Mixing
Prove that if $n$ is an integer, then $1 − n$ is even if and only if $n^2 + 1$ is even.
$begingroup$
I am practicing exam questions and have come across the following:
Prove that if n is an integer, then 1 − n is even if and only if $n^2$ + 1 is even.
The first thing that came to mind is a contrapositive proof, but how would I prove this?
discrete-mathematics logic proof-writing proof-explanation
edited Jan 25 at 1:12
DanielV
18.1k42755
asked Jan 25 at 0:05
Usama GhawjiUsama Ghawji
666
$endgroup$
add a comment |
$begingroup$
I am practicing exam questions and have come across the following:
Prove that if n is an integer, then 1 − n is even if and only if $n^2$ + 1 is even.
The first thing that came to mind is a contrapositive proof, but how would I prove this?
discrete-mathematics logic proof-writing proof-explanation
edited Jan 25 at 1:12
DanielV
18.1k42755
asked Jan 25 at 0:05
Usama GhawjiUsama Ghawji
666
$endgroup$
add a comment |
$begingroup$
I am practicing exam questions and have come across the following:
Prove that if n is an integer, then 1 − n is even if and only if $n^2$ + 1 is even.
The first thing that came to mind is a contrapositive proof, but how would I prove this?
discrete-mathematics logic proof-writing proof-explanation
edited Jan 25 at 1:12
DanielV
18.1k42755
asked Jan 25 at 0:05
Usama GhawjiUsama Ghawji
666
$endgroup$
I am practicing exam questions and have come across the following:
Prove that if n is an integer, then 1 − n is even if and only if $n^2$ + 1 is even.
The first thing that came to mind is a contrapositive proof, but how would I prove this?
discrete-mathematics logic proof-writing proof-explanation
discrete-mathematics logic proof-writing proof-explanation
edited Jan 25 at 1:12
DanielV
18.1k42755
asked Jan 25 at 0:05
Usama GhawjiUsama Ghawji
666
edited Jan 25 at 1:12
DanielV
18.1k42755
asked Jan 25 at 0:05
Usama GhawjiUsama Ghawji
666
edited Jan 25 at 1:12
DanielV
18.1k42755
edited Jan 25 at 1:12
DanielV
18.1k42755
edited Jan 25 at 1:12
DanielV
18.1k42755
18.1k42755
asked Jan 25 at 0:05
Usama GhawjiUsama Ghawji
666
asked Jan 25 at 0:05
Usama GhawjiUsama Ghawji
666
asked Jan 25 at 0:05
Usama GhawjiUsama Ghawji
666
666
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
For the forward direction:
$1-n$ being even would imply that $n$ is odd. Thus, $n^2$ is odd, so $n^2+1$ is even.
Backwards direction:
If $n^2+1$ is even, then $n^2$ is odd. Thus, $n$ is odd. Hence, $1-n$ is even.
For further clarity:
$n$ being odd means that $n = 2k+1$ for some integer $k$.
$n$ being even means that $n = 2k$ for some integer $k$.
You can proceed with a more formal argument using the above definitions.
answered Jan 25 at 0:12
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
$n^2 + 1 = (n-1)(n+1) + 2$
If $n-1$ is even, $n-1+2 = n+1$ is even, their product is even and an even number plus 2 is even.
Going the other way, if $n^2 + 1$ is even, then so is $n^2 - 1.$ one of the factors $(n-1)(n+1)$ must be even. But if one is even the other is even.
answered Jan 25 at 0:17
Doug MDoug M
45.3k31954
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add a comment |
$begingroup$
Hint: $, 2mid n(n!+!1) =, n^2, +, 1, -, (1,-,n)phantom{I_{I_{I_{I_1}}}},$
which implies that $,2mid n^2!+!1!iff! 2mid 1!-!n $ [here $ amid b $ means $,a,$ divides $,b,$]
Generally: $ $ if $,dmid y-x,$ then $ dmid yiff dmid x$
i.e. $bmod d!:, $ if $ yequiv x,$ then $,yequiv 0iff xequiv 0$
i.e. on a line $,yequiv x$ we have $,yequiv 0iff xequiv 0$
answered Jan 25 at 0:33
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
$1 - n$ is even iff $n - 1$ is even iff $n$ is odd iff $n^2$ is odd iff $n^2 + 1$ is even.
answered Jan 25 at 1:11
DanielVDanielV
18.1k42755
$endgroup$
add a comment |
$begingroup$
You can do it straight forwardly in cases.
Either $n$ is even or odd.
Case 1: $n = 2m$ is even so $1-n =1- 2m = 2(-m) + 1$ is odd and $n^2 + 1 = 4m^2 + 1 = 2(2m^2) + 1$ is odd.
Neither $1-n$ nor $n^2 + 1$ is even.
Case 2: $n = 2m + 1$ is odd. So $1-n= 1-(2m + 1) = -2m = 2(-m)$ is even, and $n^2 + 1 = (2m+1)^2 + 1 = 4m^2 + 4m + 1 + 1 = 2(2m^2 + 2m +1)$ is even.
So both $1-n$ and $n^2 + 1$ are even.
So $1-n$ is even precisely and only when $n^2 + 1$ is even.
edited Jan 25 at 1:40
J. W. Tanner
3,3451320
answered Jan 25 at 0:20
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
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5 Answers
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5 Answers
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$begingroup$
For the forward direction:
$1-n$ being even would imply that $n$ is odd. Thus, $n^2$ is odd, so $n^2+1$ is even.
Backwards direction:
If $n^2+1$ is even, then $n^2$ is odd. Thus, $n$ is odd. Hence, $1-n$ is even.
For further clarity:
$n$ being odd means that $n = 2k+1$ for some integer $k$.
$n$ being even means that $n = 2k$ for some integer $k$.
You can proceed with a more formal argument using the above definitions.
answered Jan 25 at 0:12
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
For the forward direction:
$1-n$ being even would imply that $n$ is odd. Thus, $n^2$ is odd, so $n^2+1$ is even.
Backwards direction:
If $n^2+1$ is even, then $n^2$ is odd. Thus, $n$ is odd. Hence, $1-n$ is even.
For further clarity:
$n$ being odd means that $n = 2k+1$ for some integer $k$.
$n$ being even means that $n = 2k$ for some integer $k$.
You can proceed with a more formal argument using the above definitions.
answered Jan 25 at 0:12
MetricMetric
1,25159
$endgroup$
add a comment |
$begingroup$
For the forward direction:
$1-n$ being even would imply that $n$ is odd. Thus, $n^2$ is odd, so $n^2+1$ is even.
Backwards direction:
If $n^2+1$ is even, then $n^2$ is odd. Thus, $n$ is odd. Hence, $1-n$ is even.
For further clarity:
$n$ being odd means that $n = 2k+1$ for some integer $k$.
$n$ being even means that $n = 2k$ for some integer $k$.
You can proceed with a more formal argument using the above definitions.
answered Jan 25 at 0:12
MetricMetric
1,25159
$endgroup$
For the forward direction:
$1-n$ being even would imply that $n$ is odd. Thus, $n^2$ is odd, so $n^2+1$ is even.
Backwards direction:
If $n^2+1$ is even, then $n^2$ is odd. Thus, $n$ is odd. Hence, $1-n$ is even.
For further clarity:
$n$ being odd means that $n = 2k+1$ for some integer $k$.
$n$ being even means that $n = 2k$ for some integer $k$.
You can proceed with a more formal argument using the above definitions.
answered Jan 25 at 0:12
MetricMetric
1,25159
answered Jan 25 at 0:12
MetricMetric
1,25159
answered Jan 25 at 0:12
MetricMetric
1,25159
answered Jan 25 at 0:12
MetricMetric
1,25159
1,25159
add a comment |
add a comment |
$begingroup$
$n^2 + 1 = (n-1)(n+1) + 2$
If $n-1$ is even, $n-1+2 = n+1$ is even, their product is even and an even number plus 2 is even.
Going the other way, if $n^2 + 1$ is even, then so is $n^2 - 1.$ one of the factors $(n-1)(n+1)$ must be even. But if one is even the other is even.
answered Jan 25 at 0:17
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$n^2 + 1 = (n-1)(n+1) + 2$
If $n-1$ is even, $n-1+2 = n+1$ is even, their product is even and an even number plus 2 is even.
Going the other way, if $n^2 + 1$ is even, then so is $n^2 - 1.$ one of the factors $(n-1)(n+1)$ must be even. But if one is even the other is even.
answered Jan 25 at 0:17
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
$n^2 + 1 = (n-1)(n+1) + 2$
If $n-1$ is even, $n-1+2 = n+1$ is even, their product is even and an even number plus 2 is even.
Going the other way, if $n^2 + 1$ is even, then so is $n^2 - 1.$ one of the factors $(n-1)(n+1)$ must be even. But if one is even the other is even.
answered Jan 25 at 0:17
Doug MDoug M
45.3k31954
$endgroup$
$n^2 + 1 = (n-1)(n+1) + 2$
If $n-1$ is even, $n-1+2 = n+1$ is even, their product is even and an even number plus 2 is even.
Going the other way, if $n^2 + 1$ is even, then so is $n^2 - 1.$ one of the factors $(n-1)(n+1)$ must be even. But if one is even the other is even.
answered Jan 25 at 0:17
Doug MDoug M
45.3k31954
answered Jan 25 at 0:17
Doug MDoug M
45.3k31954
answered Jan 25 at 0:17
Doug MDoug M
45.3k31954
answered Jan 25 at 0:17
Doug MDoug M
45.3k31954
45.3k31954
add a comment |
add a comment |
$begingroup$
Hint: $, 2mid n(n!+!1) =, n^2, +, 1, -, (1,-,n)phantom{I_{I_{I_{I_1}}}},$
which implies that $,2mid n^2!+!1!iff! 2mid 1!-!n $ [here $ amid b $ means $,a,$ divides $,b,$]
Generally: $ $ if $,dmid y-x,$ then $ dmid yiff dmid x$
i.e. $bmod d!:, $ if $ yequiv x,$ then $,yequiv 0iff xequiv 0$
i.e. on a line $,yequiv x$ we have $,yequiv 0iff xequiv 0$
answered Jan 25 at 0:33
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
Hint: $, 2mid n(n!+!1) =, n^2, +, 1, -, (1,-,n)phantom{I_{I_{I_{I_1}}}},$
which implies that $,2mid n^2!+!1!iff! 2mid 1!-!n $ [here $ amid b $ means $,a,$ divides $,b,$]
Generally: $ $ if $,dmid y-x,$ then $ dmid yiff dmid x$
i.e. $bmod d!:, $ if $ yequiv x,$ then $,yequiv 0iff xequiv 0$
i.e. on a line $,yequiv x$ we have $,yequiv 0iff xequiv 0$
answered Jan 25 at 0:33
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
add a comment |
$begingroup$
Hint: $, 2mid n(n!+!1) =, n^2, +, 1, -, (1,-,n)phantom{I_{I_{I_{I_1}}}},$
which implies that $,2mid n^2!+!1!iff! 2mid 1!-!n $ [here $ amid b $ means $,a,$ divides $,b,$]
Generally: $ $ if $,dmid y-x,$ then $ dmid yiff dmid x$
i.e. $bmod d!:, $ if $ yequiv x,$ then $,yequiv 0iff xequiv 0$
i.e. on a line $,yequiv x$ we have $,yequiv 0iff xequiv 0$
answered Jan 25 at 0:33
Bill DubuqueBill Dubuque
212k29195654
$endgroup$
Hint: $, 2mid n(n!+!1) =, n^2, +, 1, -, (1,-,n)phantom{I_{I_{I_{I_1}}}},$
which implies that $,2mid n^2!+!1!iff! 2mid 1!-!n $ [here $ amid b $ means $,a,$ divides $,b,$]
Generally: $ $ if $,dmid y-x,$ then $ dmid yiff dmid x$
i.e. $bmod d!:, $ if $ yequiv x,$ then $,yequiv 0iff xequiv 0$
i.e. on a line $,yequiv x$ we have $,yequiv 0iff xequiv 0$
answered Jan 25 at 0:33
Bill DubuqueBill Dubuque
212k29195654
edited Jan 25 at 0:43
answered Jan 25 at 0:33
Bill DubuqueBill Dubuque
212k29195654
answered Jan 25 at 0:33
Bill DubuqueBill Dubuque
212k29195654
answered Jan 25 at 0:33
Bill DubuqueBill Dubuque
212k29195654
212k29195654
add a comment |
add a comment |
$begingroup$
$1 - n$ is even iff $n - 1$ is even iff $n$ is odd iff $n^2$ is odd iff $n^2 + 1$ is even.
answered Jan 25 at 1:11
DanielVDanielV
18.1k42755
$endgroup$
add a comment |
$begingroup$
$1 - n$ is even iff $n - 1$ is even iff $n$ is odd iff $n^2$ is odd iff $n^2 + 1$ is even.
answered Jan 25 at 1:11
DanielVDanielV
18.1k42755
$endgroup$
add a comment |
$begingroup$
$1 - n$ is even iff $n - 1$ is even iff $n$ is odd iff $n^2$ is odd iff $n^2 + 1$ is even.
answered Jan 25 at 1:11
DanielVDanielV
18.1k42755
$endgroup$
$1 - n$ is even iff $n - 1$ is even iff $n$ is odd iff $n^2$ is odd iff $n^2 + 1$ is even.
answered Jan 25 at 1:11
DanielVDanielV
18.1k42755
answered Jan 25 at 1:11
DanielVDanielV
18.1k42755
answered Jan 25 at 1:11
DanielVDanielV
18.1k42755
answered Jan 25 at 1:11
DanielVDanielV
18.1k42755
18.1k42755
add a comment |
add a comment |
$begingroup$
You can do it straight forwardly in cases.
Either $n$ is even or odd.
Case 1: $n = 2m$ is even so $1-n =1- 2m = 2(-m) + 1$ is odd and $n^2 + 1 = 4m^2 + 1 = 2(2m^2) + 1$ is odd.
Neither $1-n$ nor $n^2 + 1$ is even.
Case 2: $n = 2m + 1$ is odd. So $1-n= 1-(2m + 1) = -2m = 2(-m)$ is even, and $n^2 + 1 = (2m+1)^2 + 1 = 4m^2 + 4m + 1 + 1 = 2(2m^2 + 2m +1)$ is even.
So both $1-n$ and $n^2 + 1$ are even.
So $1-n$ is even precisely and only when $n^2 + 1$ is even.
edited Jan 25 at 1:40
J. W. Tanner
3,3451320
answered Jan 25 at 0:20
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
$begingroup$
You can do it straight forwardly in cases.
Either $n$ is even or odd.
Case 1: $n = 2m$ is even so $1-n =1- 2m = 2(-m) + 1$ is odd and $n^2 + 1 = 4m^2 + 1 = 2(2m^2) + 1$ is odd.
Neither $1-n$ nor $n^2 + 1$ is even.
Case 2: $n = 2m + 1$ is odd. So $1-n= 1-(2m + 1) = -2m = 2(-m)$ is even, and $n^2 + 1 = (2m+1)^2 + 1 = 4m^2 + 4m + 1 + 1 = 2(2m^2 + 2m +1)$ is even.
So both $1-n$ and $n^2 + 1$ are even.
So $1-n$ is even precisely and only when $n^2 + 1$ is even.
edited Jan 25 at 1:40
J. W. Tanner
3,3451320
answered Jan 25 at 0:20
fleabloodfleablood
72.7k22788
$endgroup$
add a comment |
$begingroup$
You can do it straight forwardly in cases.
Either $n$ is even or odd.
Case 1: $n = 2m$ is even so $1-n =1- 2m = 2(-m) + 1$ is odd and $n^2 + 1 = 4m^2 + 1 = 2(2m^2) + 1$ is odd.
Neither $1-n$ nor $n^2 + 1$ is even.
Case 2: $n = 2m + 1$ is odd. So $1-n= 1-(2m + 1) = -2m = 2(-m)$ is even, and $n^2 + 1 = (2m+1)^2 + 1 = 4m^2 + 4m + 1 + 1 = 2(2m^2 + 2m +1)$ is even.
So both $1-n$ and $n^2 + 1$ are even.
So $1-n$ is even precisely and only when $n^2 + 1$ is even.
edited Jan 25 at 1:40
J. W. Tanner
3,3451320
answered Jan 25 at 0:20
fleabloodfleablood
72.7k22788
$endgroup$
You can do it straight forwardly in cases.
Either $n$ is even or odd.
Case 1: $n = 2m$ is even so $1-n =1- 2m = 2(-m) + 1$ is odd and $n^2 + 1 = 4m^2 + 1 = 2(2m^2) + 1$ is odd.
Neither $1-n$ nor $n^2 + 1$ is even.
Case 2: $n = 2m + 1$ is odd. So $1-n= 1-(2m + 1) = -2m = 2(-m)$ is even, and $n^2 + 1 = (2m+1)^2 + 1 = 4m^2 + 4m + 1 + 1 = 2(2m^2 + 2m +1)$ is even.
So both $1-n$ and $n^2 + 1$ are even.
So $1-n$ is even precisely and only when $n^2 + 1$ is even.
edited Jan 25 at 1:40
J. W. Tanner
3,3451320
answered Jan 25 at 0:20
fleabloodfleablood
72.7k22788
edited Jan 25 at 1:40
J. W. Tanner
3,3451320
edited Jan 25 at 1:40
J. W. Tanner
3,3451320
edited Jan 25 at 1:40
J. W. Tanner
3,3451320
3,3451320
answered Jan 25 at 0:20
fleabloodfleablood
72.7k22788
answered Jan 25 at 0:20
fleabloodfleablood
72.7k22788
answered Jan 25 at 0:20
fleabloodfleablood
72.7k22788
72.7k22788
add a comment |
add a comment |
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