Prove that for any integer $n > 3$ If $n$ is a prime number then $n$ $in$ $[1]_6$ or $n$ $in$ $[5]_6$












0














I am not sure how to proceed with the proof. That is, I’m trying to prove the contapostive, but I don’t know how to prove that a composite number greater than 3 will not be in either set. Insight would be appreciated.






proof-writing proof-explanation







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  • 1




    Prove that $[0]_6, [2]_6,[3]_6, [4]_6$ can't contain any primes greater than $3$.
    – fleablood
    Dec 2 at 0:32










  • Apply the division algorithm.
    – dcheuk
    Dec 2 at 0:34










  • "but I don’t know how to prove that a composositve number greater than 3 will not be in either set" That is not the contrapositive and it is false. ( $35in [6]_6$ and $25in [1]_6$)$ The contrapositive is that the other sets don't have any primes greater than $3$.
    – fleablood
    Dec 2 at 0:34
















0














I am not sure how to proceed with the proof. That is, I’m trying to prove the contapostive, but I don’t know how to prove that a composite number greater than 3 will not be in either set. Insight would be appreciated.






proof-writing proof-explanation







share|cite|improve this question




















  • 1




    Prove that $[0]_6, [2]_6,[3]_6, [4]_6$ can't contain any primes greater than $3$.
    – fleablood
    Dec 2 at 0:32










  • Apply the division algorithm.
    – dcheuk
    Dec 2 at 0:34










  • "but I don’t know how to prove that a composositve number greater than 3 will not be in either set" That is not the contrapositive and it is false. ( $35in [6]_6$ and $25in [1]_6$)$ The contrapositive is that the other sets don't have any primes greater than $3$.
    – fleablood
    Dec 2 at 0:34














0












0








0







I am not sure how to proceed with the proof. That is, I’m trying to prove the contapostive, but I don’t know how to prove that a composite number greater than 3 will not be in either set. Insight would be appreciated.






proof-writing proof-explanation







share|cite|improve this question















I am not sure how to proceed with the proof. That is, I’m trying to prove the contapostive, but I don’t know how to prove that a composite number greater than 3 will not be in either set. Insight would be appreciated.






proof-writing proof-explanation




proof-writing proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 at 0:33

























asked Dec 2 at 0:28









darylnak

158111




158111








  • 1




    Prove that $[0]_6, [2]_6,[3]_6, [4]_6$ can't contain any primes greater than $3$.
    – fleablood
    Dec 2 at 0:32










  • Apply the division algorithm.
    – dcheuk
    Dec 2 at 0:34










  • "but I don’t know how to prove that a composositve number greater than 3 will not be in either set" That is not the contrapositive and it is false. ( $35in [6]_6$ and $25in [1]_6$)$ The contrapositive is that the other sets don't have any primes greater than $3$.
    – fleablood
    Dec 2 at 0:34














  • 1




    Prove that $[0]_6, [2]_6,[3]_6, [4]_6$ can't contain any primes greater than $3$.
    – fleablood
    Dec 2 at 0:32










  • Apply the division algorithm.
    – dcheuk
    Dec 2 at 0:34










  • "but I don’t know how to prove that a composositve number greater than 3 will not be in either set" That is not the contrapositive and it is false. ( $35in [6]_6$ and $25in [1]_6$)$ The contrapositive is that the other sets don't have any primes greater than $3$.
    – fleablood
    Dec 2 at 0:34








1




1




Prove that $[0]_6, [2]_6,[3]_6, [4]_6$ can't contain any primes greater than $3$.
– fleablood
Dec 2 at 0:32




Prove that $[0]_6, [2]_6,[3]_6, [4]_6$ can't contain any primes greater than $3$.
– fleablood
Dec 2 at 0:32












Apply the division algorithm.
– dcheuk
Dec 2 at 0:34




Apply the division algorithm.
– dcheuk
Dec 2 at 0:34












"but I don’t know how to prove that a composositve number greater than 3 will not be in either set" That is not the contrapositive and it is false. ( $35in [6]_6$ and $25in [1]_6$)$ The contrapositive is that the other sets don't have any primes greater than $3$.
– fleablood
Dec 2 at 0:34




"but I don’t know how to prove that a composositve number greater than 3 will not be in either set" That is not the contrapositive and it is false. ( $35in [6]_6$ and $25in [1]_6$)$ The contrapositive is that the other sets don't have any primes greater than $3$.
– fleablood
Dec 2 at 0:34










2 Answers
2






active

oldest

votes


















2














" I don’t know how to prove that a composite number greater than 3 will not be in either set. "



That isn't the contrapositive and it isn't true.



The contra positive of: If $p$ is prime and $p > 3$ $implies$ $pin[1]_6$ or $pin [5]_6$. then the contra positive is



$pnot in [1]_6$ and $pnot in [5]_6$ $implies$ $p$ is not prime or $p le 3$.



So prove that if $pin [0]_6, [2]_6, [3]_6, [4]_6$ then eithe $p$ is composite of $p le 3$.



.....



Might be worth noting. If $m in [k]_6$ then $gcd(k,6)|m$. Do you see why that would be true?






share|cite|improve this answer





















  • I’m still not sure how to proceed with the proof. Would you mind walking me through it a bit?
    – darylnak
    Dec 2 at 1:23










  • What prime numbers, if any, are in $[0]_6$? Why or why not.
    – fleablood
    Dec 2 at 3:19










  • What sort of numbers are in $[0]_6$? Why can't they be prime? What sort of munbers are in $[2]_6$? Why is $2$ the only prime in it. What about $[3]_6$? Why is $3$ the only prime in it.
    – fleablood
    Dec 2 at 3:22










  • I’ve seen the light
    – darylnak
    Dec 2 at 3:31



















2














Well:



$$6 | 6n$$
$$2|(6n+2)$$
$$3|(6n+3)$$
$$2|(6n+4)$$






share|cite|improve this answer

















  • 1




    I don’t understand this. Could you please explain?
    – darylnak
    Dec 2 at 1:03










  • Any integer is of one of the following forms: $6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5$. My answer shows that four of those forms are composite, so the other two are the only ones primes can fall into. The $a|b$ notation means a divides b, in other words, $b$ is a multiple of $a$. In the case of $6n+2$, for example, we have $$6n+2=2(3n+1)$$
    – Rhys Hughes
    Dec 2 at 1:08











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














" I don’t know how to prove that a composite number greater than 3 will not be in either set. "



That isn't the contrapositive and it isn't true.



The contra positive of: If $p$ is prime and $p > 3$ $implies$ $pin[1]_6$ or $pin [5]_6$. then the contra positive is



$pnot in [1]_6$ and $pnot in [5]_6$ $implies$ $p$ is not prime or $p le 3$.



So prove that if $pin [0]_6, [2]_6, [3]_6, [4]_6$ then eithe $p$ is composite of $p le 3$.



.....



Might be worth noting. If $m in [k]_6$ then $gcd(k,6)|m$. Do you see why that would be true?






share|cite|improve this answer





















  • I’m still not sure how to proceed with the proof. Would you mind walking me through it a bit?
    – darylnak
    Dec 2 at 1:23










  • What prime numbers, if any, are in $[0]_6$? Why or why not.
    – fleablood
    Dec 2 at 3:19










  • What sort of numbers are in $[0]_6$? Why can't they be prime? What sort of munbers are in $[2]_6$? Why is $2$ the only prime in it. What about $[3]_6$? Why is $3$ the only prime in it.
    – fleablood
    Dec 2 at 3:22










  • I’ve seen the light
    – darylnak
    Dec 2 at 3:31
















2














" I don’t know how to prove that a composite number greater than 3 will not be in either set. "



That isn't the contrapositive and it isn't true.



The contra positive of: If $p$ is prime and $p > 3$ $implies$ $pin[1]_6$ or $pin [5]_6$. then the contra positive is



$pnot in [1]_6$ and $pnot in [5]_6$ $implies$ $p$ is not prime or $p le 3$.



So prove that if $pin [0]_6, [2]_6, [3]_6, [4]_6$ then eithe $p$ is composite of $p le 3$.



.....



Might be worth noting. If $m in [k]_6$ then $gcd(k,6)|m$. Do you see why that would be true?






share|cite|improve this answer





















  • I’m still not sure how to proceed with the proof. Would you mind walking me through it a bit?
    – darylnak
    Dec 2 at 1:23










  • What prime numbers, if any, are in $[0]_6$? Why or why not.
    – fleablood
    Dec 2 at 3:19










  • What sort of numbers are in $[0]_6$? Why can't they be prime? What sort of munbers are in $[2]_6$? Why is $2$ the only prime in it. What about $[3]_6$? Why is $3$ the only prime in it.
    – fleablood
    Dec 2 at 3:22










  • I’ve seen the light
    – darylnak
    Dec 2 at 3:31














2












2








2






" I don’t know how to prove that a composite number greater than 3 will not be in either set. "



That isn't the contrapositive and it isn't true.



The contra positive of: If $p$ is prime and $p > 3$ $implies$ $pin[1]_6$ or $pin [5]_6$. then the contra positive is



$pnot in [1]_6$ and $pnot in [5]_6$ $implies$ $p$ is not prime or $p le 3$.



So prove that if $pin [0]_6, [2]_6, [3]_6, [4]_6$ then eithe $p$ is composite of $p le 3$.



.....



Might be worth noting. If $m in [k]_6$ then $gcd(k,6)|m$. Do you see why that would be true?






share|cite|improve this answer












" I don’t know how to prove that a composite number greater than 3 will not be in either set. "



That isn't the contrapositive and it isn't true.



The contra positive of: If $p$ is prime and $p > 3$ $implies$ $pin[1]_6$ or $pin [5]_6$. then the contra positive is



$pnot in [1]_6$ and $pnot in [5]_6$ $implies$ $p$ is not prime or $p le 3$.



So prove that if $pin [0]_6, [2]_6, [3]_6, [4]_6$ then eithe $p$ is composite of $p le 3$.



.....



Might be worth noting. If $m in [k]_6$ then $gcd(k,6)|m$. Do you see why that would be true?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 at 0:40









fleablood

68.1k22684




68.1k22684












  • I’m still not sure how to proceed with the proof. Would you mind walking me through it a bit?
    – darylnak
    Dec 2 at 1:23










  • What prime numbers, if any, are in $[0]_6$? Why or why not.
    – fleablood
    Dec 2 at 3:19










  • What sort of numbers are in $[0]_6$? Why can't they be prime? What sort of munbers are in $[2]_6$? Why is $2$ the only prime in it. What about $[3]_6$? Why is $3$ the only prime in it.
    – fleablood
    Dec 2 at 3:22










  • I’ve seen the light
    – darylnak
    Dec 2 at 3:31


















  • I’m still not sure how to proceed with the proof. Would you mind walking me through it a bit?
    – darylnak
    Dec 2 at 1:23










  • What prime numbers, if any, are in $[0]_6$? Why or why not.
    – fleablood
    Dec 2 at 3:19










  • What sort of numbers are in $[0]_6$? Why can't they be prime? What sort of munbers are in $[2]_6$? Why is $2$ the only prime in it. What about $[3]_6$? Why is $3$ the only prime in it.
    – fleablood
    Dec 2 at 3:22










  • I’ve seen the light
    – darylnak
    Dec 2 at 3:31
















I’m still not sure how to proceed with the proof. Would you mind walking me through it a bit?
– darylnak
Dec 2 at 1:23




I’m still not sure how to proceed with the proof. Would you mind walking me through it a bit?
– darylnak
Dec 2 at 1:23












What prime numbers, if any, are in $[0]_6$? Why or why not.
– fleablood
Dec 2 at 3:19




What prime numbers, if any, are in $[0]_6$? Why or why not.
– fleablood
Dec 2 at 3:19












What sort of numbers are in $[0]_6$? Why can't they be prime? What sort of munbers are in $[2]_6$? Why is $2$ the only prime in it. What about $[3]_6$? Why is $3$ the only prime in it.
– fleablood
Dec 2 at 3:22




What sort of numbers are in $[0]_6$? Why can't they be prime? What sort of munbers are in $[2]_6$? Why is $2$ the only prime in it. What about $[3]_6$? Why is $3$ the only prime in it.
– fleablood
Dec 2 at 3:22












I’ve seen the light
– darylnak
Dec 2 at 3:31




I’ve seen the light
– darylnak
Dec 2 at 3:31











2














Well:



$$6 | 6n$$
$$2|(6n+2)$$
$$3|(6n+3)$$
$$2|(6n+4)$$






share|cite|improve this answer

















  • 1




    I don’t understand this. Could you please explain?
    – darylnak
    Dec 2 at 1:03










  • Any integer is of one of the following forms: $6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5$. My answer shows that four of those forms are composite, so the other two are the only ones primes can fall into. The $a|b$ notation means a divides b, in other words, $b$ is a multiple of $a$. In the case of $6n+2$, for example, we have $$6n+2=2(3n+1)$$
    – Rhys Hughes
    Dec 2 at 1:08
















2














Well:



$$6 | 6n$$
$$2|(6n+2)$$
$$3|(6n+3)$$
$$2|(6n+4)$$






share|cite|improve this answer

















  • 1




    I don’t understand this. Could you please explain?
    – darylnak
    Dec 2 at 1:03










  • Any integer is of one of the following forms: $6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5$. My answer shows that four of those forms are composite, so the other two are the only ones primes can fall into. The $a|b$ notation means a divides b, in other words, $b$ is a multiple of $a$. In the case of $6n+2$, for example, we have $$6n+2=2(3n+1)$$
    – Rhys Hughes
    Dec 2 at 1:08














2












2








2






Well:



$$6 | 6n$$
$$2|(6n+2)$$
$$3|(6n+3)$$
$$2|(6n+4)$$






share|cite|improve this answer












Well:



$$6 | 6n$$
$$2|(6n+2)$$
$$3|(6n+3)$$
$$2|(6n+4)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 at 0:43









Rhys Hughes

4,7761327




4,7761327








  • 1




    I don’t understand this. Could you please explain?
    – darylnak
    Dec 2 at 1:03










  • Any integer is of one of the following forms: $6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5$. My answer shows that four of those forms are composite, so the other two are the only ones primes can fall into. The $a|b$ notation means a divides b, in other words, $b$ is a multiple of $a$. In the case of $6n+2$, for example, we have $$6n+2=2(3n+1)$$
    – Rhys Hughes
    Dec 2 at 1:08














  • 1




    I don’t understand this. Could you please explain?
    – darylnak
    Dec 2 at 1:03










  • Any integer is of one of the following forms: $6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5$. My answer shows that four of those forms are composite, so the other two are the only ones primes can fall into. The $a|b$ notation means a divides b, in other words, $b$ is a multiple of $a$. In the case of $6n+2$, for example, we have $$6n+2=2(3n+1)$$
    – Rhys Hughes
    Dec 2 at 1:08








1




1




I don’t understand this. Could you please explain?
– darylnak
Dec 2 at 1:03




I don’t understand this. Could you please explain?
– darylnak
Dec 2 at 1:03












Any integer is of one of the following forms: $6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5$. My answer shows that four of those forms are composite, so the other two are the only ones primes can fall into. The $a|b$ notation means a divides b, in other words, $b$ is a multiple of $a$. In the case of $6n+2$, for example, we have $$6n+2=2(3n+1)$$
– Rhys Hughes
Dec 2 at 1:08




Any integer is of one of the following forms: $6n, 6n+1, 6n+2, 6n+3, 6n+4, 6n+5$. My answer shows that four of those forms are composite, so the other two are the only ones primes can fall into. The $a|b$ notation means a divides b, in other words, $b$ is a multiple of $a$. In the case of $6n+2$, for example, we have $$6n+2=2(3n+1)$$
– Rhys Hughes
Dec 2 at 1:08


















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