If $n$ is divisible by 6, then $n$ is divisible by 3












1












$begingroup$


There are various questions that are asking me to find out which of the following statements are true and to explain briefly (no proof required, although id be interested to see what it was). I've tried finding similar proofs online that could at least point me in the right direction as to figuring out which are true and false but didn't find anything.




  1. If $n$ is divisibale by 6, then $n$ is divisible by 3


  2. If $n$ is divisible by 3 then $n$ is divisible by 6.


  3. If $n$ is divisible by 2 and $n$ is divisible by 3, then $n$ is divisible by 6



How should I approach a problem like this? (This is my first class on mathematical reasoning)






proof-verification proof-explanation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    First, just try some numbers. Can you think of a number divisible by 3?
    $endgroup$
    – Bram28
    Jan 14 at 0:25










  • $begingroup$
    Sarcasm? - "can i think of a number divisible by 3" Anyway, just because you can find a number that is divisible by 6 and 3, is that sufficient in proving the statement true or false? Arent you suppose to show that it is always the case that if a number is divisible by 6 then it is also divisible by 3?
    $endgroup$
    – Forextrader
    Jan 14 at 0:29










  • $begingroup$
    If a number $n$ is divisible by 6, then it can be written as $n=6k$ for some integer $k$, and thus $n$ can also be written as ..., which shows that $n$ is divisible by 3.
    $endgroup$
    – twnly
    Jan 14 at 0:30












  • $begingroup$
    @Forextrader I was hoping you'd think of $3$, and realize that $3$ is not divisible by $6$, thus providing you with the answer to question 2: "No"
    $endgroup$
    – Bram28
    Jan 14 at 14:43
















1












$begingroup$


There are various questions that are asking me to find out which of the following statements are true and to explain briefly (no proof required, although id be interested to see what it was). I've tried finding similar proofs online that could at least point me in the right direction as to figuring out which are true and false but didn't find anything.




  1. If $n$ is divisibale by 6, then $n$ is divisible by 3


  2. If $n$ is divisible by 3 then $n$ is divisible by 6.


  3. If $n$ is divisible by 2 and $n$ is divisible by 3, then $n$ is divisible by 6



How should I approach a problem like this? (This is my first class on mathematical reasoning)






proof-verification proof-explanation







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    First, just try some numbers. Can you think of a number divisible by 3?
    $endgroup$
    – Bram28
    Jan 14 at 0:25










  • $begingroup$
    Sarcasm? - "can i think of a number divisible by 3" Anyway, just because you can find a number that is divisible by 6 and 3, is that sufficient in proving the statement true or false? Arent you suppose to show that it is always the case that if a number is divisible by 6 then it is also divisible by 3?
    $endgroup$
    – Forextrader
    Jan 14 at 0:29










  • $begingroup$
    If a number $n$ is divisible by 6, then it can be written as $n=6k$ for some integer $k$, and thus $n$ can also be written as ..., which shows that $n$ is divisible by 3.
    $endgroup$
    – twnly
    Jan 14 at 0:30












  • $begingroup$
    @Forextrader I was hoping you'd think of $3$, and realize that $3$ is not divisible by $6$, thus providing you with the answer to question 2: "No"
    $endgroup$
    – Bram28
    Jan 14 at 14:43














1












1








1





$begingroup$


There are various questions that are asking me to find out which of the following statements are true and to explain briefly (no proof required, although id be interested to see what it was). I've tried finding similar proofs online that could at least point me in the right direction as to figuring out which are true and false but didn't find anything.




  1. If $n$ is divisibale by 6, then $n$ is divisible by 3


  2. If $n$ is divisible by 3 then $n$ is divisible by 6.


  3. If $n$ is divisible by 2 and $n$ is divisible by 3, then $n$ is divisible by 6



How should I approach a problem like this? (This is my first class on mathematical reasoning)






proof-verification proof-explanation







share|cite|improve this question











$endgroup$




There are various questions that are asking me to find out which of the following statements are true and to explain briefly (no proof required, although id be interested to see what it was). I've tried finding similar proofs online that could at least point me in the right direction as to figuring out which are true and false but didn't find anything.




  1. If $n$ is divisibale by 6, then $n$ is divisible by 3


  2. If $n$ is divisible by 3 then $n$ is divisible by 6.


  3. If $n$ is divisible by 2 and $n$ is divisible by 3, then $n$ is divisible by 6



How should I approach a problem like this? (This is my first class on mathematical reasoning)






proof-verification proof-explanation




proof-verification proof-explanation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 0:29









David C. Ullrich

61.6k43994




61.6k43994










asked Jan 14 at 0:24









ForextraderForextrader

968




968








  • 1




    $begingroup$
    First, just try some numbers. Can you think of a number divisible by 3?
    $endgroup$
    – Bram28
    Jan 14 at 0:25










  • $begingroup$
    Sarcasm? - "can i think of a number divisible by 3" Anyway, just because you can find a number that is divisible by 6 and 3, is that sufficient in proving the statement true or false? Arent you suppose to show that it is always the case that if a number is divisible by 6 then it is also divisible by 3?
    $endgroup$
    – Forextrader
    Jan 14 at 0:29










  • $begingroup$
    If a number $n$ is divisible by 6, then it can be written as $n=6k$ for some integer $k$, and thus $n$ can also be written as ..., which shows that $n$ is divisible by 3.
    $endgroup$
    – twnly
    Jan 14 at 0:30












  • $begingroup$
    @Forextrader I was hoping you'd think of $3$, and realize that $3$ is not divisible by $6$, thus providing you with the answer to question 2: "No"
    $endgroup$
    – Bram28
    Jan 14 at 14:43














  • 1




    $begingroup$
    First, just try some numbers. Can you think of a number divisible by 3?
    $endgroup$
    – Bram28
    Jan 14 at 0:25










  • $begingroup$
    Sarcasm? - "can i think of a number divisible by 3" Anyway, just because you can find a number that is divisible by 6 and 3, is that sufficient in proving the statement true or false? Arent you suppose to show that it is always the case that if a number is divisible by 6 then it is also divisible by 3?
    $endgroup$
    – Forextrader
    Jan 14 at 0:29










  • $begingroup$
    If a number $n$ is divisible by 6, then it can be written as $n=6k$ for some integer $k$, and thus $n$ can also be written as ..., which shows that $n$ is divisible by 3.
    $endgroup$
    – twnly
    Jan 14 at 0:30












  • $begingroup$
    @Forextrader I was hoping you'd think of $3$, and realize that $3$ is not divisible by $6$, thus providing you with the answer to question 2: "No"
    $endgroup$
    – Bram28
    Jan 14 at 14:43








1




1




$begingroup$
First, just try some numbers. Can you think of a number divisible by 3?
$endgroup$
– Bram28
Jan 14 at 0:25




$begingroup$
First, just try some numbers. Can you think of a number divisible by 3?
$endgroup$
– Bram28
Jan 14 at 0:25












$begingroup$
Sarcasm? - "can i think of a number divisible by 3" Anyway, just because you can find a number that is divisible by 6 and 3, is that sufficient in proving the statement true or false? Arent you suppose to show that it is always the case that if a number is divisible by 6 then it is also divisible by 3?
$endgroup$
– Forextrader
Jan 14 at 0:29




$begingroup$
Sarcasm? - "can i think of a number divisible by 3" Anyway, just because you can find a number that is divisible by 6 and 3, is that sufficient in proving the statement true or false? Arent you suppose to show that it is always the case that if a number is divisible by 6 then it is also divisible by 3?
$endgroup$
– Forextrader
Jan 14 at 0:29












$begingroup$
If a number $n$ is divisible by 6, then it can be written as $n=6k$ for some integer $k$, and thus $n$ can also be written as ..., which shows that $n$ is divisible by 3.
$endgroup$
– twnly
Jan 14 at 0:30






$begingroup$
If a number $n$ is divisible by 6, then it can be written as $n=6k$ for some integer $k$, and thus $n$ can also be written as ..., which shows that $n$ is divisible by 3.
$endgroup$
– twnly
Jan 14 at 0:30














$begingroup$
@Forextrader I was hoping you'd think of $3$, and realize that $3$ is not divisible by $6$, thus providing you with the answer to question 2: "No"
$endgroup$
– Bram28
Jan 14 at 14:43




$begingroup$
@Forextrader I was hoping you'd think of $3$, and realize that $3$ is not divisible by $6$, thus providing you with the answer to question 2: "No"
$endgroup$
– Bram28
Jan 14 at 14:43










2 Answers
2






active

oldest

votes


















3












$begingroup$

For the first question, you can prove it like this. Suppose that $n$ is divisible by $6$. Then, $n = m*6$ where $m$ is an integer. Note that $6 = 2*3$ so that $n = m*(2*3) = (2m)*3.$ Therefore, $n$ is also a multiple of $3$.



Also, remember that to show something is false, you just need to provide one counterexample. This would apply to question 2.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, that really helps!
    $endgroup$
    – Forextrader
    Jan 14 at 0:41










  • $begingroup$
    The fact that the rearrangement of parentheses keeps the product the same is called the associative property of multiplication. This is usually assumed as an axiom. It's important to know what assumptions you are working with before you attempt to prove something. Good luck @Forextrader
    $endgroup$
    – Zubin Mukerjee
    Jan 14 at 2:02



















0












$begingroup$

Partial answer:



3) $n$ is divisible by $3$ and $2$ , then $n$ is divisible by $6$.



$2|n$ implies $n=2k$;



Euclid's lemma: If $p$, prime, divides $ab$, then $p$ divides $a$ or $p$ divides $b$.



$n=2k$; Also: $3$ divides $n=2k$;



Euclid's lemma: $3| k$, i.e. $k=3l$;



Combining :



$n= 2k=2(3l)=(2)(3)l=6l$,



hence $6$ divides $n$.






share|cite|improve this answer









$endgroup$













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    2 Answers
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    2 Answers
    2






    active

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    active

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    active

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    3












    $begingroup$

    For the first question, you can prove it like this. Suppose that $n$ is divisible by $6$. Then, $n = m*6$ where $m$ is an integer. Note that $6 = 2*3$ so that $n = m*(2*3) = (2m)*3.$ Therefore, $n$ is also a multiple of $3$.



    Also, remember that to show something is false, you just need to provide one counterexample. This would apply to question 2.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, that really helps!
      $endgroup$
      – Forextrader
      Jan 14 at 0:41










    • $begingroup$
      The fact that the rearrangement of parentheses keeps the product the same is called the associative property of multiplication. This is usually assumed as an axiom. It's important to know what assumptions you are working with before you attempt to prove something. Good luck @Forextrader
      $endgroup$
      – Zubin Mukerjee
      Jan 14 at 2:02
















    3












    $begingroup$

    For the first question, you can prove it like this. Suppose that $n$ is divisible by $6$. Then, $n = m*6$ where $m$ is an integer. Note that $6 = 2*3$ so that $n = m*(2*3) = (2m)*3.$ Therefore, $n$ is also a multiple of $3$.



    Also, remember that to show something is false, you just need to provide one counterexample. This would apply to question 2.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you, that really helps!
      $endgroup$
      – Forextrader
      Jan 14 at 0:41










    • $begingroup$
      The fact that the rearrangement of parentheses keeps the product the same is called the associative property of multiplication. This is usually assumed as an axiom. It's important to know what assumptions you are working with before you attempt to prove something. Good luck @Forextrader
      $endgroup$
      – Zubin Mukerjee
      Jan 14 at 2:02














    3












    3








    3





    $begingroup$

    For the first question, you can prove it like this. Suppose that $n$ is divisible by $6$. Then, $n = m*6$ where $m$ is an integer. Note that $6 = 2*3$ so that $n = m*(2*3) = (2m)*3.$ Therefore, $n$ is also a multiple of $3$.



    Also, remember that to show something is false, you just need to provide one counterexample. This would apply to question 2.






    share|cite|improve this answer









    $endgroup$



    For the first question, you can prove it like this. Suppose that $n$ is divisible by $6$. Then, $n = m*6$ where $m$ is an integer. Note that $6 = 2*3$ so that $n = m*(2*3) = (2m)*3.$ Therefore, $n$ is also a multiple of $3$.



    Also, remember that to show something is false, you just need to provide one counterexample. This would apply to question 2.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 14 at 0:31









    D.B.D.B.

    1,27518




    1,27518












    • $begingroup$
      Thank you, that really helps!
      $endgroup$
      – Forextrader
      Jan 14 at 0:41










    • $begingroup$
      The fact that the rearrangement of parentheses keeps the product the same is called the associative property of multiplication. This is usually assumed as an axiom. It's important to know what assumptions you are working with before you attempt to prove something. Good luck @Forextrader
      $endgroup$
      – Zubin Mukerjee
      Jan 14 at 2:02


















    • $begingroup$
      Thank you, that really helps!
      $endgroup$
      – Forextrader
      Jan 14 at 0:41










    • $begingroup$
      The fact that the rearrangement of parentheses keeps the product the same is called the associative property of multiplication. This is usually assumed as an axiom. It's important to know what assumptions you are working with before you attempt to prove something. Good luck @Forextrader
      $endgroup$
      – Zubin Mukerjee
      Jan 14 at 2:02
















    $begingroup$
    Thank you, that really helps!
    $endgroup$
    – Forextrader
    Jan 14 at 0:41




    $begingroup$
    Thank you, that really helps!
    $endgroup$
    – Forextrader
    Jan 14 at 0:41












    $begingroup$
    The fact that the rearrangement of parentheses keeps the product the same is called the associative property of multiplication. This is usually assumed as an axiom. It's important to know what assumptions you are working with before you attempt to prove something. Good luck @Forextrader
    $endgroup$
    – Zubin Mukerjee
    Jan 14 at 2:02




    $begingroup$
    The fact that the rearrangement of parentheses keeps the product the same is called the associative property of multiplication. This is usually assumed as an axiom. It's important to know what assumptions you are working with before you attempt to prove something. Good luck @Forextrader
    $endgroup$
    – Zubin Mukerjee
    Jan 14 at 2:02











    0












    $begingroup$

    Partial answer:



    3) $n$ is divisible by $3$ and $2$ , then $n$ is divisible by $6$.



    $2|n$ implies $n=2k$;



    Euclid's lemma: If $p$, prime, divides $ab$, then $p$ divides $a$ or $p$ divides $b$.



    $n=2k$; Also: $3$ divides $n=2k$;



    Euclid's lemma: $3| k$, i.e. $k=3l$;



    Combining :



    $n= 2k=2(3l)=(2)(3)l=6l$,



    hence $6$ divides $n$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Partial answer:



      3) $n$ is divisible by $3$ and $2$ , then $n$ is divisible by $6$.



      $2|n$ implies $n=2k$;



      Euclid's lemma: If $p$, prime, divides $ab$, then $p$ divides $a$ or $p$ divides $b$.



      $n=2k$; Also: $3$ divides $n=2k$;



      Euclid's lemma: $3| k$, i.e. $k=3l$;



      Combining :



      $n= 2k=2(3l)=(2)(3)l=6l$,



      hence $6$ divides $n$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Partial answer:



        3) $n$ is divisible by $3$ and $2$ , then $n$ is divisible by $6$.



        $2|n$ implies $n=2k$;



        Euclid's lemma: If $p$, prime, divides $ab$, then $p$ divides $a$ or $p$ divides $b$.



        $n=2k$; Also: $3$ divides $n=2k$;



        Euclid's lemma: $3| k$, i.e. $k=3l$;



        Combining :



        $n= 2k=2(3l)=(2)(3)l=6l$,



        hence $6$ divides $n$.






        share|cite|improve this answer









        $endgroup$



        Partial answer:



        3) $n$ is divisible by $3$ and $2$ , then $n$ is divisible by $6$.



        $2|n$ implies $n=2k$;



        Euclid's lemma: If $p$, prime, divides $ab$, then $p$ divides $a$ or $p$ divides $b$.



        $n=2k$; Also: $3$ divides $n=2k$;



        Euclid's lemma: $3| k$, i.e. $k=3l$;



        Combining :



        $n= 2k=2(3l)=(2)(3)l=6l$,



        hence $6$ divides $n$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 7:24









        Peter SzilasPeter Szilas

        11.6k2822




        11.6k2822






























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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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