct proof by induction is so much clearer:
Take an integer $N$. If $N$ is prime, there is nothing to prove. Otherwise, we must have $N = mn$, where $1 < m, n < N$. By induction, since $m, n$ are smaller than $N$, they must each be a product of primes. Then so is $N = mn$. Done.
answered 3 hours ago
lhflhf
166k11172402
$endgroup$
add a comment |
$begingroup$
Although the proof by contradiction is correct, your feeling of unease is fine, because the direct proof by induction is so much clearer:
Take an integer $N$. If $N$ is prime, there is nothing to prove. Otherwise, we must have $N = mn$, where $1 < m, n < N$. By induction, since $m, n$ are smaller than $N$, they must each be a product of primes. Then so is $N = mn$. Done.
answered 3 hours ago
lhflhf
166k11172402
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Although the proof by contradiction is correct, your feeling of unease is fine, because the direct proof by induction is so much clearer:
Take an integer $N$. If $N$ is prime, there is nothing to prove. Otherwise, we must have $N = mn$, where $1 < m, n < N$. By induction, since $m, n$ are smaller than $N$, they must each be a product of primes. Then so is $N = mn$. Done.
answered 3 hours ago
lhflhf
166k11172402
answered 3 hours ago
lhflhf
166k11172402
answered 3 hours ago
lhflhf
166k11172402
answered 3 hours ago
lhflhf
166k11172402
166k11172402
add a comment |
add a comment |
$begingroup$
A proof by induction has base case(s), Let m and n be said base cases. if it's true for m and true for n (not necessarily distinct), then because it's a product, it follows for mn. All the proof supposes, is N=mn for some base case ( primes or prime powers to start, in these cases) with m and n proved. Then it follows for N, which by saying N which
originally was consider the least element of a set of counterexamples, has one, it eliminates all possible least elements for the set we originally supposed existed.
answered 1 hour ago
Roddy MacPheeRoddy MacPhee
537118
$endgroup$
add a comment |
$begingroup$
A proof by induction has base case(s), Let m and n be said base cases. if it's true for m and true for n (not necessarily distinct), then because it's a product, it follows for mn. All the proof supposes, is N=mn for some base case ( primes or prime powers to start, in these cases) with m and n proved. Then it follows for N, which by saying N which
originally was consider the least element of a set of counterexamples, has one, it eliminates all possible least elements for the set we originally supposed existed.
answered 1 hour ago
Roddy MacPheeRoddy MacPhee
537118
$endgroup$
add a comment |
$begingroup$
A proof by induction has base case(s), Let m and n be said base cases. if it's true for m and true for n (not necessarily distinct), then because it's a product, it follows for mn. All the proof supposes, is N=mn for some base case ( primes or prime powers to start, in these cases) with m and n proved. Then it follows for N, which by saying N which
originally was consider the least element of a set of counterexamples, has one, it eliminates all possible least elements for the set we originally supposed existed.
answered 1 hour ago
Roddy MacPheeRoddy MacPhee
537118
$endgroup$
A proof by induction has base case(s), Let m and n be said base cases. if it's true for m and true for n (not necessarily distinct), then because it's a product, it follows for mn. All the proof supposes, is N=mn for some base case ( primes or prime powers to start, in these cases) with m and n proved. Then it follows for N, which by saying N which
originally was consider the least element of a set of counterexamples, has one, it eliminates all possible least elements for the set we originally supposed existed.
answered 1 hour ago
Roddy MacPheeRoddy MacPhee
537118
answered 1 hour ago
Roddy MacPheeRoddy MacPhee
537118
answered 1 hour ago
Roddy MacPheeRoddy MacPhee
537118
answered 1 hour ago
Roddy MacPheeRoddy MacPhee
537118
537118
add a comment |
add a comment |
Alena Gusakov is a new contributor. Be nice, and check out our Code of Conduct.
Alena Gusakov is a new contributor. Be nice, and check out our Code of Conduct.
Alena Gusakov is a new contributor. Be nice, and check out our Code of Conduct.
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That argument is by induction. the result is easy to check for small numbers, so assume it holds up to $N-1$. Then $N$ is either prime, in which case we are done, or it factors as $atimes b$ with $1<a≤b<N-1$ and you can apply the inductive hypothesis to $a,b$. Same argument.
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– lulu
3 hours ago
1
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There is nothing missing in this proof. It is just fine. And why “two primes”?
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– José Carlos Santos
3 hours ago
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@JoséCarlosSantos Typo. Fixed.
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– Alena Gusakov
3 hours ago
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It's not circular, but it could be a lot clearer. It's not strictly necessary to say $n > 1$, since $m$ is positive and $mn$ is also positive.
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– Robert Soupe
2 hours ago
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It's a valid proof by infinite descent (a.k.a. minimal criminal), the contrapositive of induction - see the Remark here. You should master both this (negative) and the normal (positive) form of induction.
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– Bill Dubuque
46 mins ago