Expanding $frac{a}{a+b}$ into two terms
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I'm unsure how I arrived at this assumption but I assumed :
$$
frac{a}{a+b} = frac{a}{a} + frac{a}{b}
$$
testing with values $a = 3$ and $b = 4$ this is not true as
$$
frac{3}{3+4} neq frac{3}{3} + frac{3}{4}
$$
Can $frac{a}{a+b}$ be expanded so that $a$ exclusively is contained in its own term and $b$ is contained in its own term.
algebra-precalculus fractions
edited Nov 23 at 12:00
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Nov 23 at 11:56
blue-sky
97011222
add a comment |
up vote
0
down vote
favorite
I'm unsure how I arrived at this assumption but I assumed :
$$
frac{a}{a+b} = frac{a}{a} + frac{a}{b}
$$
testing with values $a = 3$ and $b = 4$ this is not true as
$$
frac{3}{3+4} neq frac{3}{3} + frac{3}{4}
$$
Can $frac{a}{a+b}$ be expanded so that $a$ exclusively is contained in its own term and $b$ is contained in its own term.
algebra-precalculus fractions
edited Nov 23 at 12:00
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Nov 23 at 11:56
blue-sky
97011222
1
Well, if you assume false statements, you're going to run into more false statements..
– DaveNine
Nov 23 at 11:59
For future reference, for similar problems, one hint is to think about when the expression is defined. In order for two expressions to be the same, their domains must be the same. In this case it means that the cases where the denominator is zero, have to be the same for both sides. I at least find this hint quite useful.
– Matti P.
Nov 23 at 12:04
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm unsure how I arrived at this assumption but I assumed :
$$
frac{a}{a+b} = frac{a}{a} + frac{a}{b}
$$
testing with values $a = 3$ and $b = 4$ this is not true as
$$
frac{3}{3+4} neq frac{3}{3} + frac{3}{4}
$$
Can $frac{a}{a+b}$ be expanded so that $a$ exclusively is contained in its own term and $b$ is contained in its own term.
algebra-precalculus fractions
edited Nov 23 at 12:00
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Nov 23 at 11:56
blue-sky
97011222
I'm unsure how I arrived at this assumption but I assumed :
$$
frac{a}{a+b} = frac{a}{a} + frac{a}{b}
$$
testing with values $a = 3$ and $b = 4$ this is not true as
$$
frac{3}{3+4} neq frac{3}{3} + frac{3}{4}
$$
Can $frac{a}{a+b}$ be expanded so that $a$ exclusively is contained in its own term and $b$ is contained in its own term.
algebra-precalculus fractions
algebra-precalculus fractions
edited Nov 23 at 12:00
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Nov 23 at 11:56
blue-sky
97011222
edited Nov 23 at 12:00
GNUSupporter 8964民主女神 地下教會
12.6k72445
asked Nov 23 at 11:56
blue-sky
97011222
edited Nov 23 at 12:00
GNUSupporter 8964民主女神 地下教會
12.6k72445
edited Nov 23 at 12:00
GNUSupporter 8964民主女神 地下教會
12.6k72445
edited Nov 23 at 12:00
GNUSupporter 8964民主女神 地下教會
12.6k72445
12.6k72445
asked Nov 23 at 11:56
blue-sky
97011222
asked Nov 23 at 11:56
blue-sky
97011222
asked Nov 23 at 11:56
blue-sky
97011222
97011222
1
Well, if you assume false statements, you're going to run into more false statements..
– DaveNine
Nov 23 at 11:59
For future reference, for similar problems, one hint is to think about when the expression is defined. In order for two expressions to be the same, their domains must be the same. In this case it means that the cases where the denominator is zero, have to be the same for both sides. I at least find this hint quite useful.
– Matti P.
Nov 23 at 12:04
add a comment |
1
Well, if you assume false statements, you're going to run into more false statements..
– DaveNine
Nov 23 at 11:59
For future reference, for similar problems, one hint is to think about when the expression is defined. In order for two expressions to be the same, their domains must be the same. In this case it means that the cases where the denominator is zero, have to be the same for both sides. I at least find this hint quite useful.
– Matti P.
Nov 23 at 12:04
1
1
Well, if you assume false statements, you're going to run into more false statements..
– DaveNine
Nov 23 at 11:59
Well, if you assume false statements, you're going to run into more false statements..
– DaveNine
Nov 23 at 11:59
For future reference, for similar problems, one hint is to think about when the expression is defined. In order for two expressions to be the same, their domains must be the same. In this case it means that the cases where the denominator is zero, have to be the same for both sides. I at least find this hint quite useful.
– Matti P.
Nov 23 at 12:04
For future reference, for similar problems, one hint is to think about when the expression is defined. In order for two expressions to be the same, their domains must be the same. In this case it means that the cases where the denominator is zero, have to be the same for both sides. I at least find this hint quite useful.
– Matti P.
Nov 23 at 12:04
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
Your assumption, as you note, is not true at all. In particular, if you add those two fractions together and simplify a little, you get $frac{a + b}{b}$. No such decomposition is possible.
In particular, you can never hope to express $frac{a}{a+b}$ (or any multiple thereof as a sum of something with an $a$ in the denominator and something with a $b$ in the denominator, because $a+b$ is not a factor of $ab$ in general, but $frac{c}{a} +frac{d}{b} = frac{bc+ad}{ab}$, so the denominator of it must be some factor of $ab$.
answered Nov 23 at 12:02
user3482749
2,032413
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Your assumption, as you note, is not true at all. In particular, if you add those two fractions together and simplify a little, you get $frac{a + b}{b}$. No such decomposition is possible.
In particular, you can never hope to express $frac{a}{a+b}$ (or any multiple thereof as a sum of something with an $a$ in the denominator and something with a $b$ in the denominator, because $a+b$ is not a factor of $ab$ in general, but $frac{c}{a} +frac{d}{b} = frac{bc+ad}{ab}$, so the denominator of it must be some factor of $ab$.
answered Nov 23 at 12:02
user3482749
2,032413
add a comment |
up vote
2
down vote
Your assumption, as you note, is not true at all. In particular, if you add those two fractions together and simplify a little, you get $frac{a + b}{b}$. No such decomposition is possible.
In particular, you can never hope to express $frac{a}{a+b}$ (or any multiple thereof as a sum of something with an $a$ in the denominator and something with a $b$ in the denominator, because $a+b$ is not a factor of $ab$ in general, but $frac{c}{a} +frac{d}{b} = frac{bc+ad}{ab}$, so the denominator of it must be some factor of $ab$.
answered Nov 23 at 12:02
user3482749
2,032413
add a comment |
up vote
2
down vote
up vote
2
down vote
Your assumption, as you note, is not true at all. In particular, if you add those two fractions together and simplify a little, you get $frac{a + b}{b}$. No such decomposition is possible.
In particular, you can never hope to express $frac{a}{a+b}$ (or any multiple thereof as a sum of something with an $a$ in the denominator and something with a $b$ in the denominator, because $a+b$ is not a factor of $ab$ in general, but $frac{c}{a} +frac{d}{b} = frac{bc+ad}{ab}$, so the denominator of it must be some factor of $ab$.
answered Nov 23 at 12:02
user3482749
2,032413
Your assumption, as you note, is not true at all. In particular, if you add those two fractions together and simplify a little, you get $frac{a + b}{b}$. No such decomposition is possible.
In particular, you can never hope to express $frac{a}{a+b}$ (or any multiple thereof as a sum of something with an $a$ in the denominator and something with a $b$ in the denominator, because $a+b$ is not a factor of $ab$ in general, but $frac{c}{a} +frac{d}{b} = frac{bc+ad}{ab}$, so the denominator of it must be some factor of $ab$.
answered Nov 23 at 12:02
user3482749
2,032413
answered Nov 23 at 12:02
user3482749
2,032413
answered Nov 23 at 12:02
user3482749
2,032413
answered Nov 23 at 12:02
user3482749
2,032413
2,032413
add a comment |
add a comment |
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Well, if you assume false statements, you're going to run into more false statements..
– DaveNine
Nov 23 at 11:59
For future reference, for similar problems, one hint is to think about when the expression is defined. In order for two expressions to be the same, their domains must be the same. In this case it means that the cases where the denominator is zero, have to be the same for both sides. I at least find this hint quite useful.
– Matti P.
Nov 23 at 12:04