If $n$ is divisible by 6, then $n$ is divisible by 3
$begingroup$
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.
If $n$ is divisibale by 6, then $n$ is divisible by 3
If $n$ is divisible by 3 then $n$ is divisible by 6.
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
edited Jan 14 at 0:29
David C. Ullrich
61.8k44095
asked Jan 14 at 0:24
ForextraderForextrader
988
$endgroup$
add a comment |
$begingroup$
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.
If $n$ is divisibale by 6, then $n$ is divisible by 3
If $n$ is divisible by 3 then $n$ is divisible by 6.
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
edited Jan 14 at 0:29
David C. Ullrich
61.8k44095
asked Jan 14 at 0:24
ForextraderForextrader
988
$endgroup$
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
add a comment |
$begingroup$
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.
If $n$ is divisibale by 6, then $n$ is divisible by 3
If $n$ is divisible by 3 then $n$ is divisible by 6.
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
edited Jan 14 at 0:29
David C. Ullrich
61.8k44095
asked Jan 14 at 0:24
ForextraderForextrader
988
$endgroup$
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.
If $n$ is divisibale by 6, then $n$ is divisible by 3
If $n$ is divisible by 3 then $n$ is divisible by 6.
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
edited Jan 14 at 0:29
David C. Ullrich
61.8k44095
asked Jan 14 at 0:24
ForextraderForextrader
988
edited Jan 14 at 0:29
David C. Ullrich
61.8k44095
asked Jan 14 at 0:24
ForextraderForextrader
988
edited Jan 14 at 0:29
David C. Ullrich
61.8k44095
edited Jan 14 at 0:29
David C. Ullrich
61.8k44095
edited Jan 14 at 0:29
David C. Ullrich
61.8k44095
61.8k44095
asked Jan 14 at 0:24
ForextraderForextrader
988
asked Jan 14 at 0:24
ForextraderForextrader
988
asked Jan 14 at 0:24
ForextraderForextrader
988
988
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Jan 14 at 0:31
D.B.D.B.
1,61029
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $n$.
answered Jan 14 at 7:24
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
answered Jan 14 at 0:31
D.B.D.B.
1,61029
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 14 at 0:31
D.B.D.B.
1,61029
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 14 at 0:31
D.B.D.B.
1,61029
$endgroup$
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.
answered Jan 14 at 0:31
D.B.D.B.
1,61029
answered Jan 14 at 0:31
D.B.D.B.
1,61029
answered Jan 14 at 0:31
D.B.D.B.
1,61029
answered Jan 14 at 0:31
D.B.D.B.
1,61029
1,61029
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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 $n$.
answered Jan 14 at 7:24
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
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 $n$.
answered Jan 14 at 7:24
Peter SzilasPeter Szilas
12k2822
$endgroup$
add a comment |
$begingroup$
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 $n$.
answered Jan 14 at 7:24
Peter SzilasPeter Szilas
12k2822
$endgroup$
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 $n$.
answered Jan 14 at 7:24
Peter SzilasPeter Szilas
12k2822
answered Jan 14 at 7:24
Peter SzilasPeter Szilas
12k2822
answered Jan 14 at 7:24
Peter SzilasPeter Szilas
12k2822
answered Jan 14 at 7:24
Peter SzilasPeter Szilas
12k2822
12k2822
add a comment |
add a comment |
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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