Mixing
Need help with the proof {needed for my algorithm}
$begingroup$
I am not from pure math background.
I am working on an algorithm which works good for all the practical reasons
based on the following assumption.
that,
if
ab = cd
and a+b = c+d
then, either a = c & b = d or
a = d & b = c. where, $a,b,c,d$ $in$ $mathbb{Z}$
I am not sure if this is true in the all conditions, but if it's can anyone provide proof, why ?
Thanks in advance.
Addition -1 :
Is this also true for pair of N numbers ?
Addition -2 (Answer for the above)
- I think I got it, @saulspatz's proof can be generalized (already sufficient),
we can say,
After proof since, $a = c $ & $b = d$
then, by saying
$b = b1 + b2$ and $c = c1 + c2$ proof can be further generalized for the addition and multiplication of pair of $N$ such numbers.
$a1, a2,...aN in mathbb{Z}$
$b1, b2,....bN in mathbb{Z} $
real-analysis real-numbers integers
asked Jan 14 at 18:51
rahulbrahulb
1334
$endgroup$
add a comment |
$begingroup$
I am not from pure math background.
I am working on an algorithm which works good for all the practical reasons
based on the following assumption.
that,
if
ab = cd
and a+b = c+d
then, either a = c & b = d or
a = d & b = c. where, $a,b,c,d$ $in$ $mathbb{Z}$
I am not sure if this is true in the all conditions, but if it's can anyone provide proof, why ?
Thanks in advance.
Addition -1 :
Is this also true for pair of N numbers ?
Addition -2 (Answer for the above)
- I think I got it, @saulspatz's proof can be generalized (already sufficient),
we can say,
After proof since, $a = c $ & $b = d$
then, by saying
$b = b1 + b2$ and $c = c1 + c2$ proof can be further generalized for the addition and multiplication of pair of $N$ such numbers.
$a1, a2,...aN in mathbb{Z}$
$b1, b2,....bN in mathbb{Z} $
real-analysis real-numbers integers
asked Jan 14 at 18:51
rahulbrahulb
1334
$endgroup$
$begingroup$
What do you mean by "Is this also true for pair of N numbers ?"? What constraints do you have in mind?
$endgroup$
– lightxbulb
Jan 14 at 19:14
$begingroup$
in this case its just 2 numbers a + b = c+ d ... in case where its a1 + a2 + ... +an = b1 + b2 +b3 ...bn how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:15
$begingroup$
I updated my answer to give an example with more values. Also can you explain what your algorithm aims to achieve? @rahulb
$endgroup$
– lightxbulb
Jan 15 at 10:06
add a comment |
$begingroup$
I am not from pure math background.
I am working on an algorithm which works good for all the practical reasons
based on the following assumption.
that,
if
ab = cd
and a+b = c+d
then, either a = c & b = d or
a = d & b = c. where, $a,b,c,d$ $in$ $mathbb{Z}$
I am not sure if this is true in the all conditions, but if it's can anyone provide proof, why ?
Thanks in advance.
Addition -1 :
Is this also true for pair of N numbers ?
Addition -2 (Answer for the above)
- I think I got it, @saulspatz's proof can be generalized (already sufficient),
we can say,
After proof since, $a = c $ & $b = d$
then, by saying
$b = b1 + b2$ and $c = c1 + c2$ proof can be further generalized for the addition and multiplication of pair of $N$ such numbers.
$a1, a2,...aN in mathbb{Z}$
$b1, b2,....bN in mathbb{Z} $
real-analysis real-numbers integers
asked Jan 14 at 18:51
rahulbrahulb
1334
$endgroup$
I am not from pure math background.
I am working on an algorithm which works good for all the practical reasons
based on the following assumption.
that,
if
ab = cd
and a+b = c+d
then, either a = c & b = d or
a = d & b = c. where, $a,b,c,d$ $in$ $mathbb{Z}$
I am not sure if this is true in the all conditions, but if it's can anyone provide proof, why ?
Thanks in advance.
Addition -1 :
Is this also true for pair of N numbers ?
Addition -2 (Answer for the above)
- I think I got it, @saulspatz's proof can be generalized (already sufficient),
we can say,
After proof since, $a = c $ & $b = d$
then, by saying
$b = b1 + b2$ and $c = c1 + c2$ proof can be further generalized for the addition and multiplication of pair of $N$ such numbers.
$a1, a2,...aN in mathbb{Z}$
$b1, b2,....bN in mathbb{Z} $
real-analysis real-numbers integers
real-analysis real-numbers integers
asked Jan 14 at 18:51
rahulbrahulb
1334
asked Jan 14 at 18:51
rahulbrahulb
1334
edited Jan 15 at 19:19
rahulb
asked Jan 14 at 18:51
rahulbrahulb
1334
asked Jan 14 at 18:51
rahulbrahulb
1334
asked Jan 14 at 18:51
rahulbrahulb
1334
1334
$begingroup$
What do you mean by "Is this also true for pair of N numbers ?"? What constraints do you have in mind?
$endgroup$
– lightxbulb
Jan 14 at 19:14
$begingroup$
in this case its just 2 numbers a + b = c+ d ... in case where its a1 + a2 + ... +an = b1 + b2 +b3 ...bn how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:15
$begingroup$
I updated my answer to give an example with more values. Also can you explain what your algorithm aims to achieve? @rahulb
$endgroup$
– lightxbulb
Jan 15 at 10:06
add a comment |
$begingroup$
What do you mean by "Is this also true for pair of N numbers ?"? What constraints do you have in mind?
$endgroup$
– lightxbulb
Jan 14 at 19:14
$begingroup$
in this case its just 2 numbers a + b = c+ d ... in case where its a1 + a2 + ... +an = b1 + b2 +b3 ...bn how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:15
$begingroup$
I updated my answer to give an example with more values. Also can you explain what your algorithm aims to achieve? @rahulb
$endgroup$
– lightxbulb
Jan 15 at 10:06
$begingroup$
What do you mean by "Is this also true for pair of N numbers ?"? What constraints do you have in mind?
$endgroup$
– lightxbulb
Jan 14 at 19:14
$begingroup$
What do you mean by "Is this also true for pair of N numbers ?"? What constraints do you have in mind?
$endgroup$
– lightxbulb
Jan 14 at 19:14
$begingroup$
in this case its just 2 numbers a + b = c+ d ... in case where its a1 + a2 + ... +an = b1 + b2 +b3 ...bn how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:15
$begingroup$
in this case its just 2 numbers a + b = c+ d ... in case where its a1 + a2 + ... +an = b1 + b2 +b3 ...bn how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:15
$begingroup$
I updated my answer to give an example with more values. Also can you explain what your algorithm aims to achieve? @rahulb
$endgroup$
– lightxbulb
Jan 15 at 10:06
$begingroup$
I updated my answer to give an example with more values. Also can you explain what your algorithm aims to achieve? @rahulb
$endgroup$
– lightxbulb
Jan 15 at 10:06
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let $a+b=n=c+d$. Then $$ab=cdimplies a(n-a)=c(n-c)$$ so that $$an-a^2=cn-c^2,$$ or$$(a-c)n=a^2-c^2=(a-c)(a+c)$$
Either $a-c,$ and we are done, or $a-cne 0$ so we can cancel the $a-c$ to get $$n=a+c$$
Together with $n=aa+b$ this gives $b=c.$
In short, your assumption is correct.
answered Jan 14 at 19:00
saulspatzsaulspatz
15.3k31331
$endgroup$
$begingroup$
thanks @saulspatz .. in case where its a1 + a2 + ... +aN = b1 + b2 +b3+ ...+bN and a1* a2 * a3*... aN = b1 * b2 *b3 ...*bN how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:16
add a comment |
$begingroup$
Only an idea: if $$a+b=c+d$$ we get by squaring
$$a^2+b^2=c^2+d^2$$ since $$ab=cd$$ so we get
$$a^2-c^2=d^2-b^2$$ or $$a^2-d^2=b^2-c^2$$
answered Jan 14 at 18:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
add a comment |
$begingroup$
This is true.
In general, $r$ and $s$ are the two roots of the quadratic equation
$$
x^2 -(r+s)x + rs = (x-r)(x-s).
$$
so knowing the sum and the product determines the values (up to swapping them).
answered Jan 14 at 19:02
Ethan BolkerEthan Bolker
43.4k551116
$endgroup$
add a comment |
$begingroup$
Let $exists delta_1, delta_2 in mathbb{Z}$ such that $c = a + delta_1, d = b + delta_2$ and $(delta_1,delta_2) neq (0,0)$. However from $a+b=c+d$ it follows that $delta_1 = -delta_2$. Now $ab = cd = (a+delta_1)(b-delta_1)$, then $delta_1^2 + (a-b)delta_1 = delta_1(delta_1+a-b) = 0$. Since we assumed $delta_1 neq 0$ it follows that $delta_1 = b-a$, however then we get $c = b, d=a$. Thus it necessarily holds that for the given constraints either $(a,b) = (c,d)$ which corresponds to $delta_1 = 0$, or $(a,b)=(d,c)$ which corresponds to $delta_1 = b-a$. Note that this is true for $a,b,c,d in mathbb{R}$.
Edit:
Since the original question was extended through an edit, I'll address that also.
If you have more variables you need more equations, for example:
$$a,b,c,d,e,f in mathbb{R}$$ $$a+b+c = d+e+f$$ $$ab + bc + ac = de + ef + df$$ $$abc=def$$
answered Jan 14 at 19:10
lightxbulblightxbulb
900211
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a+b=n=c+d$. Then $$ab=cdimplies a(n-a)=c(n-c)$$ so that $$an-a^2=cn-c^2,$$ or$$(a-c)n=a^2-c^2=(a-c)(a+c)$$
Either $a-c,$ and we are done, or $a-cne 0$ so we can cancel the $a-c$ to get $$n=a+c$$
Together with $n=aa+b$ this gives $b=c.$
In short, your assumption is correct.
answered Jan 14 at 19:00
saulspatzsaulspatz
15.3k31331
$endgroup$
$begingroup$
thanks @saulspatz .. in case where its a1 + a2 + ... +aN = b1 + b2 +b3+ ...+bN and a1* a2 * a3*... aN = b1 * b2 *b3 ...*bN how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:16
add a comment |
$begingroup$
Let $a+b=n=c+d$. Then $$ab=cdimplies a(n-a)=c(n-c)$$ so that $$an-a^2=cn-c^2,$$ or$$(a-c)n=a^2-c^2=(a-c)(a+c)$$
Either $a-c,$ and we are done, or $a-cne 0$ so we can cancel the $a-c$ to get $$n=a+c$$
Together with $n=aa+b$ this gives $b=c.$
In short, your assumption is correct.
answered Jan 14 at 19:00
saulspatzsaulspatz
15.3k31331
$endgroup$
$begingroup$
thanks @saulspatz .. in case where its a1 + a2 + ... +aN = b1 + b2 +b3+ ...+bN and a1* a2 * a3*... aN = b1 * b2 *b3 ...*bN how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:16
add a comment |
$begingroup$
Let $a+b=n=c+d$. Then $$ab=cdimplies a(n-a)=c(n-c)$$ so that $$an-a^2=cn-c^2,$$ or$$(a-c)n=a^2-c^2=(a-c)(a+c)$$
Either $a-c,$ and we are done, or $a-cne 0$ so we can cancel the $a-c$ to get $$n=a+c$$
Together with $n=aa+b$ this gives $b=c.$
In short, your assumption is correct.
answered Jan 14 at 19:00
saulspatzsaulspatz
15.3k31331
$endgroup$
Let $a+b=n=c+d$. Then $$ab=cdimplies a(n-a)=c(n-c)$$ so that $$an-a^2=cn-c^2,$$ or$$(a-c)n=a^2-c^2=(a-c)(a+c)$$
Either $a-c,$ and we are done, or $a-cne 0$ so we can cancel the $a-c$ to get $$n=a+c$$
Together with $n=aa+b$ this gives $b=c.$
In short, your assumption is correct.
answered Jan 14 at 19:00
saulspatzsaulspatz
15.3k31331
edited Jan 15 at 1:21
answered Jan 14 at 19:00
saulspatzsaulspatz
15.3k31331
answered Jan 14 at 19:00
saulspatzsaulspatz
15.3k31331
answered Jan 14 at 19:00
saulspatzsaulspatz
15.3k31331
15.3k31331
$begingroup$
thanks @saulspatz .. in case where its a1 + a2 + ... +aN = b1 + b2 +b3+ ...+bN and a1* a2 * a3*... aN = b1 * b2 *b3 ...*bN how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:16
add a comment |
$begingroup$
thanks @saulspatz .. in case where its a1 + a2 + ... +aN = b1 + b2 +b3+ ...+bN and a1* a2 * a3*... aN = b1 * b2 *b3 ...*bN how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:16
$begingroup$
thanks @saulspatz .. in case where its a1 + a2 + ... +aN = b1 + b2 +b3+ ...+bN and a1* a2 * a3*... aN = b1 * b2 *b3 ...*bN how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:16
$begingroup$
thanks @saulspatz .. in case where its a1 + a2 + ... +aN = b1 + b2 +b3+ ...+bN and a1* a2 * a3*... aN = b1 * b2 *b3 ...*bN how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:16
add a comment |
$begingroup$
Only an idea: if $$a+b=c+d$$ we get by squaring
$$a^2+b^2=c^2+d^2$$ since $$ab=cd$$ so we get
$$a^2-c^2=d^2-b^2$$ or $$a^2-d^2=b^2-c^2$$
answered Jan 14 at 18:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
add a comment |
$begingroup$
Only an idea: if $$a+b=c+d$$ we get by squaring
$$a^2+b^2=c^2+d^2$$ since $$ab=cd$$ so we get
$$a^2-c^2=d^2-b^2$$ or $$a^2-d^2=b^2-c^2$$
answered Jan 14 at 18:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
add a comment |
$begingroup$
Only an idea: if $$a+b=c+d$$ we get by squaring
$$a^2+b^2=c^2+d^2$$ since $$ab=cd$$ so we get
$$a^2-c^2=d^2-b^2$$ or $$a^2-d^2=b^2-c^2$$
answered Jan 14 at 18:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
$endgroup$
Only an idea: if $$a+b=c+d$$ we get by squaring
$$a^2+b^2=c^2+d^2$$ since $$ab=cd$$ so we get
$$a^2-c^2=d^2-b^2$$ or $$a^2-d^2=b^2-c^2$$
answered Jan 14 at 18:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
answered Jan 14 at 18:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
answered Jan 14 at 18:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
answered Jan 14 at 18:58
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.6k42866
75.6k42866
add a comment |
add a comment |
$begingroup$
This is true.
In general, $r$ and $s$ are the two roots of the quadratic equation
$$
x^2 -(r+s)x + rs = (x-r)(x-s).
$$
so knowing the sum and the product determines the values (up to swapping them).
answered Jan 14 at 19:02
Ethan BolkerEthan Bolker
43.4k551116
$endgroup$
add a comment |
$begingroup$
This is true.
In general, $r$ and $s$ are the two roots of the quadratic equation
$$
x^2 -(r+s)x + rs = (x-r)(x-s).
$$
so knowing the sum and the product determines the values (up to swapping them).
answered Jan 14 at 19:02
Ethan BolkerEthan Bolker
43.4k551116
$endgroup$
add a comment |
$begingroup$
This is true.
In general, $r$ and $s$ are the two roots of the quadratic equation
$$
x^2 -(r+s)x + rs = (x-r)(x-s).
$$
so knowing the sum and the product determines the values (up to swapping them).
answered Jan 14 at 19:02
Ethan BolkerEthan Bolker
43.4k551116
$endgroup$
This is true.
In general, $r$ and $s$ are the two roots of the quadratic equation
$$
x^2 -(r+s)x + rs = (x-r)(x-s).
$$
so knowing the sum and the product determines the values (up to swapping them).
answered Jan 14 at 19:02
Ethan BolkerEthan Bolker
43.4k551116
answered Jan 14 at 19:02
Ethan BolkerEthan Bolker
43.4k551116
answered Jan 14 at 19:02
Ethan BolkerEthan Bolker
43.4k551116
answered Jan 14 at 19:02
Ethan BolkerEthan Bolker
43.4k551116
43.4k551116
add a comment |
add a comment |
$begingroup$
Let $exists delta_1, delta_2 in mathbb{Z}$ such that $c = a + delta_1, d = b + delta_2$ and $(delta_1,delta_2) neq (0,0)$. However from $a+b=c+d$ it follows that $delta_1 = -delta_2$. Now $ab = cd = (a+delta_1)(b-delta_1)$, then $delta_1^2 + (a-b)delta_1 = delta_1(delta_1+a-b) = 0$. Since we assumed $delta_1 neq 0$ it follows that $delta_1 = b-a$, however then we get $c = b, d=a$. Thus it necessarily holds that for the given constraints either $(a,b) = (c,d)$ which corresponds to $delta_1 = 0$, or $(a,b)=(d,c)$ which corresponds to $delta_1 = b-a$. Note that this is true for $a,b,c,d in mathbb{R}$.
Edit:
Since the original question was extended through an edit, I'll address that also.
If you have more variables you need more equations, for example:
$$a,b,c,d,e,f in mathbb{R}$$ $$a+b+c = d+e+f$$ $$ab + bc + ac = de + ef + df$$ $$abc=def$$
answered Jan 14 at 19:10
lightxbulblightxbulb
900211
$endgroup$
add a comment |
$begingroup$
Let $exists delta_1, delta_2 in mathbb{Z}$ such that $c = a + delta_1, d = b + delta_2$ and $(delta_1,delta_2) neq (0,0)$. However from $a+b=c+d$ it follows that $delta_1 = -delta_2$. Now $ab = cd = (a+delta_1)(b-delta_1)$, then $delta_1^2 + (a-b)delta_1 = delta_1(delta_1+a-b) = 0$. Since we assumed $delta_1 neq 0$ it follows that $delta_1 = b-a$, however then we get $c = b, d=a$. Thus it necessarily holds that for the given constraints either $(a,b) = (c,d)$ which corresponds to $delta_1 = 0$, or $(a,b)=(d,c)$ which corresponds to $delta_1 = b-a$. Note that this is true for $a,b,c,d in mathbb{R}$.
Edit:
Since the original question was extended through an edit, I'll address that also.
If you have more variables you need more equations, for example:
$$a,b,c,d,e,f in mathbb{R}$$ $$a+b+c = d+e+f$$ $$ab + bc + ac = de + ef + df$$ $$abc=def$$
answered Jan 14 at 19:10
lightxbulblightxbulb
900211
$endgroup$
add a comment |
$begingroup$
Let $exists delta_1, delta_2 in mathbb{Z}$ such that $c = a + delta_1, d = b + delta_2$ and $(delta_1,delta_2) neq (0,0)$. However from $a+b=c+d$ it follows that $delta_1 = -delta_2$. Now $ab = cd = (a+delta_1)(b-delta_1)$, then $delta_1^2 + (a-b)delta_1 = delta_1(delta_1+a-b) = 0$. Since we assumed $delta_1 neq 0$ it follows that $delta_1 = b-a$, however then we get $c = b, d=a$. Thus it necessarily holds that for the given constraints either $(a,b) = (c,d)$ which corresponds to $delta_1 = 0$, or $(a,b)=(d,c)$ which corresponds to $delta_1 = b-a$. Note that this is true for $a,b,c,d in mathbb{R}$.
Edit:
Since the original question was extended through an edit, I'll address that also.
If you have more variables you need more equations, for example:
$$a,b,c,d,e,f in mathbb{R}$$ $$a+b+c = d+e+f$$ $$ab + bc + ac = de + ef + df$$ $$abc=def$$
answered Jan 14 at 19:10
lightxbulblightxbulb
900211
$endgroup$
Let $exists delta_1, delta_2 in mathbb{Z}$ such that $c = a + delta_1, d = b + delta_2$ and $(delta_1,delta_2) neq (0,0)$. However from $a+b=c+d$ it follows that $delta_1 = -delta_2$. Now $ab = cd = (a+delta_1)(b-delta_1)$, then $delta_1^2 + (a-b)delta_1 = delta_1(delta_1+a-b) = 0$. Since we assumed $delta_1 neq 0$ it follows that $delta_1 = b-a$, however then we get $c = b, d=a$. Thus it necessarily holds that for the given constraints either $(a,b) = (c,d)$ which corresponds to $delta_1 = 0$, or $(a,b)=(d,c)$ which corresponds to $delta_1 = b-a$. Note that this is true for $a,b,c,d in mathbb{R}$.
Edit:
Since the original question was extended through an edit, I'll address that also.
If you have more variables you need more equations, for example:
$$a,b,c,d,e,f in mathbb{R}$$ $$a+b+c = d+e+f$$ $$ab + bc + ac = de + ef + df$$ $$abc=def$$
answered Jan 14 at 19:10
lightxbulblightxbulb
900211
edited Jan 15 at 10:06
answered Jan 14 at 19:10
lightxbulblightxbulb
900211
answered Jan 14 at 19:10
lightxbulblightxbulb
900211
answered Jan 14 at 19:10
lightxbulblightxbulb
900211
900211
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$begingroup$
What do you mean by "Is this also true for pair of N numbers ?"? What constraints do you have in mind?
$endgroup$
– lightxbulb
Jan 14 at 19:14
$begingroup$
in this case its just 2 numbers a + b = c+ d ... in case where its a1 + a2 + ... +an = b1 + b2 +b3 ...bn how can the proof be extended ?
$endgroup$
– rahulb
Jan 15 at 1:15
$begingroup$
I updated my answer to give an example with more values. Also can you explain what your algorithm aims to achieve? @rahulb
$endgroup$
– lightxbulb
Jan 15 at 10:06