Show that if n is an integer and 3n+ 2 is even, then n is even using contradiction
$begingroup$
Show that if $n$ is an integer and $3n+ 2$ is even, then $n$ is even, using a proof by contradiction.
That's the question. So since we're using contradiction, I need to show that N is odd and prove a contradiction exists, right?
$N = 2k + 1$
$3(2k + 1) + 1=6k + 5quad$ ?
I think I'm off base, or else I don't know how to proceed.
logic
edited Sep 19 '14 at 14:15
amWhy
1
asked Sep 18 '14 at 0:55
user3032755user3032755
8218
$endgroup$
add a comment |
$begingroup$
Show that if $n$ is an integer and $3n+ 2$ is even, then $n$ is even, using a proof by contradiction.
That's the question. So since we're using contradiction, I need to show that N is odd and prove a contradiction exists, right?
$N = 2k + 1$
$3(2k + 1) + 1=6k + 5quad$ ?
I think I'm off base, or else I don't know how to proceed.
logic
edited Sep 19 '14 at 14:15
amWhy
1
asked Sep 18 '14 at 0:55
user3032755user3032755
8218
$endgroup$
$begingroup$
$3(2k+1)+1$ should be $6k + 4$ and so is even.
$endgroup$
– E W H Lee
Sep 18 '14 at 1:04
add a comment |
$begingroup$
Show that if $n$ is an integer and $3n+ 2$ is even, then $n$ is even, using a proof by contradiction.
That's the question. So since we're using contradiction, I need to show that N is odd and prove a contradiction exists, right?
$N = 2k + 1$
$3(2k + 1) + 1=6k + 5quad$ ?
I think I'm off base, or else I don't know how to proceed.
logic
edited Sep 19 '14 at 14:15
amWhy
1
asked Sep 18 '14 at 0:55
user3032755user3032755
8218
$endgroup$
Show that if $n$ is an integer and $3n+ 2$ is even, then $n$ is even, using a proof by contradiction.
That's the question. So since we're using contradiction, I need to show that N is odd and prove a contradiction exists, right?
$N = 2k + 1$
$3(2k + 1) + 1=6k + 5quad$ ?
I think I'm off base, or else I don't know how to proceed.
logic
logic
edited Sep 19 '14 at 14:15
amWhy
1
asked Sep 18 '14 at 0:55
user3032755user3032755
8218
edited Sep 19 '14 at 14:15
amWhy
1
asked Sep 18 '14 at 0:55
user3032755user3032755
8218
edited Sep 19 '14 at 14:15
amWhy
1
edited Sep 19 '14 at 14:15
amWhy
1
edited Sep 19 '14 at 14:15
amWhy
1
1
asked Sep 18 '14 at 0:55
user3032755user3032755
8218
asked Sep 18 '14 at 0:55
user3032755user3032755
8218
asked Sep 18 '14 at 0:55
user3032755user3032755
8218
8218
$begingroup$
$3(2k+1)+1$ should be $6k + 4$ and so is even.
$endgroup$
– E W H Lee
Sep 18 '14 at 1:04
add a comment |
$begingroup$
$3(2k+1)+1$ should be $6k + 4$ and so is even.
$endgroup$
– E W H Lee
Sep 18 '14 at 1:04
$begingroup$
$3(2k+1)+1$ should be $6k + 4$ and so is even.
$endgroup$
– E W H Lee
Sep 18 '14 at 1:04
$begingroup$
$3(2k+1)+1$ should be $6k + 4$ and so is even.
$endgroup$
– E W H Lee
Sep 18 '14 at 1:04
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
That's exactly right! Since:
$$
3n + 2 = 6k + 5 = 2(underbrace{3k + 2}_{in ~ mathbb Z}) + 1
$$
it follows that $3n + 2$ is odd, contradicting the fact that $3n + 2$ is even.
answered Sep 18 '14 at 1:01
AdrianoAdriano
36.3k33071
$endgroup$
add a comment |
$begingroup$
Use these facts:
$mbox{odd}timesmbox{odd}=mbox{odd}$
$mbox{odd}+mbox{even}=mbox{odd}$
Thus, if $n$ was odd, then:
$3n+2=mbox{odd}!times!mbox{odd}+mbox{even}=mbox{odd}+mbox{even}=mbox{odd}$
But we know that $3n+2$ is even - thus, $n$ can't be odd.
Another way of doing it: Say $n$ is odd. Thus, there exists a $k$ such that $n=2k+1$. Then:
$$3n+2overset{n=2k+1}=3(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+1$$
Anything of the form $2j+1$ is odd. (In this case, $j=3k+2$.) Thus, we have that $3n+2$ is odd.
But we know that $3n+2$ is even - thus, $n$ can't be odd.
answered Sep 18 '14 at 1:49
Akiva WeinbergerAkiva Weinberger
13.8k12168
$endgroup$
add a comment |
$begingroup$
Just write $3n + 2 = n + 2(n +1)$ or $(3n + 2) - 2(n+1) = n$ and then note that the difference and sum of even numbers must be even.
answered Sep 18 '14 at 1:08
akechakech
2,652618
$endgroup$
add a comment |
$begingroup$
You've got a fine start, but it is good, in such a proof, to state clearly what is given, and what we are assuming for the sake of contradiction.
Let $n$ be an integer such that $3n+2$ is even.
Suppose, for the sake of contradiction, that $n$ is odd.
Then there exists an integer $k$ such that $n = 2k+1$.
So, $$3n + 2 = 3(2k+1) + 2 = 6k + 3 + 2 = 6k + 5 = 6k +4 + 1 = 2(3k+2) + 1$$
Since $k$ is an integer, so is $3k+2 = mathcal l$. This means that $3n+2$ is of the form $2mathcal l + 1$, and hence is odd.
But this contradicts the fact that we are given $3n+2$ is even.
Hence, $n$ cannot be odd, as we supposed. Thus, we have proven that in fact, $n$ must be even if $2n+3$ is even.
answered Sep 19 '14 at 14:11
amWhyamWhy
1
$endgroup$
add a comment |
$begingroup$
$3n+2=2k$
$n=2k-2n-2$
$n=2(k-n-1)$
$n=2b$
Hence $3n+2$ is even when $n$ is even therefore also odd when $n$ is odd.
edited Dec 7 '18 at 8:36
Brahadeesh
6,35442363
answered Dec 7 '18 at 8:12
Byron odhiamboByron odhiambo
1
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
That's exactly right! Since:
$$
3n + 2 = 6k + 5 = 2(underbrace{3k + 2}_{in ~ mathbb Z}) + 1
$$
it follows that $3n + 2$ is odd, contradicting the fact that $3n + 2$ is even.
answered Sep 18 '14 at 1:01
AdrianoAdriano
36.3k33071
$endgroup$
add a comment |
$begingroup$
That's exactly right! Since:
$$
3n + 2 = 6k + 5 = 2(underbrace{3k + 2}_{in ~ mathbb Z}) + 1
$$
it follows that $3n + 2$ is odd, contradicting the fact that $3n + 2$ is even.
answered Sep 18 '14 at 1:01
AdrianoAdriano
36.3k33071
$endgroup$
add a comment |
$begingroup$
That's exactly right! Since:
$$
3n + 2 = 6k + 5 = 2(underbrace{3k + 2}_{in ~ mathbb Z}) + 1
$$
it follows that $3n + 2$ is odd, contradicting the fact that $3n + 2$ is even.
answered Sep 18 '14 at 1:01
AdrianoAdriano
36.3k33071
$endgroup$
That's exactly right! Since:
$$
3n + 2 = 6k + 5 = 2(underbrace{3k + 2}_{in ~ mathbb Z}) + 1
$$
it follows that $3n + 2$ is odd, contradicting the fact that $3n + 2$ is even.
answered Sep 18 '14 at 1:01
AdrianoAdriano
36.3k33071
answered Sep 18 '14 at 1:01
AdrianoAdriano
36.3k33071
answered Sep 18 '14 at 1:01
AdrianoAdriano
36.3k33071
answered Sep 18 '14 at 1:01
AdrianoAdriano
36.3k33071
36.3k33071
add a comment |
add a comment |
$begingroup$
Use these facts:
$mbox{odd}timesmbox{odd}=mbox{odd}$
$mbox{odd}+mbox{even}=mbox{odd}$
Thus, if $n$ was odd, then:
$3n+2=mbox{odd}!times!mbox{odd}+mbox{even}=mbox{odd}+mbox{even}=mbox{odd}$
But we know that $3n+2$ is even - thus, $n$ can't be odd.
Another way of doing it: Say $n$ is odd. Thus, there exists a $k$ such that $n=2k+1$. Then:
$$3n+2overset{n=2k+1}=3(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+1$$
Anything of the form $2j+1$ is odd. (In this case, $j=3k+2$.) Thus, we have that $3n+2$ is odd.
But we know that $3n+2$ is even - thus, $n$ can't be odd.
answered Sep 18 '14 at 1:49
Akiva WeinbergerAkiva Weinberger
13.8k12168
$endgroup$
add a comment |
$begingroup$
Use these facts:
$mbox{odd}timesmbox{odd}=mbox{odd}$
$mbox{odd}+mbox{even}=mbox{odd}$
Thus, if $n$ was odd, then:
$3n+2=mbox{odd}!times!mbox{odd}+mbox{even}=mbox{odd}+mbox{even}=mbox{odd}$
But we know that $3n+2$ is even - thus, $n$ can't be odd.
Another way of doing it: Say $n$ is odd. Thus, there exists a $k$ such that $n=2k+1$. Then:
$$3n+2overset{n=2k+1}=3(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+1$$
Anything of the form $2j+1$ is odd. (In this case, $j=3k+2$.) Thus, we have that $3n+2$ is odd.
But we know that $3n+2$ is even - thus, $n$ can't be odd.
answered Sep 18 '14 at 1:49
Akiva WeinbergerAkiva Weinberger
13.8k12168
$endgroup$
add a comment |
$begingroup$
Use these facts:
$mbox{odd}timesmbox{odd}=mbox{odd}$
$mbox{odd}+mbox{even}=mbox{odd}$
Thus, if $n$ was odd, then:
$3n+2=mbox{odd}!times!mbox{odd}+mbox{even}=mbox{odd}+mbox{even}=mbox{odd}$
But we know that $3n+2$ is even - thus, $n$ can't be odd.
Another way of doing it: Say $n$ is odd. Thus, there exists a $k$ such that $n=2k+1$. Then:
$$3n+2overset{n=2k+1}=3(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+1$$
Anything of the form $2j+1$ is odd. (In this case, $j=3k+2$.) Thus, we have that $3n+2$ is odd.
But we know that $3n+2$ is even - thus, $n$ can't be odd.
answered Sep 18 '14 at 1:49
Akiva WeinbergerAkiva Weinberger
13.8k12168
$endgroup$
Use these facts:
$mbox{odd}timesmbox{odd}=mbox{odd}$
$mbox{odd}+mbox{even}=mbox{odd}$
Thus, if $n$ was odd, then:
$3n+2=mbox{odd}!times!mbox{odd}+mbox{even}=mbox{odd}+mbox{even}=mbox{odd}$
But we know that $3n+2$ is even - thus, $n$ can't be odd.
Another way of doing it: Say $n$ is odd. Thus, there exists a $k$ such that $n=2k+1$. Then:
$$3n+2overset{n=2k+1}=3(2k+1)+2=6k+3+2=6k+5=6k+4+1=2(3k+2)+1$$
Anything of the form $2j+1$ is odd. (In this case, $j=3k+2$.) Thus, we have that $3n+2$ is odd.
But we know that $3n+2$ is even - thus, $n$ can't be odd.
answered Sep 18 '14 at 1:49
Akiva WeinbergerAkiva Weinberger
13.8k12168
answered Sep 18 '14 at 1:49
Akiva WeinbergerAkiva Weinberger
13.8k12168
answered Sep 18 '14 at 1:49
Akiva WeinbergerAkiva Weinberger
13.8k12168
answered Sep 18 '14 at 1:49
Akiva WeinbergerAkiva Weinberger
13.8k12168
13.8k12168
add a comment |
add a comment |
$begingroup$
Just write $3n + 2 = n + 2(n +1)$ or $(3n + 2) - 2(n+1) = n$ and then note that the difference and sum of even numbers must be even.
answered Sep 18 '14 at 1:08
akechakech
2,652618
$endgroup$
add a comment |
$begingroup$
Just write $3n + 2 = n + 2(n +1)$ or $(3n + 2) - 2(n+1) = n$ and then note that the difference and sum of even numbers must be even.
answered Sep 18 '14 at 1:08
akechakech
2,652618
$endgroup$
add a comment |
$begingroup$
Just write $3n + 2 = n + 2(n +1)$ or $(3n + 2) - 2(n+1) = n$ and then note that the difference and sum of even numbers must be even.
answered Sep 18 '14 at 1:08
akechakech
2,652618
$endgroup$
Just write $3n + 2 = n + 2(n +1)$ or $(3n + 2) - 2(n+1) = n$ and then note that the difference and sum of even numbers must be even.
answered Sep 18 '14 at 1:08
akechakech
2,652618
answered Sep 18 '14 at 1:08
akechakech
2,652618
answered Sep 18 '14 at 1:08
akechakech
2,652618
answered Sep 18 '14 at 1:08
akechakech
2,652618
2,652618
add a comment |
add a comment |
$begingroup$
You've got a fine start, but it is good, in such a proof, to state clearly what is given, and what we are assuming for the sake of contradiction.
Let $n$ be an integer such that $3n+2$ is even.
Suppose, for the sake of contradiction, that $n$ is odd.
Then there exists an integer $k$ such that $n = 2k+1$.
So, $$3n + 2 = 3(2k+1) + 2 = 6k + 3 + 2 = 6k + 5 = 6k +4 + 1 = 2(3k+2) + 1$$
Since $k$ is an integer, so is $3k+2 = mathcal l$. This means that $3n+2$ is of the form $2mathcal l + 1$, and hence is odd.
But this contradicts the fact that we are given $3n+2$ is even.
Hence, $n$ cannot be odd, as we supposed. Thus, we have proven that in fact, $n$ must be even if $2n+3$ is even.
answered Sep 19 '14 at 14:11
amWhyamWhy
1
$endgroup$
add a comment |
$begingroup$
You've got a fine start, but it is good, in such a proof, to state clearly what is given, and what we are assuming for the sake of contradiction.
Let $n$ be an integer such that $3n+2$ is even.
Suppose, for the sake of contradiction, that $n$ is odd.
Then there exists an integer $k$ such that $n = 2k+1$.
So, $$3n + 2 = 3(2k+1) + 2 = 6k + 3 + 2 = 6k + 5 = 6k +4 + 1 = 2(3k+2) + 1$$
Since $k$ is an integer, so is $3k+2 = mathcal l$. This means that $3n+2$ is of the form $2mathcal l + 1$, and hence is odd.
But this contradicts the fact that we are given $3n+2$ is even.
Hence, $n$ cannot be odd, as we supposed. Thus, we have proven that in fact, $n$ must be even if $2n+3$ is even.
answered Sep 19 '14 at 14:11
amWhyamWhy
1
$endgroup$
add a comment |
$begingroup$
You've got a fine start, but it is good, in such a proof, to state clearly what is given, and what we are assuming for the sake of contradiction.
Let $n$ be an integer such that $3n+2$ is even.
Suppose, for the sake of contradiction, that $n$ is odd.
Then there exists an integer $k$ such that $n = 2k+1$.
So, $$3n + 2 = 3(2k+1) + 2 = 6k + 3 + 2 = 6k + 5 = 6k +4 + 1 = 2(3k+2) + 1$$
Since $k$ is an integer, so is $3k+2 = mathcal l$. This means that $3n+2$ is of the form $2mathcal l + 1$, and hence is odd.
But this contradicts the fact that we are given $3n+2$ is even.
Hence, $n$ cannot be odd, as we supposed. Thus, we have proven that in fact, $n$ must be even if $2n+3$ is even.
answered Sep 19 '14 at 14:11
amWhyamWhy
1
$endgroup$
You've got a fine start, but it is good, in such a proof, to state clearly what is given, and what we are assuming for the sake of contradiction.
Let $n$ be an integer such that $3n+2$ is even.
Suppose, for the sake of contradiction, that $n$ is odd.
Then there exists an integer $k$ such that $n = 2k+1$.
So, $$3n + 2 = 3(2k+1) + 2 = 6k + 3 + 2 = 6k + 5 = 6k +4 + 1 = 2(3k+2) + 1$$
Since $k$ is an integer, so is $3k+2 = mathcal l$. This means that $3n+2$ is of the form $2mathcal l + 1$, and hence is odd.
But this contradicts the fact that we are given $3n+2$ is even.
Hence, $n$ cannot be odd, as we supposed. Thus, we have proven that in fact, $n$ must be even if $2n+3$ is even.
answered Sep 19 '14 at 14:11
amWhyamWhy
1
answered Sep 19 '14 at 14:11
amWhyamWhy
1
answered Sep 19 '14 at 14:11
amWhyamWhy
1
answered Sep 19 '14 at 14:11
amWhyamWhy
1
1
add a comment |
add a comment |
$begingroup$
$3n+2=2k$
$n=2k-2n-2$
$n=2(k-n-1)$
$n=2b$
Hence $3n+2$ is even when $n$ is even therefore also odd when $n$ is odd.
edited Dec 7 '18 at 8:36
Brahadeesh
6,35442363
answered Dec 7 '18 at 8:12
Byron odhiamboByron odhiambo
1
$endgroup$
add a comment |
$begingroup$
$3n+2=2k$
$n=2k-2n-2$
$n=2(k-n-1)$
$n=2b$
Hence $3n+2$ is even when $n$ is even therefore also odd when $n$ is odd.
edited Dec 7 '18 at 8:36
Brahadeesh
6,35442363
answered Dec 7 '18 at 8:12
Byron odhiamboByron odhiambo
1
$endgroup$
add a comment |
$begingroup$
$3n+2=2k$
$n=2k-2n-2$
$n=2(k-n-1)$
$n=2b$
Hence $3n+2$ is even when $n$ is even therefore also odd when $n$ is odd.
edited Dec 7 '18 at 8:36
Brahadeesh
6,35442363
answered Dec 7 '18 at 8:12
Byron odhiamboByron odhiambo
1
$endgroup$
$3n+2=2k$
$n=2k-2n-2$
$n=2(k-n-1)$
$n=2b$
Hence $3n+2$ is even when $n$ is even therefore also odd when $n$ is odd.
edited Dec 7 '18 at 8:36
Brahadeesh
6,35442363
answered Dec 7 '18 at 8:12
Byron odhiamboByron odhiambo
1
edited Dec 7 '18 at 8:36
Brahadeesh
6,35442363
edited Dec 7 '18 at 8:36
Brahadeesh
6,35442363
edited Dec 7 '18 at 8:36
Brahadeesh
6,35442363
6,35442363
answered Dec 7 '18 at 8:12
Byron odhiamboByron odhiambo
1
answered Dec 7 '18 at 8:12
Byron odhiamboByron odhiambo
1
answered Dec 7 '18 at 8:12
Byron odhiamboByron odhiambo
1
1
add a comment |
add a comment |
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$begingroup$
$3(2k+1)+1$ should be $6k + 4$ and so is even.
$endgroup$
– E W H Lee
Sep 18 '14 at 1:04