contradiction proof on divides
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Suppose a,b ∈ Z. If 4 | $(a^2 + b^2)$ then a and b are not both odd.
So, assuming that 4 | $(a^2 + b^2)$ and $a$ and $b$ are odd
this gives $4k=(2l+1)^2+(2u+1)^2$ for some $k,l,uin z$
eventually leading to $4k=4(l^2+l+u)+2(u^2+1)$
The RHS is not a multiple of 4 when $u=2$ contradiction.
Is this valid, thanks.
proof-verification divisibility
edited 14 hours ago
GNUSupporter 8964民主女神 地下教會
12.3k72243
asked 14 hours ago
Carlos Bacca
18210
add a comment |
up vote
2
down vote
favorite
Suppose a,b ∈ Z. If 4 | $(a^2 + b^2)$ then a and b are not both odd.
So, assuming that 4 | $(a^2 + b^2)$ and $a$ and $b$ are odd
this gives $4k=(2l+1)^2+(2u+1)^2$ for some $k,l,uin z$
eventually leading to $4k=4(l^2+l+u)+2(u^2+1)$
The RHS is not a multiple of 4 when $u=2$ contradiction.
Is this valid, thanks.
proof-verification divisibility
edited 14 hours ago
GNUSupporter 8964民主女神 地下教會
12.3k72243
asked 14 hours ago
Carlos Bacca
18210
You are trying to prove the claim in the first line? Not sure why you can set $u=2$ towards the end...isn't $u$ an unknown parameter?
– lulu
14 hours ago
As a hint; show that $m$ odd implies $m^2=4k+1$ for some $k$.
– lulu
14 hours ago
thanks i see my mistake, should i leave the question as it is or correct it?
– Carlos Bacca
13 hours ago
@CarlosBacca You asked where the mistake is. We told you where the mistake is. The logical next step is to accept one of the answers. Don't edit out your mistake.
– 5xum
13 hours ago
True that would be pointless
– Carlos Bacca
13 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose a,b ∈ Z. If 4 | $(a^2 + b^2)$ then a and b are not both odd.
So, assuming that 4 | $(a^2 + b^2)$ and $a$ and $b$ are odd
this gives $4k=(2l+1)^2+(2u+1)^2$ for some $k,l,uin z$
eventually leading to $4k=4(l^2+l+u)+2(u^2+1)$
The RHS is not a multiple of 4 when $u=2$ contradiction.
Is this valid, thanks.
proof-verification divisibility
edited 14 hours ago
GNUSupporter 8964民主女神 地下教會
12.3k72243
asked 14 hours ago
Carlos Bacca
18210
Suppose a,b ∈ Z. If 4 | $(a^2 + b^2)$ then a and b are not both odd.
So, assuming that 4 | $(a^2 + b^2)$ and $a$ and $b$ are odd
this gives $4k=(2l+1)^2+(2u+1)^2$ for some $k,l,uin z$
eventually leading to $4k=4(l^2+l+u)+2(u^2+1)$
The RHS is not a multiple of 4 when $u=2$ contradiction.
Is this valid, thanks.
proof-verification divisibility
proof-verification divisibility
edited 14 hours ago
GNUSupporter 8964民主女神 地下教會
12.3k72243
asked 14 hours ago
Carlos Bacca
18210
edited 14 hours ago
GNUSupporter 8964民主女神 地下教會
12.3k72243
asked 14 hours ago
Carlos Bacca
18210
edited 14 hours ago
GNUSupporter 8964民主女神 地下教會
12.3k72243
edited 14 hours ago
GNUSupporter 8964民主女神 地下教會
12.3k72243
edited 14 hours ago
GNUSupporter 8964民主女神 地下教會
12.3k72243
12.3k72243
asked 14 hours ago
Carlos Bacca
18210
asked 14 hours ago
Carlos Bacca
18210
asked 14 hours ago
Carlos Bacca
18210
18210
You are trying to prove the claim in the first line? Not sure why you can set $u=2$ towards the end...isn't $u$ an unknown parameter?
– lulu
14 hours ago
As a hint; show that $m$ odd implies $m^2=4k+1$ for some $k$.
– lulu
14 hours ago
thanks i see my mistake, should i leave the question as it is or correct it?
– Carlos Bacca
13 hours ago
@CarlosBacca You asked where the mistake is. We told you where the mistake is. The logical next step is to accept one of the answers. Don't edit out your mistake.
– 5xum
13 hours ago
True that would be pointless
– Carlos Bacca
13 hours ago
add a comment |
You are trying to prove the claim in the first line? Not sure why you can set $u=2$ towards the end...isn't $u$ an unknown parameter?
– lulu
14 hours ago
As a hint; show that $m$ odd implies $m^2=4k+1$ for some $k$.
– lulu
14 hours ago
thanks i see my mistake, should i leave the question as it is or correct it?
– Carlos Bacca
13 hours ago
@CarlosBacca You asked where the mistake is. We told you where the mistake is. The logical next step is to accept one of the answers. Don't edit out your mistake.
– 5xum
13 hours ago
True that would be pointless
– Carlos Bacca
13 hours ago
You are trying to prove the claim in the first line? Not sure why you can set $u=2$ towards the end...isn't $u$ an unknown parameter?
– lulu
14 hours ago
You are trying to prove the claim in the first line? Not sure why you can set $u=2$ towards the end...isn't $u$ an unknown parameter?
– lulu
14 hours ago
As a hint; show that $m$ odd implies $m^2=4k+1$ for some $k$.
– lulu
14 hours ago
As a hint; show that $m$ odd implies $m^2=4k+1$ for some $k$.
– lulu
14 hours ago
thanks i see my mistake, should i leave the question as it is or correct it?
– Carlos Bacca
13 hours ago
thanks i see my mistake, should i leave the question as it is or correct it?
– Carlos Bacca
13 hours ago
@CarlosBacca You asked where the mistake is. We told you where the mistake is. The logical next step is to accept one of the answers. Don't edit out your mistake.
– 5xum
13 hours ago
@CarlosBacca You asked where the mistake is. We told you where the mistake is. The logical next step is to accept one of the answers. Don't edit out your mistake.
– 5xum
13 hours ago
True that would be pointless
– Carlos Bacca
13 hours ago
True that would be pointless
– Carlos Bacca
13 hours ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
It's not yet valid, because you haven't shown why the RHS cannot be a multiple of $4$. You cannot simply set $u=2$, because the $u$ you have is already determined by $b$, since $b=2u+1$.
To correct your proof, re-think how you got from
$$4k=(2l+1)^2 + (2u+1)^2$$
to
$$4k = 4(l^2+l+u) + 2(u^2+1)$$
because I think you were a bit sloppy here.
answered 14 hours ago
5xum
87.6k391159
add a comment |
up vote
1
down vote
This is invalid. You have done a mistake in calculation.
RHS will be $4(l^2+u^2+l+u)+2$, which is not divisible by 4.
Hope it helps:)
answered 14 hours ago
Crazy for maths
1424
add a comment |
up vote
0
down vote
If you multiply correctly, the RHS leaves a rest of 2 modulo 4. So your idea was correct, you just need to recalculate.
answered 13 hours ago
Nodt Greenish
266
New contributor
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
It's not yet valid, because you haven't shown why the RHS cannot be a multiple of $4$. You cannot simply set $u=2$, because the $u$ you have is already determined by $b$, since $b=2u+1$.
To correct your proof, re-think how you got from
$$4k=(2l+1)^2 + (2u+1)^2$$
to
$$4k = 4(l^2+l+u) + 2(u^2+1)$$
because I think you were a bit sloppy here.
answered 14 hours ago
5xum
87.6k391159
add a comment |
up vote
3
down vote
accepted
It's not yet valid, because you haven't shown why the RHS cannot be a multiple of $4$. You cannot simply set $u=2$, because the $u$ you have is already determined by $b$, since $b=2u+1$.
To correct your proof, re-think how you got from
$$4k=(2l+1)^2 + (2u+1)^2$$
to
$$4k = 4(l^2+l+u) + 2(u^2+1)$$
because I think you were a bit sloppy here.
answered 14 hours ago
5xum
87.6k391159
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
It's not yet valid, because you haven't shown why the RHS cannot be a multiple of $4$. You cannot simply set $u=2$, because the $u$ you have is already determined by $b$, since $b=2u+1$.
To correct your proof, re-think how you got from
$$4k=(2l+1)^2 + (2u+1)^2$$
to
$$4k = 4(l^2+l+u) + 2(u^2+1)$$
because I think you were a bit sloppy here.
answered 14 hours ago
5xum
87.6k391159
It's not yet valid, because you haven't shown why the RHS cannot be a multiple of $4$. You cannot simply set $u=2$, because the $u$ you have is already determined by $b$, since $b=2u+1$.
To correct your proof, re-think how you got from
$$4k=(2l+1)^2 + (2u+1)^2$$
to
$$4k = 4(l^2+l+u) + 2(u^2+1)$$
because I think you were a bit sloppy here.
answered 14 hours ago
5xum
87.6k391159
answered 14 hours ago
5xum
87.6k391159
answered 14 hours ago
5xum
87.6k391159
answered 14 hours ago
5xum
87.6k391159
87.6k391159
add a comment |
add a comment |
up vote
1
down vote
This is invalid. You have done a mistake in calculation.
RHS will be $4(l^2+u^2+l+u)+2$, which is not divisible by 4.
Hope it helps:)
answered 14 hours ago
Crazy for maths
1424
add a comment |
up vote
1
down vote
This is invalid. You have done a mistake in calculation.
RHS will be $4(l^2+u^2+l+u)+2$, which is not divisible by 4.
Hope it helps:)
answered 14 hours ago
Crazy for maths
1424
add a comment |
up vote
1
down vote
up vote
1
down vote
This is invalid. You have done a mistake in calculation.
RHS will be $4(l^2+u^2+l+u)+2$, which is not divisible by 4.
Hope it helps:)
answered 14 hours ago
Crazy for maths
1424
This is invalid. You have done a mistake in calculation.
RHS will be $4(l^2+u^2+l+u)+2$, which is not divisible by 4.
Hope it helps:)
answered 14 hours ago
Crazy for maths
1424
answered 14 hours ago
Crazy for maths
1424
answered 14 hours ago
Crazy for maths
1424
answered 14 hours ago
Crazy for maths
1424
1424
add a comment |
add a comment |
up vote
0
down vote
If you multiply correctly, the RHS leaves a rest of 2 modulo 4. So your idea was correct, you just need to recalculate.
answered 13 hours ago
Nodt Greenish
266
New contributor
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
If you multiply correctly, the RHS leaves a rest of 2 modulo 4. So your idea was correct, you just need to recalculate.
answered 13 hours ago
Nodt Greenish
266
New contributor
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
If you multiply correctly, the RHS leaves a rest of 2 modulo 4. So your idea was correct, you just need to recalculate.
answered 13 hours ago
Nodt Greenish
266
New contributor
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If you multiply correctly, the RHS leaves a rest of 2 modulo 4. So your idea was correct, you just need to recalculate.
answered 13 hours ago
Nodt Greenish
266
New contributor
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 13 hours ago
Nodt Greenish
266
New contributor
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 13 hours ago
Nodt Greenish
266
answered 13 hours ago
Nodt Greenish
266
266
New contributor
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nodt Greenish is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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You are trying to prove the claim in the first line? Not sure why you can set $u=2$ towards the end...isn't $u$ an unknown parameter?
– lulu
14 hours ago
As a hint; show that $m$ odd implies $m^2=4k+1$ for some $k$.
– lulu
14 hours ago
thanks i see my mistake, should i leave the question as it is or correct it?
– Carlos Bacca
13 hours ago
@CarlosBacca You asked where the mistake is. We told you where the mistake is. The logical next step is to accept one of the answers. Don't edit out your mistake.
– 5xum
13 hours ago
True that would be pointless
– Carlos Bacca
13 hours ago