Is this proof valid? $P|a^k implies P|a$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that if $p$ is prime and $pmid a_1 a_2cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, …, n$.Least Element $implies ngeq 2$ Has Prime Factorization: An Analysis of Strong InductionProving this property of a tournament graph by induction?$a_1a_2cdots a_n = 1 implies a_1 + a_2 + cdots + a_n geq n$ if $a_1, a_2, dots, a_n > 0$Proof by Induction $3^n > n^3$Proof by contradiction and mathematical inductionFibonacci Sequence: Prove $f_1+f_3+dots+f_2n-1=f_2n$ by Induction.Is This Proof Valid? - $sum_k=1^nmu(k!)=1$ for $n geq 3$Is this a valid proof for $n! gt 2^n$ for $n ge 4$How to use induction hypothesis correctly in a proofCan someone please check if my reasoning for this proof is valid or not?
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Is this proof valid? $P|a^k implies P|a$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that if $p$ is prime and $pmid a_1 a_2cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, …, n$.Least Element $implies ngeq 2$ Has Prime Factorization: An Analysis of Strong InductionProving this property of a tournament graph by induction?$a_1a_2cdots a_n = 1 implies a_1 + a_2 + cdots + a_n geq n$ if $a_1, a_2, dots, a_n > 0$Proof by Induction $3^n > n^3$Proof by contradiction and mathematical inductionFibonacci Sequence: Prove $f_1+f_3+dots+f_2n-1=f_2n$ by Induction.Is This Proof Valid? - $sum_k=1^nmu(k!)=1$ for $n geq 3$Is this a valid proof for $n! gt 2^n$ for $n ge 4$How to use induction hypothesis correctly in a proofCan someone please check if my reasoning for this proof is valid or not?
$begingroup$
I'm trying to prove $P|a^k implies P|a$
$p$ is a prime,
$a in mathbbZ^+ \$,
$n geq 2 in mathbbZ$
I know that: $P|ab implies P|a text or P|b dots [1]$
a is a positive integer,
b is a positive integer
Base:
$P|a^2 implies P|a$
$P|aa implies P|a$ using [1]
Induction hypothesis:
Assume $P|a^k implies P|a$
Check: $P|a^k+1 implies P|a$
$P|aa^k implies P|a$ using [1]
therefore $P|a^k implies P|a$
discrete-mathematics proof-verification induction divisibility
edited Apr 1 at 16:05
Eric Toporek
10910
asked Mar 1 at 17:41
DonaldDonald
1556
$endgroup$
add a comment |
$begingroup$
I'm trying to prove $P|a^k implies P|a$
$p$ is a prime,
$a in mathbbZ^+ \$,
$n geq 2 in mathbbZ$
I know that: $P|ab implies P|a text or P|b dots [1]$
a is a positive integer,
b is a positive integer
Base:
$P|a^2 implies P|a$
$P|aa implies P|a$ using [1]
Induction hypothesis:
Assume $P|a^k implies P|a$
Check: $P|a^k+1 implies P|a$
$P|aa^k implies P|a$ using [1]
therefore $P|a^k implies P|a$
discrete-mathematics proof-verification induction divisibility
edited Apr 1 at 16:05
Eric Toporek
10910
asked Mar 1 at 17:41
DonaldDonald
1556
$endgroup$
$begingroup$
Duplicate of Prove that if $p$ is prime and $pmid a_1 a_2cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, ...., n$.
$endgroup$
– Bill Dubuque
Mar 1 at 18:04
$begingroup$
Bill Dubuque its not a duplicate. I use a different method and different base case
$endgroup$
– Donald
Mar 1 at 18:20
$begingroup$
Au contraire, it is essentially the same (you should use that simpler base case, but the base case plays no role in the essence of your question). A good answer would have explained that and much more (but it's unlikely you'll get one after accepting the fastest answer - a common newbie mistake)
$endgroup$
– Bill Dubuque
Mar 1 at 18:29
add a comment |
$begingroup$
I'm trying to prove $P|a^k implies P|a$
$p$ is a prime,
$a in mathbbZ^+ \$,
$n geq 2 in mathbbZ$
I know that: $P|ab implies P|a text or P|b dots [1]$
a is a positive integer,
b is a positive integer
Base:
$P|a^2 implies P|a$
$P|aa implies P|a$ using [1]
Induction hypothesis:
Assume $P|a^k implies P|a$
Check: $P|a^k+1 implies P|a$
$P|aa^k implies P|a$ using [1]
therefore $P|a^k implies P|a$
discrete-mathematics proof-verification induction divisibility
edited Apr 1 at 16:05
Eric Toporek
10910
asked Mar 1 at 17:41
DonaldDonald
1556
$endgroup$
I'm trying to prove $P|a^k implies P|a$
$p$ is a prime,
$a in mathbbZ^+ \$,
$n geq 2 in mathbbZ$
I know that: $P|ab implies P|a text or P|b dots [1]$
a is a positive integer,
b is a positive integer
Base:
$P|a^2 implies P|a$
$P|aa implies P|a$ using [1]
Induction hypothesis:
Assume $P|a^k implies P|a$
Check: $P|a^k+1 implies P|a$
$P|aa^k implies P|a$ using [1]
therefore $P|a^k implies P|a$
discrete-mathematics proof-verification induction divisibility
discrete-mathematics proof-verification induction divisibility
edited Apr 1 at 16:05
Eric Toporek
10910
asked Mar 1 at 17:41
DonaldDonald
1556
edited Apr 1 at 16:05
Eric Toporek
10910
asked Mar 1 at 17:41
DonaldDonald
1556
edited Apr 1 at 16:05
Eric Toporek
10910
edited Apr 1 at 16:05
Eric Toporek
10910
edited Apr 1 at 16:05
Eric Toporek
10910
10910
asked Mar 1 at 17:41
DonaldDonald
1556
asked Mar 1 at 17:41
DonaldDonald
1556
asked Mar 1 at 17:41
DonaldDonald
1556
1556
$begingroup$
Duplicate of Prove that if $p$ is prime and $pmid a_1 a_2cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, ...., n$.
$endgroup$
– Bill Dubuque
Mar 1 at 18:04
$begingroup$
Bill Dubuque its not a duplicate. I use a different method and different base case
$endgroup$
– Donald
Mar 1 at 18:20
$begingroup$
Au contraire, it is essentially the same (you should use that simpler base case, but the base case plays no role in the essence of your question). A good answer would have explained that and much more (but it's unlikely you'll get one after accepting the fastest answer - a common newbie mistake)
$endgroup$
– Bill Dubuque
Mar 1 at 18:29
add a comment |
$begingroup$
Duplicate of Prove that if $p$ is prime and $pmid a_1 a_2cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, ...., n$.
$endgroup$
– Bill Dubuque
Mar 1 at 18:04
$begingroup$
Bill Dubuque its not a duplicate. I use a different method and different base case
$endgroup$
– Donald
Mar 1 at 18:20
$begingroup$
Au contraire, it is essentially the same (you should use that simpler base case, but the base case plays no role in the essence of your question). A good answer would have explained that and much more (but it's unlikely you'll get one after accepting the fastest answer - a common newbie mistake)
$endgroup$
– Bill Dubuque
Mar 1 at 18:29
$begingroup$
Duplicate of Prove that if $p$ is prime and $pmid a_1 a_2cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, ...., n$.
$endgroup$
– Bill Dubuque
Mar 1 at 18:04
$begingroup$
Duplicate of Prove that if $p$ is prime and $pmid a_1 a_2cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, ...., n$.
$endgroup$
– Bill Dubuque
Mar 1 at 18:04
$begingroup$
Bill Dubuque its not a duplicate. I use a different method and different base case
$endgroup$
– Donald
Mar 1 at 18:20
$begingroup$
Bill Dubuque its not a duplicate. I use a different method and different base case
$endgroup$
– Donald
Mar 1 at 18:20
$begingroup$
Au contraire, it is essentially the same (you should use that simpler base case, but the base case plays no role in the essence of your question). A good answer would have explained that and much more (but it's unlikely you'll get one after accepting the fastest answer - a common newbie mistake)
$endgroup$
– Bill Dubuque
Mar 1 at 18:29
$begingroup$
Au contraire, it is essentially the same (you should use that simpler base case, but the base case plays no role in the essence of your question). A good answer would have explained that and much more (but it's unlikely you'll get one after accepting the fastest answer - a common newbie mistake)
$endgroup$
– Bill Dubuque
Mar 1 at 18:29
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It is not correct. From $Pmid a^k+1=atimes a^k$, what you can deduce is that $Pmid a$ or that $Pmid a^k$. And then you can use the induction hypothesis.
answered Mar 1 at 17:45
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
So,P|aXa^k -> P|a or P|a^k then, P|a*a^k -> P|a or (P|a^k -> P|a) ?
$endgroup$
– Donald
Mar 1 at 17:49
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Mar 1 at 17:52
add a comment |
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
It is not correct. From $Pmid a^k+1=atimes a^k$, what you can deduce is that $Pmid a$ or that $Pmid a^k$. And then you can use the induction hypothesis.
answered Mar 1 at 17:45
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
So,P|aXa^k -> P|a or P|a^k then, P|a*a^k -> P|a or (P|a^k -> P|a) ?
$endgroup$
– Donald
Mar 1 at 17:49
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Mar 1 at 17:52
add a comment |
$begingroup$
It is not correct. From $Pmid a^k+1=atimes a^k$, what you can deduce is that $Pmid a$ or that $Pmid a^k$. And then you can use the induction hypothesis.
answered Mar 1 at 17:45
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
$begingroup$
So,P|aXa^k -> P|a or P|a^k then, P|a*a^k -> P|a or (P|a^k -> P|a) ?
$endgroup$
– Donald
Mar 1 at 17:49
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Mar 1 at 17:52
add a comment |
$begingroup$
It is not correct. From $Pmid a^k+1=atimes a^k$, what you can deduce is that $Pmid a$ or that $Pmid a^k$. And then you can use the induction hypothesis.
answered Mar 1 at 17:45
José Carlos SantosJosé Carlos Santos
175k24134243
$endgroup$
It is not correct. From $Pmid a^k+1=atimes a^k$, what you can deduce is that $Pmid a$ or that $Pmid a^k$. And then you can use the induction hypothesis.
answered Mar 1 at 17:45
José Carlos SantosJosé Carlos Santos
175k24134243
edited Apr 1 at 18:33
answered Mar 1 at 17:45
José Carlos SantosJosé Carlos Santos
175k24134243
answered Mar 1 at 17:45
José Carlos SantosJosé Carlos Santos
175k24134243
answered Mar 1 at 17:45
José Carlos SantosJosé Carlos Santos
175k24134243
175k24134243
$begingroup$
So,P|aXa^k -> P|a or P|a^k then, P|a*a^k -> P|a or (P|a^k -> P|a) ?
$endgroup$
– Donald
Mar 1 at 17:49
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Mar 1 at 17:52
add a comment |
$begingroup$
So,P|aXa^k -> P|a or P|a^k then, P|a*a^k -> P|a or (P|a^k -> P|a) ?
$endgroup$
– Donald
Mar 1 at 17:49
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Mar 1 at 17:52
$begingroup$
So,P|aXa^k -> P|a or P|a^k then, P|a*a^k -> P|a or (P|a^k -> P|a) ?
$endgroup$
– Donald
Mar 1 at 17:49
$begingroup$
So,P|aXa^k -> P|a or P|a^k then, P|a*a^k -> P|a or (P|a^k -> P|a) ?
$endgroup$
– Donald
Mar 1 at 17:49
1
1
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Mar 1 at 17:52
$begingroup$
Yes, that is correct.
$endgroup$
– José Carlos Santos
Mar 1 at 17:52
add a comment |
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$begingroup$
Duplicate of Prove that if $p$ is prime and $pmid a_1 a_2cdots a_n$ then $ p| a_i$ for some $ i = 1, 2, ...., n$.
$endgroup$
– Bill Dubuque
Mar 1 at 18:04
$begingroup$
Bill Dubuque its not a duplicate. I use a different method and different base case
$endgroup$
– Donald
Mar 1 at 18:20
$begingroup$
Au contraire, it is essentially the same (you should use that simpler base case, but the base case plays no role in the essence of your question). A good answer would have explained that and much more (but it's unlikely you'll get one after accepting the fastest answer - a common newbie mistake)
$endgroup$
– Bill Dubuque
Mar 1 at 18:29