If the quantity $P=frac{1}{a}+frac{1}{b} $ is conserved for different combinations of the two variables, is...
$begingroup$
If the quantity $P=frac{1}{a}+frac{1}{b} $ must always be the same for different combinations of the values of the two variables, then is it correct to state that the quantity $Q=a+b$ is also conserved?
linear-algebra algebra-precalculus
edited Dec 30 '18 at 3:44
Martund
1,415212
asked Dec 30 '18 at 2:17
Gonzalo FerrandoGonzalo Ferrando
111
$endgroup$
add a comment |
$begingroup$
If the quantity $P=frac{1}{a}+frac{1}{b} $ must always be the same for different combinations of the values of the two variables, then is it correct to state that the quantity $Q=a+b$ is also conserved?
linear-algebra algebra-precalculus
edited Dec 30 '18 at 3:44
Martund
1,415212
asked Dec 30 '18 at 2:17
Gonzalo FerrandoGonzalo Ferrando
111
$endgroup$
add a comment |
$begingroup$
If the quantity $P=frac{1}{a}+frac{1}{b} $ must always be the same for different combinations of the values of the two variables, then is it correct to state that the quantity $Q=a+b$ is also conserved?
linear-algebra algebra-precalculus
edited Dec 30 '18 at 3:44
Martund
1,415212
asked Dec 30 '18 at 2:17
Gonzalo FerrandoGonzalo Ferrando
111
$endgroup$
If the quantity $P=frac{1}{a}+frac{1}{b} $ must always be the same for different combinations of the values of the two variables, then is it correct to state that the quantity $Q=a+b$ is also conserved?
linear-algebra algebra-precalculus
linear-algebra algebra-precalculus
edited Dec 30 '18 at 3:44
Martund
1,415212
asked Dec 30 '18 at 2:17
Gonzalo FerrandoGonzalo Ferrando
111
edited Dec 30 '18 at 3:44
Martund
1,415212
asked Dec 30 '18 at 2:17
Gonzalo FerrandoGonzalo Ferrando
111
edited Dec 30 '18 at 3:44
Martund
1,415212
edited Dec 30 '18 at 3:44
Martund
1,415212
edited Dec 30 '18 at 3:44
Martund
1,415212
1,415212
asked Dec 30 '18 at 2:17
Gonzalo FerrandoGonzalo Ferrando
111
asked Dec 30 '18 at 2:17
Gonzalo FerrandoGonzalo Ferrando
111
asked Dec 30 '18 at 2:17
Gonzalo FerrandoGonzalo Ferrando
111
111
add a comment |
add a comment |
3 Answers
3
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oldest
votes
$begingroup$
No. For example, take the pairs $(a, b)=(1,1)$ and $(a,b)=(2,frac 2 3)$. Both pairs satisfy $P=frac{1}{a}+frac{1}{b}=2$, but for the first pair $Q=1+1=2$ while for the second pair, $Q=2+frac 2 3=frac 8 3$.
answered Dec 30 '18 at 2:20
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
add a comment |
$begingroup$
No, suppose that it is conserved. Then, $$ab=frac{Q}{P}$$
is also conserved. But, in that case, $a$ and $b$ are also fixed, being the roots of $$x^2-(a+b)x+ab=0$$
i.e. $$x^2-Qx+frac{Q}{P}=0$$
Hope it is helpful
answered Dec 30 '18 at 3:39
MartundMartund
1,415212
$endgroup$
add a comment |
$begingroup$
Let the quantity
$$frac{1}{a}+frac{1}{b}=frac{a+b}{ab}=X$$
Then
$$a+b=abX$$
If $a+b$ is to be conserved, then we must have
$$a+b=abX:=-BRightarrow ab=frac{-B}{X}:=C$$
And so $ab$ is also conserved.
Thus $a$ and $b$ are roots of the quadratic equation
$$x^2+Bx+C=0$$
where $B$ and $C$ are fixed constants.
Thus $a$ and $b$ are fixed numbers and not variable. We have a contradiction.
answered Dec 30 '18 at 4:39
Hugh EntwistleHugh Entwistle
841217
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No. For example, take the pairs $(a, b)=(1,1)$ and $(a,b)=(2,frac 2 3)$. Both pairs satisfy $P=frac{1}{a}+frac{1}{b}=2$, but for the first pair $Q=1+1=2$ while for the second pair, $Q=2+frac 2 3=frac 8 3$.
answered Dec 30 '18 at 2:20
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
add a comment |
$begingroup$
No. For example, take the pairs $(a, b)=(1,1)$ and $(a,b)=(2,frac 2 3)$. Both pairs satisfy $P=frac{1}{a}+frac{1}{b}=2$, but for the first pair $Q=1+1=2$ while for the second pair, $Q=2+frac 2 3=frac 8 3$.
answered Dec 30 '18 at 2:20
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
add a comment |
$begingroup$
No. For example, take the pairs $(a, b)=(1,1)$ and $(a,b)=(2,frac 2 3)$. Both pairs satisfy $P=frac{1}{a}+frac{1}{b}=2$, but for the first pair $Q=1+1=2$ while for the second pair, $Q=2+frac 2 3=frac 8 3$.
answered Dec 30 '18 at 2:20
Noble MushtakNoble Mushtak
15.2k1735
$endgroup$
No. For example, take the pairs $(a, b)=(1,1)$ and $(a,b)=(2,frac 2 3)$. Both pairs satisfy $P=frac{1}{a}+frac{1}{b}=2$, but for the first pair $Q=1+1=2$ while for the second pair, $Q=2+frac 2 3=frac 8 3$.
answered Dec 30 '18 at 2:20
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 30 '18 at 2:20
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 30 '18 at 2:20
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 30 '18 at 2:20
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
add a comment |
add a comment |
$begingroup$
No, suppose that it is conserved. Then, $$ab=frac{Q}{P}$$
is also conserved. But, in that case, $a$ and $b$ are also fixed, being the roots of $$x^2-(a+b)x+ab=0$$
i.e. $$x^2-Qx+frac{Q}{P}=0$$
Hope it is helpful
answered Dec 30 '18 at 3:39
MartundMartund
1,415212
$endgroup$
add a comment |
$begingroup$
No, suppose that it is conserved. Then, $$ab=frac{Q}{P}$$
is also conserved. But, in that case, $a$ and $b$ are also fixed, being the roots of $$x^2-(a+b)x+ab=0$$
i.e. $$x^2-Qx+frac{Q}{P}=0$$
Hope it is helpful
answered Dec 30 '18 at 3:39
MartundMartund
1,415212
$endgroup$
add a comment |
$begingroup$
No, suppose that it is conserved. Then, $$ab=frac{Q}{P}$$
is also conserved. But, in that case, $a$ and $b$ are also fixed, being the roots of $$x^2-(a+b)x+ab=0$$
i.e. $$x^2-Qx+frac{Q}{P}=0$$
Hope it is helpful
answered Dec 30 '18 at 3:39
MartundMartund
1,415212
$endgroup$
No, suppose that it is conserved. Then, $$ab=frac{Q}{P}$$
is also conserved. But, in that case, $a$ and $b$ are also fixed, being the roots of $$x^2-(a+b)x+ab=0$$
i.e. $$x^2-Qx+frac{Q}{P}=0$$
Hope it is helpful
answered Dec 30 '18 at 3:39
MartundMartund
1,415212
answered Dec 30 '18 at 3:39
MartundMartund
1,415212
answered Dec 30 '18 at 3:39
MartundMartund
1,415212
answered Dec 30 '18 at 3:39
MartundMartund
1,415212
1,415212
add a comment |
add a comment |
$begingroup$
Let the quantity
$$frac{1}{a}+frac{1}{b}=frac{a+b}{ab}=X$$
Then
$$a+b=abX$$
If $a+b$ is to be conserved, then we must have
$$a+b=abX:=-BRightarrow ab=frac{-B}{X}:=C$$
And so $ab$ is also conserved.
Thus $a$ and $b$ are roots of the quadratic equation
$$x^2+Bx+C=0$$
where $B$ and $C$ are fixed constants.
Thus $a$ and $b$ are fixed numbers and not variable. We have a contradiction.
answered Dec 30 '18 at 4:39
Hugh EntwistleHugh Entwistle
841217
$endgroup$
add a comment |
$begingroup$
Let the quantity
$$frac{1}{a}+frac{1}{b}=frac{a+b}{ab}=X$$
Then
$$a+b=abX$$
If $a+b$ is to be conserved, then we must have
$$a+b=abX:=-BRightarrow ab=frac{-B}{X}:=C$$
And so $ab$ is also conserved.
Thus $a$ and $b$ are roots of the quadratic equation
$$x^2+Bx+C=0$$
where $B$ and $C$ are fixed constants.
Thus $a$ and $b$ are fixed numbers and not variable. We have a contradiction.
answered Dec 30 '18 at 4:39
Hugh EntwistleHugh Entwistle
841217
$endgroup$
add a comment |
$begingroup$
Let the quantity
$$frac{1}{a}+frac{1}{b}=frac{a+b}{ab}=X$$
Then
$$a+b=abX$$
If $a+b$ is to be conserved, then we must have
$$a+b=abX:=-BRightarrow ab=frac{-B}{X}:=C$$
And so $ab$ is also conserved.
Thus $a$ and $b$ are roots of the quadratic equation
$$x^2+Bx+C=0$$
where $B$ and $C$ are fixed constants.
Thus $a$ and $b$ are fixed numbers and not variable. We have a contradiction.
answered Dec 30 '18 at 4:39
Hugh EntwistleHugh Entwistle
841217
$endgroup$
Let the quantity
$$frac{1}{a}+frac{1}{b}=frac{a+b}{ab}=X$$
Then
$$a+b=abX$$
If $a+b$ is to be conserved, then we must have
$$a+b=abX:=-BRightarrow ab=frac{-B}{X}:=C$$
And so $ab$ is also conserved.
Thus $a$ and $b$ are roots of the quadratic equation
$$x^2+Bx+C=0$$
where $B$ and $C$ are fixed constants.
Thus $a$ and $b$ are fixed numbers and not variable. We have a contradiction.
answered Dec 30 '18 at 4:39
Hugh EntwistleHugh Entwistle
841217
answered Dec 30 '18 at 4:39
Hugh EntwistleHugh Entwistle
841217
answered Dec 30 '18 at 4:39
Hugh EntwistleHugh Entwistle
841217
answered Dec 30 '18 at 4:39
Hugh EntwistleHugh Entwistle
841217
841217
add a comment |
add a comment |
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