What is the correct way to prove that for all $m in Z$, if $m$ is even then then $mn$ is even?
$begingroup$
As the title says, prove:
$m in Z$, if $m$ is even then then $mn$ is even
My attempt:
By definition of an even number, $2i$, and definition of an odd number, $2j+1$ I did the following by cases. I thought it was logical to create cases for if $n$ was even or if it was odd.
Case 1: Even
$$mn=(2i)(2j)$$
$$=4(ij)$$
Case 2: Odd
$$mn=(2i)(2j+1)$$
$$= (2i)(2j)+2i$$
$$=4(ij)+2i$$
However, after my attempt the books answer was the following:
"Suppose $x,y in Z$ and $x$ is even. Then $x=2a$ for some integer a, by definition of an even number. Thus $xy = (2a)(y) = 2(ay).$ Therefore, $xy=2b$ where $b$ is the integer $ay$, so $xy$ is even. "
Are both methods acceptable? Any suggestions to the way I did this are more then welcome!
proof-verification proof-writing proof-explanation
asked Jan 15 at 23:42
ForextraderForextrader
988
$endgroup$
add a comment |
$begingroup$
As the title says, prove:
$m in Z$, if $m$ is even then then $mn$ is even
My attempt:
By definition of an even number, $2i$, and definition of an odd number, $2j+1$ I did the following by cases. I thought it was logical to create cases for if $n$ was even or if it was odd.
Case 1: Even
$$mn=(2i)(2j)$$
$$=4(ij)$$
Case 2: Odd
$$mn=(2i)(2j+1)$$
$$= (2i)(2j)+2i$$
$$=4(ij)+2i$$
However, after my attempt the books answer was the following:
"Suppose $x,y in Z$ and $x$ is even. Then $x=2a$ for some integer a, by definition of an even number. Thus $xy = (2a)(y) = 2(ay).$ Therefore, $xy=2b$ where $b$ is the integer $ay$, so $xy$ is even. "
Are both methods acceptable? Any suggestions to the way I did this are more then welcome!
proof-verification proof-writing proof-explanation
asked Jan 15 at 23:42
ForextraderForextrader
988
$endgroup$
$begingroup$
I don't see any problems with the way you went about it. An integer $n$ can either be even or odd. Nothing else. Your books method simply took care of both cases with one argument
$endgroup$
– WaveX
Jan 15 at 23:48
$begingroup$
There is no need to consider cases. The book's version is better. In your case for odd numbers you performed an unnecessary multiplication and didn't finish the proof. All you needed was $mn=(2i)(2j+1)=2(i(2j+1))implies mn$ is even.
$endgroup$
– John Douma
Jan 15 at 23:48
$begingroup$
Your proof is perfectly reasonable and acceptable and neither have any error. The book is more efficient. We don't actually care if $n = 2j+1$ or $2j$ or $5j^2 +27l$ we can do those cases and it will be correct. But we don't need to do those cases. It's enough to do $m = 2k$ and $n = n$ so $mn = 2ktimes n = 2(ktimes n)$ is even. But we could do cases i)if $n=2j$ then $mn=4kj$ and ii) if $n = 2j+1$ then $mn=4kj + 2j$ and if $n=5j^2 +27l$ then $mn=10kj^2+54kl$. Nothing wrong with doing those. We just don't have to.
$endgroup$
– fleablood
Jan 16 at 0:30
$begingroup$
Thank you for the comments!
$endgroup$
– Forextrader
Jan 16 at 1:00
add a comment |
$begingroup$
As the title says, prove:
$m in Z$, if $m$ is even then then $mn$ is even
My attempt:
By definition of an even number, $2i$, and definition of an odd number, $2j+1$ I did the following by cases. I thought it was logical to create cases for if $n$ was even or if it was odd.
Case 1: Even
$$mn=(2i)(2j)$$
$$=4(ij)$$
Case 2: Odd
$$mn=(2i)(2j+1)$$
$$= (2i)(2j)+2i$$
$$=4(ij)+2i$$
However, after my attempt the books answer was the following:
"Suppose $x,y in Z$ and $x$ is even. Then $x=2a$ for some integer a, by definition of an even number. Thus $xy = (2a)(y) = 2(ay).$ Therefore, $xy=2b$ where $b$ is the integer $ay$, so $xy$ is even. "
Are both methods acceptable? Any suggestions to the way I did this are more then welcome!
proof-verification proof-writing proof-explanation
asked Jan 15 at 23:42
ForextraderForextrader
988
$endgroup$
As the title says, prove:
$m in Z$, if $m$ is even then then $mn$ is even
My attempt:
By definition of an even number, $2i$, and definition of an odd number, $2j+1$ I did the following by cases. I thought it was logical to create cases for if $n$ was even or if it was odd.
Case 1: Even
$$mn=(2i)(2j)$$
$$=4(ij)$$
Case 2: Odd
$$mn=(2i)(2j+1)$$
$$= (2i)(2j)+2i$$
$$=4(ij)+2i$$
However, after my attempt the books answer was the following:
"Suppose $x,y in Z$ and $x$ is even. Then $x=2a$ for some integer a, by definition of an even number. Thus $xy = (2a)(y) = 2(ay).$ Therefore, $xy=2b$ where $b$ is the integer $ay$, so $xy$ is even. "
Are both methods acceptable? Any suggestions to the way I did this are more then welcome!
proof-verification proof-writing proof-explanation
proof-verification proof-writing proof-explanation
asked Jan 15 at 23:42
ForextraderForextrader
988
asked Jan 15 at 23:42
ForextraderForextrader
988
asked Jan 15 at 23:42
ForextraderForextrader
988
asked Jan 15 at 23:42
ForextraderForextrader
988
asked Jan 15 at 23:42
ForextraderForextrader
988
988
$begingroup$
I don't see any problems with the way you went about it. An integer $n$ can either be even or odd. Nothing else. Your books method simply took care of both cases with one argument
$endgroup$
– WaveX
Jan 15 at 23:48
$begingroup$
There is no need to consider cases. The book's version is better. In your case for odd numbers you performed an unnecessary multiplication and didn't finish the proof. All you needed was $mn=(2i)(2j+1)=2(i(2j+1))implies mn$ is even.
$endgroup$
– John Douma
Jan 15 at 23:48
$begingroup$
Your proof is perfectly reasonable and acceptable and neither have any error. The book is more efficient. We don't actually care if $n = 2j+1$ or $2j$ or $5j^2 +27l$ we can do those cases and it will be correct. But we don't need to do those cases. It's enough to do $m = 2k$ and $n = n$ so $mn = 2ktimes n = 2(ktimes n)$ is even. But we could do cases i)if $n=2j$ then $mn=4kj$ and ii) if $n = 2j+1$ then $mn=4kj + 2j$ and if $n=5j^2 +27l$ then $mn=10kj^2+54kl$. Nothing wrong with doing those. We just don't have to.
$endgroup$
– fleablood
Jan 16 at 0:30
$begingroup$
Thank you for the comments!
$endgroup$
– Forextrader
Jan 16 at 1:00
add a comment |
$begingroup$
I don't see any problems with the way you went about it. An integer $n$ can either be even or odd. Nothing else. Your books method simply took care of both cases with one argument
$endgroup$
– WaveX
Jan 15 at 23:48
$begingroup$
There is no need to consider cases. The book's version is better. In your case for odd numbers you performed an unnecessary multiplication and didn't finish the proof. All you needed was $mn=(2i)(2j+1)=2(i(2j+1))implies mn$ is even.
$endgroup$
– John Douma
Jan 15 at 23:48
$begingroup$
Your proof is perfectly reasonable and acceptable and neither have any error. The book is more efficient. We don't actually care if $n = 2j+1$ or $2j$ or $5j^2 +27l$ we can do those cases and it will be correct. But we don't need to do those cases. It's enough to do $m = 2k$ and $n = n$ so $mn = 2ktimes n = 2(ktimes n)$ is even. But we could do cases i)if $n=2j$ then $mn=4kj$ and ii) if $n = 2j+1$ then $mn=4kj + 2j$ and if $n=5j^2 +27l$ then $mn=10kj^2+54kl$. Nothing wrong with doing those. We just don't have to.
$endgroup$
– fleablood
Jan 16 at 0:30
$begingroup$
Thank you for the comments!
$endgroup$
– Forextrader
Jan 16 at 1:00
$begingroup$
I don't see any problems with the way you went about it. An integer $n$ can either be even or odd. Nothing else. Your books method simply took care of both cases with one argument
$endgroup$
– WaveX
Jan 15 at 23:48
$begingroup$
I don't see any problems with the way you went about it. An integer $n$ can either be even or odd. Nothing else. Your books method simply took care of both cases with one argument
$endgroup$
– WaveX
Jan 15 at 23:48
$begingroup$
There is no need to consider cases. The book's version is better. In your case for odd numbers you performed an unnecessary multiplication and didn't finish the proof. All you needed was $mn=(2i)(2j+1)=2(i(2j+1))implies mn$ is even.
$endgroup$
– John Douma
Jan 15 at 23:48
$begingroup$
There is no need to consider cases. The book's version is better. In your case for odd numbers you performed an unnecessary multiplication and didn't finish the proof. All you needed was $mn=(2i)(2j+1)=2(i(2j+1))implies mn$ is even.
$endgroup$
– John Douma
Jan 15 at 23:48
$begingroup$
Your proof is perfectly reasonable and acceptable and neither have any error. The book is more efficient. We don't actually care if $n = 2j+1$ or $2j$ or $5j^2 +27l$ we can do those cases and it will be correct. But we don't need to do those cases. It's enough to do $m = 2k$ and $n = n$ so $mn = 2ktimes n = 2(ktimes n)$ is even. But we could do cases i)if $n=2j$ then $mn=4kj$ and ii) if $n = 2j+1$ then $mn=4kj + 2j$ and if $n=5j^2 +27l$ then $mn=10kj^2+54kl$. Nothing wrong with doing those. We just don't have to.
$endgroup$
– fleablood
Jan 16 at 0:30
$begingroup$
Your proof is perfectly reasonable and acceptable and neither have any error. The book is more efficient. We don't actually care if $n = 2j+1$ or $2j$ or $5j^2 +27l$ we can do those cases and it will be correct. But we don't need to do those cases. It's enough to do $m = 2k$ and $n = n$ so $mn = 2ktimes n = 2(ktimes n)$ is even. But we could do cases i)if $n=2j$ then $mn=4kj$ and ii) if $n = 2j+1$ then $mn=4kj + 2j$ and if $n=5j^2 +27l$ then $mn=10kj^2+54kl$. Nothing wrong with doing those. We just don't have to.
$endgroup$
– fleablood
Jan 16 at 0:30
$begingroup$
Thank you for the comments!
$endgroup$
– Forextrader
Jan 16 at 1:00
$begingroup$
Thank you for the comments!
$endgroup$
– Forextrader
Jan 16 at 1:00
add a comment |
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$begingroup$
I don't see any problems with the way you went about it. An integer $n$ can either be even or odd. Nothing else. Your books method simply took care of both cases with one argument
$endgroup$
– WaveX
Jan 15 at 23:48
$begingroup$
There is no need to consider cases. The book's version is better. In your case for odd numbers you performed an unnecessary multiplication and didn't finish the proof. All you needed was $mn=(2i)(2j+1)=2(i(2j+1))implies mn$ is even.
$endgroup$
– John Douma
Jan 15 at 23:48
$begingroup$
Your proof is perfectly reasonable and acceptable and neither have any error. The book is more efficient. We don't actually care if $n = 2j+1$ or $2j$ or $5j^2 +27l$ we can do those cases and it will be correct. But we don't need to do those cases. It's enough to do $m = 2k$ and $n = n$ so $mn = 2ktimes n = 2(ktimes n)$ is even. But we could do cases i)if $n=2j$ then $mn=4kj$ and ii) if $n = 2j+1$ then $mn=4kj + 2j$ and if $n=5j^2 +27l$ then $mn=10kj^2+54kl$. Nothing wrong with doing those. We just don't have to.
$endgroup$
– fleablood
Jan 16 at 0:30
$begingroup$
Thank you for the comments!
$endgroup$
– Forextrader
Jan 16 at 1:00