Is this a sufficient proof for "For all $k inBbb R$, if $k$ is odd, then $4k + 7$ is odd
2
$begingroup$
Can I just say 4k is obviously even and even + odd is always odd, so odd?
Or is that too simple, am I missing something?
elementary-number-theory
edited Dec 4 '18 at 23:41
Bill Dubuque
209k29190632
asked Dec 4 '18 at 22:43
mingming
3185
$endgroup$
add a comment |
2
$begingroup$
Can I just say 4k is obviously even and even + odd is always odd, so odd?
Or is that too simple, am I missing something?
elementary-number-theory
edited Dec 4 '18 at 23:41
Bill Dubuque
209k29190632
asked Dec 4 '18 at 22:43
mingming
3185
$endgroup$
$begingroup$
You can say as much, yeah. You could also just show that $4k$ is even by showing it's divisible by $2$, that's fine too. Or you could show it by induction maybe? Not sure, just a thought. It just all really depends on the level of formality you're expected to show and the context as well. Though note: the notion of being "even" or "odd" doesn't make sense for non-integer numbers (you said real numbers in the title). At least in any context I'm aware of.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 22:47
$begingroup$
Oh shoot you're right, so if it says "k is odd" is it just implying it's an odd integer?
$endgroup$
– ming
Dec 5 '18 at 0:54
1
$begingroup$
Yup, that's exactly what's being implied!
$endgroup$
– Eevee Trainer
Dec 5 '18 at 1:30
add a comment |
2
2
2
$begingroup$
Can I just say 4k is obviously even and even + odd is always odd, so odd?
Or is that too simple, am I missing something?
elementary-number-theory
edited Dec 4 '18 at 23:41
Bill Dubuque
209k29190632
asked Dec 4 '18 at 22:43
mingming
3185
$endgroup$
Can I just say 4k is obviously even and even + odd is always odd, so odd?
Or is that too simple, am I missing something?
elementary-number-theory
elementary-number-theory
edited Dec 4 '18 at 23:41
Bill Dubuque
209k29190632
asked Dec 4 '18 at 22:43
mingming
3185
edited Dec 4 '18 at 23:41
Bill Dubuque
209k29190632
asked Dec 4 '18 at 22:43
mingming
3185
edited Dec 4 '18 at 23:41
Bill Dubuque
209k29190632
edited Dec 4 '18 at 23:41
Bill Dubuque
209k29190632
edited Dec 4 '18 at 23:41
Bill Dubuque
209k29190632
209k29190632
asked Dec 4 '18 at 22:43
mingming
3185
asked Dec 4 '18 at 22:43
mingming
3185
asked Dec 4 '18 at 22:43
mingming
3185
3185
$begingroup$
You can say as much, yeah. You could also just show that $4k$ is even by showing it's divisible by $2$, that's fine too. Or you could show it by induction maybe? Not sure, just a thought. It just all really depends on the level of formality you're expected to show and the context as well. Though note: the notion of being "even" or "odd" doesn't make sense for non-integer numbers (you said real numbers in the title). At least in any context I'm aware of.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 22:47
$begingroup$
Oh shoot you're right, so if it says "k is odd" is it just implying it's an odd integer?
$endgroup$
– ming
Dec 5 '18 at 0:54
1
$begingroup$
Yup, that's exactly what's being implied!
$endgroup$
– Eevee Trainer
Dec 5 '18 at 1:30
add a comment |
$begingroup$
You can say as much, yeah. You could also just show that $4k$ is even by showing it's divisible by $2$, that's fine too. Or you could show it by induction maybe? Not sure, just a thought. It just all really depends on the level of formality you're expected to show and the context as well. Though note: the notion of being "even" or "odd" doesn't make sense for non-integer numbers (you said real numbers in the title). At least in any context I'm aware of.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 22:47
$begingroup$
Oh shoot you're right, so if it says "k is odd" is it just implying it's an odd integer?
$endgroup$
– ming
Dec 5 '18 at 0:54
1
$begingroup$
Yup, that's exactly what's being implied!
$endgroup$
– Eevee Trainer
Dec 5 '18 at 1:30
$begingroup$
You can say as much, yeah. You could also just show that $4k$ is even by showing it's divisible by $2$, that's fine too. Or you could show it by induction maybe? Not sure, just a thought. It just all really depends on the level of formality you're expected to show and the context as well. Though note: the notion of being "even" or "odd" doesn't make sense for non-integer numbers (you said real numbers in the title). At least in any context I'm aware of.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 22:47
$begingroup$
You can say as much, yeah. You could also just show that $4k$ is even by showing it's divisible by $2$, that's fine too. Or you could show it by induction maybe? Not sure, just a thought. It just all really depends on the level of formality you're expected to show and the context as well. Though note: the notion of being "even" or "odd" doesn't make sense for non-integer numbers (you said real numbers in the title). At least in any context I'm aware of.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 22:47
$begingroup$
Oh shoot you're right, so if it says "k is odd" is it just implying it's an odd integer?
$endgroup$
– ming
Dec 5 '18 at 0:54
$begingroup$
Oh shoot you're right, so if it says "k is odd" is it just implying it's an odd integer?
$endgroup$
– ming
Dec 5 '18 at 0:54
1
1
$begingroup$
Yup, that's exactly what's being implied!
$endgroup$
– Eevee Trainer
Dec 5 '18 at 1:30
$begingroup$
Yup, that's exactly what's being implied!
$endgroup$
– Eevee Trainer
Dec 5 '18 at 1:30
add a comment |
1 Answer
1
active
oldest
votes
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$begingroup$
Yes, its simple like that. But if you are making a test or something like that you have to be sure that you can use the fact that "even+odd is odd" without proving it... otherwise just prove it! (Not so difficult)
answered Dec 4 '18 at 22:47
RobsonRobson
769221
$endgroup$
$begingroup$
Is it just using the "for some integer $j$, $k = 2j + 1$" ?
$endgroup$
– ming
Dec 5 '18 at 0:54
$begingroup$
Yes, you have to get that "for some integer $j$, $k=2j+1$"... for this you just manipulate the equation like : $4k+7=2(2k+3)+1$ so in this case $j=2k+3$... like people said this question just make sense if $kin Bbb{Z}$
$endgroup$
– Robson
Dec 5 '18 at 13:12
add a comment |
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1 Answer
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1 Answer
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Yes, its simple like that. But if you are making a test or something like that you have to be sure that you can use the fact that "even+odd is odd" without proving it... otherwise just prove it! (Not so difficult)
answered Dec 4 '18 at 22:47
RobsonRobson
769221
$endgroup$
$begingroup$
Is it just using the "for some integer $j$, $k = 2j + 1$" ?
$endgroup$
– ming
Dec 5 '18 at 0:54
$begingroup$
Yes, you have to get that "for some integer $j$, $k=2j+1$"... for this you just manipulate the equation like : $4k+7=2(2k+3)+1$ so in this case $j=2k+3$... like people said this question just make sense if $kin Bbb{Z}$
$endgroup$
– Robson
Dec 5 '18 at 13:12
add a comment |
1
$begingroup$
Yes, its simple like that. But if you are making a test or something like that you have to be sure that you can use the fact that "even+odd is odd" without proving it... otherwise just prove it! (Not so difficult)
answered Dec 4 '18 at 22:47
RobsonRobson
769221
$endgroup$
$begingroup$
Is it just using the "for some integer $j$, $k = 2j + 1$" ?
$endgroup$
– ming
Dec 5 '18 at 0:54
$begingroup$
Yes, you have to get that "for some integer $j$, $k=2j+1$"... for this you just manipulate the equation like : $4k+7=2(2k+3)+1$ so in this case $j=2k+3$... like people said this question just make sense if $kin Bbb{Z}$
$endgroup$
– Robson
Dec 5 '18 at 13:12
add a comment |
1
1
1
$begingroup$
Yes, its simple like that. But if you are making a test or something like that you have to be sure that you can use the fact that "even+odd is odd" without proving it... otherwise just prove it! (Not so difficult)
answered Dec 4 '18 at 22:47
RobsonRobson
769221
$endgroup$
Yes, its simple like that. But if you are making a test or something like that you have to be sure that you can use the fact that "even+odd is odd" without proving it... otherwise just prove it! (Not so difficult)
answered Dec 4 '18 at 22:47
RobsonRobson
769221
answered Dec 4 '18 at 22:47
RobsonRobson
769221
answered Dec 4 '18 at 22:47
RobsonRobson
769221
answered Dec 4 '18 at 22:47
RobsonRobson
769221
769221
$begingroup$
Is it just using the "for some integer $j$, $k = 2j + 1$" ?
$endgroup$
– ming
Dec 5 '18 at 0:54
$begingroup$
Yes, you have to get that "for some integer $j$, $k=2j+1$"... for this you just manipulate the equation like : $4k+7=2(2k+3)+1$ so in this case $j=2k+3$... like people said this question just make sense if $kin Bbb{Z}$
$endgroup$
– Robson
Dec 5 '18 at 13:12
add a comment |
$begingroup$
Is it just using the "for some integer $j$, $k = 2j + 1$" ?
$endgroup$
– ming
Dec 5 '18 at 0:54
$begingroup$
Yes, you have to get that "for some integer $j$, $k=2j+1$"... for this you just manipulate the equation like : $4k+7=2(2k+3)+1$ so in this case $j=2k+3$... like people said this question just make sense if $kin Bbb{Z}$
$endgroup$
– Robson
Dec 5 '18 at 13:12
$begingroup$
Is it just using the "for some integer $j$, $k = 2j + 1$" ?
$endgroup$
– ming
Dec 5 '18 at 0:54
$begingroup$
Is it just using the "for some integer $j$, $k = 2j + 1$" ?
$endgroup$
– ming
Dec 5 '18 at 0:54
$begingroup$
Yes, you have to get that "for some integer $j$, $k=2j+1$"... for this you just manipulate the equation like : $4k+7=2(2k+3)+1$ so in this case $j=2k+3$... like people said this question just make sense if $kin Bbb{Z}$
$endgroup$
– Robson
Dec 5 '18 at 13:12
$begingroup$
Yes, you have to get that "for some integer $j$, $k=2j+1$"... for this you just manipulate the equation like : $4k+7=2(2k+3)+1$ so in this case $j=2k+3$... like people said this question just make sense if $kin Bbb{Z}$
$endgroup$
– Robson
Dec 5 '18 at 13:12
add a comment |
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$begingroup$
You can say as much, yeah. You could also just show that $4k$ is even by showing it's divisible by $2$, that's fine too. Or you could show it by induction maybe? Not sure, just a thought. It just all really depends on the level of formality you're expected to show and the context as well. Though note: the notion of being "even" or "odd" doesn't make sense for non-integer numbers (you said real numbers in the title). At least in any context I'm aware of.
$endgroup$
– Eevee Trainer
Dec 4 '18 at 22:47
$begingroup$
Oh shoot you're right, so if it says "k is odd" is it just implying it's an odd integer?
$endgroup$
– ming
Dec 5 '18 at 0:54
1
$begingroup$
Yup, that's exactly what's being implied!
$endgroup$
– Eevee Trainer
Dec 5 '18 at 1:30