Mixing
Prove that $n^2$ is even if and only if $n$ is even
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I am practising exam questions and have come across the following.
Prove that $n^2$ is even if and only if $n$ is even
A contrapositive proof comes to mind, since it must be the case that if $n$ isn't odd then $n$ is even. So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1$ which is odd, hence $n^2$ must be even.
Is this correct?
elementary-number-theory proof-verification
edited Jan 1 at 20:04
TheSimpliFire
12.6k62260
asked Jan 1 at 19:51
TravitrinketTravitrinket
194
$endgroup$
add a comment |
$begingroup$
I am practising exam questions and have come across the following.
Prove that $n^2$ is even if and only if $n$ is even
A contrapositive proof comes to mind, since it must be the case that if $n$ isn't odd then $n$ is even. So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1$ which is odd, hence $n^2$ must be even.
Is this correct?
elementary-number-theory proof-verification
edited Jan 1 at 20:04
TheSimpliFire
12.6k62260
asked Jan 1 at 19:51
TravitrinketTravitrinket
194
$endgroup$
2
$begingroup$
This shows that 'if $n$ is odd then $n^2$ is odd', which is the contrapositive of 'if $n^2$ is even then $n$ is even'. But that is only half of the question, since you're supposed to show an 'if and only if'. So, you still need to show that 'if $n$ is even, then $n^2$ is even'.
$endgroup$
– Bram28
Jan 1 at 19:59
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Are you allowed to use "even * even is even" and "odd * odd is odd", or are you meant to prove those?
$endgroup$
– timtfj
Jan 1 at 20:06
$begingroup$
I assume that I'm allowed to, but maybe I should state this?
$endgroup$
– Travitrinket
Jan 1 at 20:18
add a comment |
$begingroup$
I am practising exam questions and have come across the following.
Prove that $n^2$ is even if and only if $n$ is even
A contrapositive proof comes to mind, since it must be the case that if $n$ isn't odd then $n$ is even. So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1$ which is odd, hence $n^2$ must be even.
Is this correct?
elementary-number-theory proof-verification
edited Jan 1 at 20:04
TheSimpliFire
12.6k62260
asked Jan 1 at 19:51
TravitrinketTravitrinket
194
$endgroup$
I am practising exam questions and have come across the following.
Prove that $n^2$ is even if and only if $n$ is even
A contrapositive proof comes to mind, since it must be the case that if $n$ isn't odd then $n$ is even. So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1$ which is odd, hence $n^2$ must be even.
Is this correct?
elementary-number-theory proof-verification
elementary-number-theory proof-verification
edited Jan 1 at 20:04
TheSimpliFire
12.6k62260
asked Jan 1 at 19:51
TravitrinketTravitrinket
194
edited Jan 1 at 20:04
TheSimpliFire
12.6k62260
asked Jan 1 at 19:51
TravitrinketTravitrinket
194
edited Jan 1 at 20:04
TheSimpliFire
12.6k62260
edited Jan 1 at 20:04
TheSimpliFire
12.6k62260
edited Jan 1 at 20:04
TheSimpliFire
12.6k62260
12.6k62260
asked Jan 1 at 19:51
TravitrinketTravitrinket
194
asked Jan 1 at 19:51
TravitrinketTravitrinket
194
asked Jan 1 at 19:51
TravitrinketTravitrinket
194
194
2
$begingroup$
This shows that 'if $n$ is odd then $n^2$ is odd', which is the contrapositive of 'if $n^2$ is even then $n$ is even'. But that is only half of the question, since you're supposed to show an 'if and only if'. So, you still need to show that 'if $n$ is even, then $n^2$ is even'.
$endgroup$
– Bram28
Jan 1 at 19:59
$begingroup$
Are you allowed to use "even * even is even" and "odd * odd is odd", or are you meant to prove those?
$endgroup$
– timtfj
Jan 1 at 20:06
$begingroup$
I assume that I'm allowed to, but maybe I should state this?
$endgroup$
– Travitrinket
Jan 1 at 20:18
add a comment |
2
$begingroup$
This shows that 'if $n$ is odd then $n^2$ is odd', which is the contrapositive of 'if $n^2$ is even then $n$ is even'. But that is only half of the question, since you're supposed to show an 'if and only if'. So, you still need to show that 'if $n$ is even, then $n^2$ is even'.
$endgroup$
– Bram28
Jan 1 at 19:59
$begingroup$
Are you allowed to use "even * even is even" and "odd * odd is odd", or are you meant to prove those?
$endgroup$
– timtfj
Jan 1 at 20:06
$begingroup$
I assume that I'm allowed to, but maybe I should state this?
$endgroup$
– Travitrinket
Jan 1 at 20:18
2
2
$begingroup$
This shows that 'if $n$ is odd then $n^2$ is odd', which is the contrapositive of 'if $n^2$ is even then $n$ is even'. But that is only half of the question, since you're supposed to show an 'if and only if'. So, you still need to show that 'if $n$ is even, then $n^2$ is even'.
$endgroup$
– Bram28
Jan 1 at 19:59
$begingroup$
This shows that 'if $n$ is odd then $n^2$ is odd', which is the contrapositive of 'if $n^2$ is even then $n$ is even'. But that is only half of the question, since you're supposed to show an 'if and only if'. So, you still need to show that 'if $n$ is even, then $n^2$ is even'.
$endgroup$
– Bram28
Jan 1 at 19:59
$begingroup$
Are you allowed to use "even * even is even" and "odd * odd is odd", or are you meant to prove those?
$endgroup$
– timtfj
Jan 1 at 20:06
$begingroup$
Are you allowed to use "even * even is even" and "odd * odd is odd", or are you meant to prove those?
$endgroup$
– timtfj
Jan 1 at 20:06
$begingroup$
I assume that I'm allowed to, but maybe I should state this?
$endgroup$
– Travitrinket
Jan 1 at 20:18
$begingroup$
I assume that I'm allowed to, but maybe I should state this?
$endgroup$
– Travitrinket
Jan 1 at 20:18
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
For an if-and-only-if proof, you need to show both ways - you have only done $n$ even $implies$ $n^2$ even, which is correct. You must show that
if $n=2k$ for some integer $k$, $n^2=4k^2=2(2k^2)$ is even, which is true
if $n^2=2t$ for some integer $t$ but $n=2s+1$ is odd for some integer $s$, $$(2s+1)^2=2timplies 4s^2+4s-2t=2(2s^2+2s-t)=-1$$ is a contradiction as the parities on LHS and RHS are different
answered Jan 1 at 19:58
TheSimpliFireTheSimpliFire
12.6k62260
$endgroup$
$begingroup$
I think you can be clearer: you're answering his question about his own proof, but you're not pointing directly what he's done wrong or hasn't done at all.
$endgroup$
– Lucas Henrique
Jan 1 at 20:08
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@LucasHenrique Thanks for the reminder! Edited
$endgroup$
– TheSimpliFire
Jan 1 at 20:09
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+1. Nice answer :)
$endgroup$
– Lucas Henrique
Jan 1 at 20:12
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Why is the last step necessary?
$endgroup$
– Travitrinket
Jan 1 at 20:49
$begingroup$
You also need to show that is $n^2$ is even, so is $n$. This can be done by assuming that $n^2$ is even and $n$ is odd, reaching a contradiction.
$endgroup$
– TheSimpliFire
Jan 2 at 8:59
|
show 6 more comments
$begingroup$
You've proved that if $n^2$ is even then so is $n$. So you've proved the "only if" direction, but not the "if".
So you've just got to show that the square $n^2$ of an even number $n=2k$ is even, and you've got the "if" as well. Put them together and you're there.
answered Jan 1 at 20:16
timtfjtimtfj
1,183318
$endgroup$
$begingroup$
$( neg q Rightarrow neg p )$ $Leftrightarrow$ $( p Rightarrow q ) $. This is what you mean right?
$endgroup$
– Travitrinket
Jan 3 at 11:28
$begingroup$
@Travitrinket Yes, that exactly :-)
$endgroup$
– timtfj
Jan 3 at 19:23
add a comment |
$begingroup$
$n>1;$
$D_n:= n^2-n=n(n-1); $
$D_n$ is even, since either $n$ or $(n-1)$ is even.
1) $n$ is even $Rightarrow$
$n^2 = D_n +n$ is even;
(Sum of $2$ even numbers)
2)$n^2$ is even $Rightarrow$
$n= n^2-D_n$ is even;
(Difference of $2$ even numbers)
Used: Sum and difference of $2$ even numbers is even(Why?)
answered Jan 1 at 20:19
Peter SzilasPeter Szilas
10.9k2720
$endgroup$
$begingroup$
Quote: "This is a proof-verification question. You did not answer him."
$endgroup$
– Did
Jan 2 at 9:21
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For an if-and-only-if proof, you need to show both ways - you have only done $n$ even $implies$ $n^2$ even, which is correct. You must show that
if $n=2k$ for some integer $k$, $n^2=4k^2=2(2k^2)$ is even, which is true
if $n^2=2t$ for some integer $t$ but $n=2s+1$ is odd for some integer $s$, $$(2s+1)^2=2timplies 4s^2+4s-2t=2(2s^2+2s-t)=-1$$ is a contradiction as the parities on LHS and RHS are different
answered Jan 1 at 19:58
TheSimpliFireTheSimpliFire
12.6k62260
$endgroup$
$begingroup$
I think you can be clearer: you're answering his question about his own proof, but you're not pointing directly what he's done wrong or hasn't done at all.
$endgroup$
– Lucas Henrique
Jan 1 at 20:08
$begingroup$
@LucasHenrique Thanks for the reminder! Edited
$endgroup$
– TheSimpliFire
Jan 1 at 20:09
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+1. Nice answer :)
$endgroup$
– Lucas Henrique
Jan 1 at 20:12
$begingroup$
Why is the last step necessary?
$endgroup$
– Travitrinket
Jan 1 at 20:49
$begingroup$
You also need to show that is $n^2$ is even, so is $n$. This can be done by assuming that $n^2$ is even and $n$ is odd, reaching a contradiction.
$endgroup$
– TheSimpliFire
Jan 2 at 8:59
|
show 6 more comments
$begingroup$
For an if-and-only-if proof, you need to show both ways - you have only done $n$ even $implies$ $n^2$ even, which is correct. You must show that
if $n=2k$ for some integer $k$, $n^2=4k^2=2(2k^2)$ is even, which is true
if $n^2=2t$ for some integer $t$ but $n=2s+1$ is odd for some integer $s$, $$(2s+1)^2=2timplies 4s^2+4s-2t=2(2s^2+2s-t)=-1$$ is a contradiction as the parities on LHS and RHS are different
answered Jan 1 at 19:58
TheSimpliFireTheSimpliFire
12.6k62260
$endgroup$
$begingroup$
I think you can be clearer: you're answering his question about his own proof, but you're not pointing directly what he's done wrong or hasn't done at all.
$endgroup$
– Lucas Henrique
Jan 1 at 20:08
$begingroup$
@LucasHenrique Thanks for the reminder! Edited
$endgroup$
– TheSimpliFire
Jan 1 at 20:09
$begingroup$
+1. Nice answer :)
$endgroup$
– Lucas Henrique
Jan 1 at 20:12
$begingroup$
Why is the last step necessary?
$endgroup$
– Travitrinket
Jan 1 at 20:49
$begingroup$
You also need to show that is $n^2$ is even, so is $n$. This can be done by assuming that $n^2$ is even and $n$ is odd, reaching a contradiction.
$endgroup$
– TheSimpliFire
Jan 2 at 8:59
|
show 6 more comments
$begingroup$
For an if-and-only-if proof, you need to show both ways - you have only done $n$ even $implies$ $n^2$ even, which is correct. You must show that
if $n=2k$ for some integer $k$, $n^2=4k^2=2(2k^2)$ is even, which is true
if $n^2=2t$ for some integer $t$ but $n=2s+1$ is odd for some integer $s$, $$(2s+1)^2=2timplies 4s^2+4s-2t=2(2s^2+2s-t)=-1$$ is a contradiction as the parities on LHS and RHS are different
answered Jan 1 at 19:58
TheSimpliFireTheSimpliFire
12.6k62260
$endgroup$
For an if-and-only-if proof, you need to show both ways - you have only done $n$ even $implies$ $n^2$ even, which is correct. You must show that
if $n=2k$ for some integer $k$, $n^2=4k^2=2(2k^2)$ is even, which is true
if $n^2=2t$ for some integer $t$ but $n=2s+1$ is odd for some integer $s$, $$(2s+1)^2=2timplies 4s^2+4s-2t=2(2s^2+2s-t)=-1$$ is a contradiction as the parities on LHS and RHS are different
answered Jan 1 at 19:58
TheSimpliFireTheSimpliFire
12.6k62260
edited Jan 1 at 20:09
answered Jan 1 at 19:58
TheSimpliFireTheSimpliFire
12.6k62260
answered Jan 1 at 19:58
TheSimpliFireTheSimpliFire
12.6k62260
answered Jan 1 at 19:58
TheSimpliFireTheSimpliFire
12.6k62260
12.6k62260
$begingroup$
I think you can be clearer: you're answering his question about his own proof, but you're not pointing directly what he's done wrong or hasn't done at all.
$endgroup$
– Lucas Henrique
Jan 1 at 20:08
$begingroup$
@LucasHenrique Thanks for the reminder! Edited
$endgroup$
– TheSimpliFire
Jan 1 at 20:09
$begingroup$
+1. Nice answer :)
$endgroup$
– Lucas Henrique
Jan 1 at 20:12
$begingroup$
Why is the last step necessary?
$endgroup$
– Travitrinket
Jan 1 at 20:49
$begingroup$
You also need to show that is $n^2$ is even, so is $n$. This can be done by assuming that $n^2$ is even and $n$ is odd, reaching a contradiction.
$endgroup$
– TheSimpliFire
Jan 2 at 8:59
|
show 6 more comments
$begingroup$
I think you can be clearer: you're answering his question about his own proof, but you're not pointing directly what he's done wrong or hasn't done at all.
$endgroup$
– Lucas Henrique
Jan 1 at 20:08
$begingroup$
@LucasHenrique Thanks for the reminder! Edited
$endgroup$
– TheSimpliFire
Jan 1 at 20:09
$begingroup$
+1. Nice answer :)
$endgroup$
– Lucas Henrique
Jan 1 at 20:12
$begingroup$
Why is the last step necessary?
$endgroup$
– Travitrinket
Jan 1 at 20:49
$begingroup$
You also need to show that is $n^2$ is even, so is $n$. This can be done by assuming that $n^2$ is even and $n$ is odd, reaching a contradiction.
$endgroup$
– TheSimpliFire
Jan 2 at 8:59
$begingroup$
I think you can be clearer: you're answering his question about his own proof, but you're not pointing directly what he's done wrong or hasn't done at all.
$endgroup$
– Lucas Henrique
Jan 1 at 20:08
$begingroup$
I think you can be clearer: you're answering his question about his own proof, but you're not pointing directly what he's done wrong or hasn't done at all.
$endgroup$
– Lucas Henrique
Jan 1 at 20:08
$begingroup$
@LucasHenrique Thanks for the reminder! Edited
$endgroup$
– TheSimpliFire
Jan 1 at 20:09
$begingroup$
@LucasHenrique Thanks for the reminder! Edited
$endgroup$
– TheSimpliFire
Jan 1 at 20:09
$begingroup$
+1. Nice answer :)
$endgroup$
– Lucas Henrique
Jan 1 at 20:12
$begingroup$
+1. Nice answer :)
$endgroup$
– Lucas Henrique
Jan 1 at 20:12
$begingroup$
Why is the last step necessary?
$endgroup$
– Travitrinket
Jan 1 at 20:49
$begingroup$
Why is the last step necessary?
$endgroup$
– Travitrinket
Jan 1 at 20:49
$begingroup$
You also need to show that is $n^2$ is even, so is $n$. This can be done by assuming that $n^2$ is even and $n$ is odd, reaching a contradiction.
$endgroup$
– TheSimpliFire
Jan 2 at 8:59
$begingroup$
You also need to show that is $n^2$ is even, so is $n$. This can be done by assuming that $n^2$ is even and $n$ is odd, reaching a contradiction.
$endgroup$
– TheSimpliFire
Jan 2 at 8:59
|
show 6 more comments
$begingroup$
You've proved that if $n^2$ is even then so is $n$. So you've proved the "only if" direction, but not the "if".
So you've just got to show that the square $n^2$ of an even number $n=2k$ is even, and you've got the "if" as well. Put them together and you're there.
answered Jan 1 at 20:16
timtfjtimtfj
1,183318
$endgroup$
$begingroup$
$( neg q Rightarrow neg p )$ $Leftrightarrow$ $( p Rightarrow q ) $. This is what you mean right?
$endgroup$
– Travitrinket
Jan 3 at 11:28
$begingroup$
@Travitrinket Yes, that exactly :-)
$endgroup$
– timtfj
Jan 3 at 19:23
add a comment |
$begingroup$
You've proved that if $n^2$ is even then so is $n$. So you've proved the "only if" direction, but not the "if".
So you've just got to show that the square $n^2$ of an even number $n=2k$ is even, and you've got the "if" as well. Put them together and you're there.
answered Jan 1 at 20:16
timtfjtimtfj
1,183318
$endgroup$
$begingroup$
$( neg q Rightarrow neg p )$ $Leftrightarrow$ $( p Rightarrow q ) $. This is what you mean right?
$endgroup$
– Travitrinket
Jan 3 at 11:28
$begingroup$
@Travitrinket Yes, that exactly :-)
$endgroup$
– timtfj
Jan 3 at 19:23
add a comment |
$begingroup$
You've proved that if $n^2$ is even then so is $n$. So you've proved the "only if" direction, but not the "if".
So you've just got to show that the square $n^2$ of an even number $n=2k$ is even, and you've got the "if" as well. Put them together and you're there.
answered Jan 1 at 20:16
timtfjtimtfj
1,183318
$endgroup$
You've proved that if $n^2$ is even then so is $n$. So you've proved the "only if" direction, but not the "if".
So you've just got to show that the square $n^2$ of an even number $n=2k$ is even, and you've got the "if" as well. Put them together and you're there.
answered Jan 1 at 20:16
timtfjtimtfj
1,183318
edited Jan 1 at 20:22
answered Jan 1 at 20:16
timtfjtimtfj
1,183318
answered Jan 1 at 20:16
timtfjtimtfj
1,183318
answered Jan 1 at 20:16
timtfjtimtfj
1,183318
1,183318
$begingroup$
$( neg q Rightarrow neg p )$ $Leftrightarrow$ $( p Rightarrow q ) $. This is what you mean right?
$endgroup$
– Travitrinket
Jan 3 at 11:28
$begingroup$
@Travitrinket Yes, that exactly :-)
$endgroup$
– timtfj
Jan 3 at 19:23
add a comment |
$begingroup$
$( neg q Rightarrow neg p )$ $Leftrightarrow$ $( p Rightarrow q ) $. This is what you mean right?
$endgroup$
– Travitrinket
Jan 3 at 11:28
$begingroup$
@Travitrinket Yes, that exactly :-)
$endgroup$
– timtfj
Jan 3 at 19:23
$begingroup$
$( neg q Rightarrow neg p )$ $Leftrightarrow$ $( p Rightarrow q ) $. This is what you mean right?
$endgroup$
– Travitrinket
Jan 3 at 11:28
$begingroup$
$( neg q Rightarrow neg p )$ $Leftrightarrow$ $( p Rightarrow q ) $. This is what you mean right?
$endgroup$
– Travitrinket
Jan 3 at 11:28
$begingroup$
@Travitrinket Yes, that exactly :-)
$endgroup$
– timtfj
Jan 3 at 19:23
$begingroup$
@Travitrinket Yes, that exactly :-)
$endgroup$
– timtfj
Jan 3 at 19:23
add a comment |
$begingroup$
$n>1;$
$D_n:= n^2-n=n(n-1); $
$D_n$ is even, since either $n$ or $(n-1)$ is even.
1) $n$ is even $Rightarrow$
$n^2 = D_n +n$ is even;
(Sum of $2$ even numbers)
2)$n^2$ is even $Rightarrow$
$n= n^2-D_n$ is even;
(Difference of $2$ even numbers)
Used: Sum and difference of $2$ even numbers is even(Why?)
answered Jan 1 at 20:19
Peter SzilasPeter Szilas
10.9k2720
$endgroup$
$begingroup$
Quote: "This is a proof-verification question. You did not answer him."
$endgroup$
– Did
Jan 2 at 9:21
add a comment |
$begingroup$
$n>1;$
$D_n:= n^2-n=n(n-1); $
$D_n$ is even, since either $n$ or $(n-1)$ is even.
1) $n$ is even $Rightarrow$
$n^2 = D_n +n$ is even;
(Sum of $2$ even numbers)
2)$n^2$ is even $Rightarrow$
$n= n^2-D_n$ is even;
(Difference of $2$ even numbers)
Used: Sum and difference of $2$ even numbers is even(Why?)
answered Jan 1 at 20:19
Peter SzilasPeter Szilas
10.9k2720
$endgroup$
$begingroup$
Quote: "This is a proof-verification question. You did not answer him."
$endgroup$
– Did
Jan 2 at 9:21
add a comment |
$begingroup$
$n>1;$
$D_n:= n^2-n=n(n-1); $
$D_n$ is even, since either $n$ or $(n-1)$ is even.
1) $n$ is even $Rightarrow$
$n^2 = D_n +n$ is even;
(Sum of $2$ even numbers)
2)$n^2$ is even $Rightarrow$
$n= n^2-D_n$ is even;
(Difference of $2$ even numbers)
Used: Sum and difference of $2$ even numbers is even(Why?)
answered Jan 1 at 20:19
Peter SzilasPeter Szilas
10.9k2720
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$n>1;$
$D_n:= n^2-n=n(n-1); $
$D_n$ is even, since either $n$ or $(n-1)$ is even.
1) $n$ is even $Rightarrow$
$n^2 = D_n +n$ is even;
(Sum of $2$ even numbers)
2)$n^2$ is even $Rightarrow$
$n= n^2-D_n$ is even;
(Difference of $2$ even numbers)
Used: Sum and difference of $2$ even numbers is even(Why?)
answered Jan 1 at 20:19
Peter SzilasPeter Szilas
10.9k2720
answered Jan 1 at 20:19
Peter SzilasPeter Szilas
10.9k2720
answered Jan 1 at 20:19
Peter SzilasPeter Szilas
10.9k2720
answered Jan 1 at 20:19
Peter SzilasPeter Szilas
10.9k2720
10.9k2720
$begingroup$
Quote: "This is a proof-verification question. You did not answer him."
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– Did
Jan 2 at 9:21
add a comment |
$begingroup$
Quote: "This is a proof-verification question. You did not answer him."
$endgroup$
– Did
Jan 2 at 9:21
$begingroup$
Quote: "This is a proof-verification question. You did not answer him."
$endgroup$
– Did
Jan 2 at 9:21
$begingroup$
Quote: "This is a proof-verification question. You did not answer him."
$endgroup$
– Did
Jan 2 at 9:21
add a comment |
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2
$begingroup$
This shows that 'if $n$ is odd then $n^2$ is odd', which is the contrapositive of 'if $n^2$ is even then $n$ is even'. But that is only half of the question, since you're supposed to show an 'if and only if'. So, you still need to show that 'if $n$ is even, then $n^2$ is even'.
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– Bram28
Jan 1 at 19:59
$begingroup$
Are you allowed to use "even * even is even" and "odd * odd is odd", or are you meant to prove those?
$endgroup$
– timtfj
Jan 1 at 20:06
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I assume that I'm allowed to, but maybe I should state this?
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– Travitrinket
Jan 1 at 20:18