Mixing
A proof using the contrapositive
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
I am trying to prove the following conjecture:
Prove that if $m$ and $n$ are integers and $mn$ is even, then $m$
is even or $n$ is even.
Proof by contraposition:
Assume $m$ and $n$ are odd. Then $m = 2k + 1$ and $n = 2l + 1$. So
$$mn = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1$$
QED
Is there anything else I need to do in order to prove this conjecture? Thank you!
proof-writing
edited 1 hour ago
ArsenBerk
7,04721134
asked 1 hour ago
Lee
254
add a comment |Â
up vote
4
down vote
favorite
I am trying to prove the following conjecture:
Prove that if $m$ and $n$ are integers and $mn$ is even, then $m$
is even or $n$ is even.
Proof by contraposition:
Assume $m$ and $n$ are odd. Then $m = 2k + 1$ and $n = 2l + 1$. So
$$mn = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1$$
QED
Is there anything else I need to do in order to prove this conjecture? Thank you!
proof-writing
edited 1 hour ago
ArsenBerk
7,04721134
asked 1 hour ago
Lee
254
1
It looks fine to me.
– ArsenBerk
1 hour ago
4
This looks perfectly fine to me. You could just as well be using a contradiction argument, assuming $mn$ is even and both $m$ and $n$ are odd.
– miradulo
1 hour ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am trying to prove the following conjecture:
Prove that if $m$ and $n$ are integers and $mn$ is even, then $m$
is even or $n$ is even.
Proof by contraposition:
Assume $m$ and $n$ are odd. Then $m = 2k + 1$ and $n = 2l + 1$. So
$$mn = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1$$
QED
Is there anything else I need to do in order to prove this conjecture? Thank you!
proof-writing
edited 1 hour ago
ArsenBerk
7,04721134
asked 1 hour ago
Lee
254
I am trying to prove the following conjecture:
Prove that if $m$ and $n$ are integers and $mn$ is even, then $m$
is even or $n$ is even.
Proof by contraposition:
Assume $m$ and $n$ are odd. Then $m = 2k + 1$ and $n = 2l + 1$. So
$$mn = (2k + 1)(2l + 1) = 4kl + 2k + 2l + 1 = 2(2kl + k + l) + 1$$
QED
Is there anything else I need to do in order to prove this conjecture? Thank you!
proof-writing
proof-writing
edited 1 hour ago
ArsenBerk
7,04721134
asked 1 hour ago
Lee
254
edited 1 hour ago
ArsenBerk
7,04721134
asked 1 hour ago
Lee
254
edited 1 hour ago
ArsenBerk
7,04721134
edited 1 hour ago
ArsenBerk
7,04721134
edited 1 hour ago
ArsenBerk
7,04721134
7,04721134
asked 1 hour ago
Lee
254
asked 1 hour ago
Lee
254
asked 1 hour ago
Lee
254
254
1
It looks fine to me.
– ArsenBerk
1 hour ago
4
This looks perfectly fine to me. You could just as well be using a contradiction argument, assuming $mn$ is even and both $m$ and $n$ are odd.
– miradulo
1 hour ago
add a comment |Â
1
It looks fine to me.
– ArsenBerk
1 hour ago
4
This looks perfectly fine to me. You could just as well be using a contradiction argument, assuming $mn$ is even and both $m$ and $n$ are odd.
– miradulo
1 hour ago
1
1
It looks fine to me.
– ArsenBerk
1 hour ago
It looks fine to me.
– ArsenBerk
1 hour ago
4
4
This looks perfectly fine to me. You could just as well be using a contradiction argument, assuming $mn$ is even and both $m$ and $n$ are odd.
– miradulo
1 hour ago
This looks perfectly fine to me. You could just as well be using a contradiction argument, assuming $mn$ is even and both $m$ and $n$ are odd.
– miradulo
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
You did very well: you got to the "meat" of the proof.
I'll simply add "a side dish":
I would simply add, after demonstrating that, given $m$ and $n$ are both odd, and hence, as odd, there is an integer k such that $m= 2k+1$, and an integer $l$ such that $n = 2l+1$. Thus it follows that $$mn= 2(2kl + k + l) + 1 $$ by concluding that $$mn= 2(2kl + k + l) + 1 ,text is odd. $$
Hence, by the equivalence of the contrapositive, we have proven: "If $mn$ is even, then either $m$ or $n$ (or both) is even."
answered 1 hour ago
amWhy
191k27221434
1
Note: It is always better to be more explicit when you are unsure "how much to say" in a proof, than it is to not say enough.
– amWhy
1 hour ago
Thank you for your help!
– Lee
1 hour ago
1
You're welcome, Lee. Glad to help.
– amWhy
56 mins ago
add a comment |Â
up vote
2
down vote
Here is a slightly different way of doing it.
First note that the difference between two even numbers is even because $2a-2b=2(a-b)$ and if $m$ is odd then $m-1$ is even so $m-1=2r$ and $m=2r+1$.
Now suppose $mn$ is even and $m$ is odd, then $mn=(2r+1)n=2rn+n$ and $n=mn-2rn$ is the difference between two even numbers and hence is even.
So if $mn$ is even and $m$ is not we have shown that $n$ is even. Hence if $mn$ is even either $m$ is even or $n$ is even.
answered 1 hour ago
Mark Bennet
77.8k775176
add a comment |Â
up vote
0
down vote
The contrapositive is
If $m$ is odd and $n$ is odd, then $mn$ is odd.
If you prove this, then, since it is equivalent to your original problem, you will have solved the problem.
Which you did correctly. Perhaps, only thing you could add is at the beginning:
$m$ odd so we can find some integer $k$ so that $m=2k+1$ and we can find some integer $l$ so that $n=2k+1$
answered 1 hour ago
Jimmy Sabater
1,26713
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You did very well: you got to the "meat" of the proof.
I'll simply add "a side dish":
I would simply add, after demonstrating that, given $m$ and $n$ are both odd, and hence, as odd, there is an integer k such that $m= 2k+1$, and an integer $l$ such that $n = 2l+1$. Thus it follows that $$mn= 2(2kl + k + l) + 1 $$ by concluding that $$mn= 2(2kl + k + l) + 1 ,text is odd. $$
Hence, by the equivalence of the contrapositive, we have proven: "If $mn$ is even, then either $m$ or $n$ (or both) is even."
answered 1 hour ago
amWhy
191k27221434
1
Note: It is always better to be more explicit when you are unsure "how much to say" in a proof, than it is to not say enough.
– amWhy
1 hour ago
Thank you for your help!
– Lee
1 hour ago
1
You're welcome, Lee. Glad to help.
– amWhy
56 mins ago
add a comment |Â
up vote
3
down vote
accepted
You did very well: you got to the "meat" of the proof.
I'll simply add "a side dish":
I would simply add, after demonstrating that, given $m$ and $n$ are both odd, and hence, as odd, there is an integer k such that $m= 2k+1$, and an integer $l$ such that $n = 2l+1$. Thus it follows that $$mn= 2(2kl + k + l) + 1 $$ by concluding that $$mn= 2(2kl + k + l) + 1 ,text is odd. $$
Hence, by the equivalence of the contrapositive, we have proven: "If $mn$ is even, then either $m$ or $n$ (or both) is even."
answered 1 hour ago
amWhy
191k27221434
1
Note: It is always better to be more explicit when you are unsure "how much to say" in a proof, than it is to not say enough.
– amWhy
1 hour ago
Thank you for your help!
– Lee
1 hour ago
1
You're welcome, Lee. Glad to help.
– amWhy
56 mins ago
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You did very well: you got to the "meat" of the proof.
I'll simply add "a side dish":
I would simply add, after demonstrating that, given $m$ and $n$ are both odd, and hence, as odd, there is an integer k such that $m= 2k+1$, and an integer $l$ such that $n = 2l+1$. Thus it follows that $$mn= 2(2kl + k + l) + 1 $$ by concluding that $$mn= 2(2kl + k + l) + 1 ,text is odd. $$
Hence, by the equivalence of the contrapositive, we have proven: "If $mn$ is even, then either $m$ or $n$ (or both) is even."
answered 1 hour ago
amWhy
191k27221434
You did very well: you got to the "meat" of the proof.
I'll simply add "a side dish":
I would simply add, after demonstrating that, given $m$ and $n$ are both odd, and hence, as odd, there is an integer k such that $m= 2k+1$, and an integer $l$ such that $n = 2l+1$. Thus it follows that $$mn= 2(2kl + k + l) + 1 $$ by concluding that $$mn= 2(2kl + k + l) + 1 ,text is odd. $$
Hence, by the equivalence of the contrapositive, we have proven: "If $mn$ is even, then either $m$ or $n$ (or both) is even."
answered 1 hour ago
amWhy
191k27221434
edited 1 hour ago
answered 1 hour ago
amWhy
191k27221434
answered 1 hour ago
amWhy
191k27221434
answered 1 hour ago
amWhy
191k27221434
191k27221434
1
Note: It is always better to be more explicit when you are unsure "how much to say" in a proof, than it is to not say enough.
– amWhy
1 hour ago
Thank you for your help!
– Lee
1 hour ago
1
You're welcome, Lee. Glad to help.
– amWhy
56 mins ago
add a comment |Â
1
Note: It is always better to be more explicit when you are unsure "how much to say" in a proof, than it is to not say enough.
– amWhy
1 hour ago
Thank you for your help!
– Lee
1 hour ago
1
You're welcome, Lee. Glad to help.
– amWhy
56 mins ago
1
1
Note: It is always better to be more explicit when you are unsure "how much to say" in a proof, than it is to not say enough.
– amWhy
1 hour ago
Note: It is always better to be more explicit when you are unsure "how much to say" in a proof, than it is to not say enough.
– amWhy
1 hour ago
Thank you for your help!
– Lee
1 hour ago
Thank you for your help!
– Lee
1 hour ago
1
1
You're welcome, Lee. Glad to help.
– amWhy
56 mins ago
You're welcome, Lee. Glad to help.
– amWhy
56 mins ago
add a comment |Â
up vote
2
down vote
Here is a slightly different way of doing it.
First note that the difference between two even numbers is even because $2a-2b=2(a-b)$ and if $m$ is odd then $m-1$ is even so $m-1=2r$ and $m=2r+1$.
Now suppose $mn$ is even and $m$ is odd, then $mn=(2r+1)n=2rn+n$ and $n=mn-2rn$ is the difference between two even numbers and hence is even.
So if $mn$ is even and $m$ is not we have shown that $n$ is even. Hence if $mn$ is even either $m$ is even or $n$ is even.
answered 1 hour ago
Mark Bennet
77.8k775176
add a comment |Â
up vote
2
down vote
Here is a slightly different way of doing it.
First note that the difference between two even numbers is even because $2a-2b=2(a-b)$ and if $m$ is odd then $m-1$ is even so $m-1=2r$ and $m=2r+1$.
Now suppose $mn$ is even and $m$ is odd, then $mn=(2r+1)n=2rn+n$ and $n=mn-2rn$ is the difference between two even numbers and hence is even.
So if $mn$ is even and $m$ is not we have shown that $n$ is even. Hence if $mn$ is even either $m$ is even or $n$ is even.
answered 1 hour ago
Mark Bennet
77.8k775176
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here is a slightly different way of doing it.
First note that the difference between two even numbers is even because $2a-2b=2(a-b)$ and if $m$ is odd then $m-1$ is even so $m-1=2r$ and $m=2r+1$.
Now suppose $mn$ is even and $m$ is odd, then $mn=(2r+1)n=2rn+n$ and $n=mn-2rn$ is the difference between two even numbers and hence is even.
So if $mn$ is even and $m$ is not we have shown that $n$ is even. Hence if $mn$ is even either $m$ is even or $n$ is even.
answered 1 hour ago
Mark Bennet
77.8k775176
Here is a slightly different way of doing it.
First note that the difference between two even numbers is even because $2a-2b=2(a-b)$ and if $m$ is odd then $m-1$ is even so $m-1=2r$ and $m=2r+1$.
Now suppose $mn$ is even and $m$ is odd, then $mn=(2r+1)n=2rn+n$ and $n=mn-2rn$ is the difference between two even numbers and hence is even.
So if $mn$ is even and $m$ is not we have shown that $n$ is even. Hence if $mn$ is even either $m$ is even or $n$ is even.
answered 1 hour ago
Mark Bennet
77.8k775176
answered 1 hour ago
Mark Bennet
77.8k775176
answered 1 hour ago
Mark Bennet
77.8k775176
answered 1 hour ago
Mark Bennet
77.8k775176
77.8k775176
add a comment |Â
add a comment |Â
up vote
0
down vote
The contrapositive is
If $m$ is odd and $n$ is odd, then $mn$ is odd.
If you prove this, then, since it is equivalent to your original problem, you will have solved the problem.
Which you did correctly. Perhaps, only thing you could add is at the beginning:
$m$ odd so we can find some integer $k$ so that $m=2k+1$ and we can find some integer $l$ so that $n=2k+1$
answered 1 hour ago
Jimmy Sabater
1,26713
add a comment |Â
up vote
0
down vote
The contrapositive is
If $m$ is odd and $n$ is odd, then $mn$ is odd.
If you prove this, then, since it is equivalent to your original problem, you will have solved the problem.
Which you did correctly. Perhaps, only thing you could add is at the beginning:
$m$ odd so we can find some integer $k$ so that $m=2k+1$ and we can find some integer $l$ so that $n=2k+1$
answered 1 hour ago
Jimmy Sabater
1,26713
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The contrapositive is
If $m$ is odd and $n$ is odd, then $mn$ is odd.
If you prove this, then, since it is equivalent to your original problem, you will have solved the problem.
Which you did correctly. Perhaps, only thing you could add is at the beginning:
$m$ odd so we can find some integer $k$ so that $m=2k+1$ and we can find some integer $l$ so that $n=2k+1$
answered 1 hour ago
Jimmy Sabater
1,26713
The contrapositive is
If $m$ is odd and $n$ is odd, then $mn$ is odd.
If you prove this, then, since it is equivalent to your original problem, you will have solved the problem.
Which you did correctly. Perhaps, only thing you could add is at the beginning:
$m$ odd so we can find some integer $k$ so that $m=2k+1$ and we can find some integer $l$ so that $n=2k+1$
answered 1 hour ago
Jimmy Sabater
1,26713
answered 1 hour ago
Jimmy Sabater
1,26713
answered 1 hour ago
Jimmy Sabater
1,26713
answered 1 hour ago
Jimmy Sabater
1,26713
1,26713
add a comment |Â
add a comment |Â
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1
It looks fine to me.
– ArsenBerk
1 hour ago
4
This looks perfectly fine to me. You could just as well be using a contradiction argument, assuming $mn$ is even and both $m$ and $n$ are odd.
– miradulo
1 hour ago