Percentual change from 0
up vote
0
down vote
favorite
How to calculate the procentual change with respect to 0.
For example: The old value is 0, while the new value is 10.
How to calculate the procentual change in this case?
average
edited Dec 6 at 9:15
nippon
335212
asked Dec 6 at 7:46
Abhijit Bendigiri
71
add a comment |
up vote
0
down vote
favorite
How to calculate the procentual change with respect to 0.
For example: The old value is 0, while the new value is 10.
How to calculate the procentual change in this case?
average
edited Dec 6 at 9:15
nippon
335212
asked Dec 6 at 7:46
Abhijit Bendigiri
71
Isn't it always $100$% since the previous number is $0$? Edit: Oh, too late
– NotEinstein
Dec 6 at 7:50
1
It's undefined. There can be no relative change from $0$. The absolute change is $10$ ($10-0 = 10$).
– Eff
Dec 6 at 8:31
(I forgot about the initial value in my previous comment, so here’s the correction.) $frac{v_2-v_1}{v_1}$ is undefined for $v_1 = 0$. If you have $v_1 = 2$ and $v_2 = 3$, for instance, you get a percentage change of $50$%. In other words, $2$ added to $50$% of itself gives $3$. Can $0$ added to any percentage of itself give a non-zero number?
– KM101
Dec 6 at 8:41
The "percentage change" number isn't that useful when your number is hovering around zero anyway.
– James
Dec 6 at 13:44
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to calculate the procentual change with respect to 0.
For example: The old value is 0, while the new value is 10.
How to calculate the procentual change in this case?
average
edited Dec 6 at 9:15
nippon
335212
asked Dec 6 at 7:46
Abhijit Bendigiri
71
How to calculate the procentual change with respect to 0.
For example: The old value is 0, while the new value is 10.
How to calculate the procentual change in this case?
average
average
edited Dec 6 at 9:15
nippon
335212
asked Dec 6 at 7:46
Abhijit Bendigiri
71
edited Dec 6 at 9:15
nippon
335212
asked Dec 6 at 7:46
Abhijit Bendigiri
71
edited Dec 6 at 9:15
nippon
335212
edited Dec 6 at 9:15
nippon
335212
edited Dec 6 at 9:15
nippon
335212
335212
asked Dec 6 at 7:46
Abhijit Bendigiri
71
asked Dec 6 at 7:46
Abhijit Bendigiri
71
asked Dec 6 at 7:46
Abhijit Bendigiri
71
71
Isn't it always $100$% since the previous number is $0$? Edit: Oh, too late
– NotEinstein
Dec 6 at 7:50
1
It's undefined. There can be no relative change from $0$. The absolute change is $10$ ($10-0 = 10$).
– Eff
Dec 6 at 8:31
(I forgot about the initial value in my previous comment, so here’s the correction.) $frac{v_2-v_1}{v_1}$ is undefined for $v_1 = 0$. If you have $v_1 = 2$ and $v_2 = 3$, for instance, you get a percentage change of $50$%. In other words, $2$ added to $50$% of itself gives $3$. Can $0$ added to any percentage of itself give a non-zero number?
– KM101
Dec 6 at 8:41
The "percentage change" number isn't that useful when your number is hovering around zero anyway.
– James
Dec 6 at 13:44
add a comment |
Isn't it always $100$% since the previous number is $0$? Edit: Oh, too late
– NotEinstein
Dec 6 at 7:50
1
It's undefined. There can be no relative change from $0$. The absolute change is $10$ ($10-0 = 10$).
– Eff
Dec 6 at 8:31
(I forgot about the initial value in my previous comment, so here’s the correction.) $frac{v_2-v_1}{v_1}$ is undefined for $v_1 = 0$. If you have $v_1 = 2$ and $v_2 = 3$, for instance, you get a percentage change of $50$%. In other words, $2$ added to $50$% of itself gives $3$. Can $0$ added to any percentage of itself give a non-zero number?
– KM101
Dec 6 at 8:41
The "percentage change" number isn't that useful when your number is hovering around zero anyway.
– James
Dec 6 at 13:44
Isn't it always $100$% since the previous number is $0$? Edit: Oh, too late
– NotEinstein
Dec 6 at 7:50
Isn't it always $100$% since the previous number is $0$? Edit: Oh, too late
– NotEinstein
Dec 6 at 7:50
1
1
It's undefined. There can be no relative change from $0$. The absolute change is $10$ ($10-0 = 10$).
– Eff
Dec 6 at 8:31
It's undefined. There can be no relative change from $0$. The absolute change is $10$ ($10-0 = 10$).
– Eff
Dec 6 at 8:31
(I forgot about the initial value in my previous comment, so here’s the correction.) $frac{v_2-v_1}{v_1}$ is undefined for $v_1 = 0$. If you have $v_1 = 2$ and $v_2 = 3$, for instance, you get a percentage change of $50$%. In other words, $2$ added to $50$% of itself gives $3$. Can $0$ added to any percentage of itself give a non-zero number?
– KM101
Dec 6 at 8:41
(I forgot about the initial value in my previous comment, so here’s the correction.) $frac{v_2-v_1}{v_1}$ is undefined for $v_1 = 0$. If you have $v_1 = 2$ and $v_2 = 3$, for instance, you get a percentage change of $50$%. In other words, $2$ added to $50$% of itself gives $3$. Can $0$ added to any percentage of itself give a non-zero number?
– KM101
Dec 6 at 8:41
The "percentage change" number isn't that useful when your number is hovering around zero anyway.
– James
Dec 6 at 13:44
The "percentage change" number isn't that useful when your number is hovering around zero anyway.
– James
Dec 6 at 13:44
add a comment |
2 Answers
2
active
oldest
votes
up vote
5
down vote
When the previous value $x$ is non-zero, the formula is $$frac{y-x}{x} times 100 %$$
Now, remember that we are not supposed to divide by $0$.
It should be undefined.
If it is defined and we claim that it is $r$, a real number, given the percentage change and the current value, we can't recover $x$ to be $0$ and we might misguide ourselves to think that $x$ is some non-zero number.
$$frac{y}x =1+frac{r}{100}$$
$$x=frac{y}{1+frac{r}{100}}$$
If you have to put down something, perhaps put down a symbol to indicate that the previous number is zero.
answered Dec 6 at 7:55
Siong Thye Goh
97.4k1463116
"perhaps put down a symbol to..." - and I'd suggest to use the symbol ∞. Just remember that this is not a number, it's just a symbol that indicated the previous number is zero. Which should be clear to a mathematician, but perhaps not to a casual reader.
– Radovan Garabík
Dec 6 at 14:27
add a comment |
up vote
4
down vote
While the other answer by Siong Thye Goh is right, I want to present a slightly more intuitive approach.
The "percent change" of $x$ to $y$ is essentially saying, for some number $p$,
$$x + (p% text{ of } x) = y$$
For example, the percentage change of $2$ to $5$ can be calculated via the method Siong mentioned:
$$frac{5-2}{2} times 100% = frac{3}{2} times 100% = 150%$$
We can equivalently state what this means by
$$2 + (150% text{ of } 2) = 5$$
Since $150%$ of $2$ is given by $150% cdot 2 = 1.5 cdot 2 = 3$, this idea becomes somewhat clear, this idea being that
$$x + (p% text{ of } x) = y ;;; Leftrightarrow ;;; frac{y-x}{x} = p$$
i.e. this intuitive approach is analogous to the formula you usually get told a la what Siong noted.
So we ask: what if $x = 0$, and $y neq 0$? Well, we have
$$0 + (p% text{ of } 0) = y neq 0$$
Okay, but $0$ plus anything is just that thing, and any percentage of $0$ is $0$, and $0+0=0$. Thus, we get
$$0 = y neq 0$$
i.e. a nonsensical answer, and our $p$ is irrelevant here. No matter what $p$ is, there is always this contradiction. Similarly, if $y=0$, then we get a true statement instead but it's true for all possible values of $p$ instead.
So in that sense, we say that the percentage change starting from $0$ is nonsensical. Either you have a contradictory answer or just a plain true statement, regardless of your $p$ - $p$ becomes irrelevant in both cases. So it only makes sense to say that the percentage change $p$ is undefined.
answered Dec 6 at 8:28
Eevee Trainer
2,639221
add a comment |
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2 Answers
2
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2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
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up vote
5
down vote
When the previous value $x$ is non-zero, the formula is $$frac{y-x}{x} times 100 %$$
Now, remember that we are not supposed to divide by $0$.
It should be undefined.
If it is defined and we claim that it is $r$, a real number, given the percentage change and the current value, we can't recover $x$ to be $0$ and we might misguide ourselves to think that $x$ is some non-zero number.
$$frac{y}x =1+frac{r}{100}$$
$$x=frac{y}{1+frac{r}{100}}$$
If you have to put down something, perhaps put down a symbol to indicate that the previous number is zero.
answered Dec 6 at 7:55
Siong Thye Goh
97.4k1463116
"perhaps put down a symbol to..." - and I'd suggest to use the symbol ∞. Just remember that this is not a number, it's just a symbol that indicated the previous number is zero. Which should be clear to a mathematician, but perhaps not to a casual reader.
– Radovan Garabík
Dec 6 at 14:27
add a comment |
up vote
5
down vote
When the previous value $x$ is non-zero, the formula is $$frac{y-x}{x} times 100 %$$
Now, remember that we are not supposed to divide by $0$.
It should be undefined.
If it is defined and we claim that it is $r$, a real number, given the percentage change and the current value, we can't recover $x$ to be $0$ and we might misguide ourselves to think that $x$ is some non-zero number.
$$frac{y}x =1+frac{r}{100}$$
$$x=frac{y}{1+frac{r}{100}}$$
If you have to put down something, perhaps put down a symbol to indicate that the previous number is zero.
answered Dec 6 at 7:55
Siong Thye Goh
97.4k1463116
"perhaps put down a symbol to..." - and I'd suggest to use the symbol ∞. Just remember that this is not a number, it's just a symbol that indicated the previous number is zero. Which should be clear to a mathematician, but perhaps not to a casual reader.
– Radovan Garabík
Dec 6 at 14:27
add a comment |
up vote
5
down vote
up vote
5
down vote
When the previous value $x$ is non-zero, the formula is $$frac{y-x}{x} times 100 %$$
Now, remember that we are not supposed to divide by $0$.
It should be undefined.
If it is defined and we claim that it is $r$, a real number, given the percentage change and the current value, we can't recover $x$ to be $0$ and we might misguide ourselves to think that $x$ is some non-zero number.
$$frac{y}x =1+frac{r}{100}$$
$$x=frac{y}{1+frac{r}{100}}$$
If you have to put down something, perhaps put down a symbol to indicate that the previous number is zero.
answered Dec 6 at 7:55
Siong Thye Goh
97.4k1463116
When the previous value $x$ is non-zero, the formula is $$frac{y-x}{x} times 100 %$$
Now, remember that we are not supposed to divide by $0$.
It should be undefined.
If it is defined and we claim that it is $r$, a real number, given the percentage change and the current value, we can't recover $x$ to be $0$ and we might misguide ourselves to think that $x$ is some non-zero number.
$$frac{y}x =1+frac{r}{100}$$
$$x=frac{y}{1+frac{r}{100}}$$
If you have to put down something, perhaps put down a symbol to indicate that the previous number is zero.
answered Dec 6 at 7:55
Siong Thye Goh
97.4k1463116
answered Dec 6 at 7:55
Siong Thye Goh
97.4k1463116
answered Dec 6 at 7:55
Siong Thye Goh
97.4k1463116
answered Dec 6 at 7:55
Siong Thye Goh
97.4k1463116
97.4k1463116
"perhaps put down a symbol to..." - and I'd suggest to use the symbol ∞. Just remember that this is not a number, it's just a symbol that indicated the previous number is zero. Which should be clear to a mathematician, but perhaps not to a casual reader.
– Radovan Garabík
Dec 6 at 14:27
add a comment |
"perhaps put down a symbol to..." - and I'd suggest to use the symbol ∞. Just remember that this is not a number, it's just a symbol that indicated the previous number is zero. Which should be clear to a mathematician, but perhaps not to a casual reader.
– Radovan Garabík
Dec 6 at 14:27
"perhaps put down a symbol to..." - and I'd suggest to use the symbol ∞. Just remember that this is not a number, it's just a symbol that indicated the previous number is zero. Which should be clear to a mathematician, but perhaps not to a casual reader.
– Radovan Garabík
Dec 6 at 14:27
"perhaps put down a symbol to..." - and I'd suggest to use the symbol ∞. Just remember that this is not a number, it's just a symbol that indicated the previous number is zero. Which should be clear to a mathematician, but perhaps not to a casual reader.
– Radovan Garabík
Dec 6 at 14:27
add a comment |
up vote
4
down vote
While the other answer by Siong Thye Goh is right, I want to present a slightly more intuitive approach.
The "percent change" of $x$ to $y$ is essentially saying, for some number $p$,
$$x + (p% text{ of } x) = y$$
For example, the percentage change of $2$ to $5$ can be calculated via the method Siong mentioned:
$$frac{5-2}{2} times 100% = frac{3}{2} times 100% = 150%$$
We can equivalently state what this means by
$$2 + (150% text{ of } 2) = 5$$
Since $150%$ of $2$ is given by $150% cdot 2 = 1.5 cdot 2 = 3$, this idea becomes somewhat clear, this idea being that
$$x + (p% text{ of } x) = y ;;; Leftrightarrow ;;; frac{y-x}{x} = p$$
i.e. this intuitive approach is analogous to the formula you usually get told a la what Siong noted.
So we ask: what if $x = 0$, and $y neq 0$? Well, we have
$$0 + (p% text{ of } 0) = y neq 0$$
Okay, but $0$ plus anything is just that thing, and any percentage of $0$ is $0$, and $0+0=0$. Thus, we get
$$0 = y neq 0$$
i.e. a nonsensical answer, and our $p$ is irrelevant here. No matter what $p$ is, there is always this contradiction. Similarly, if $y=0$, then we get a true statement instead but it's true for all possible values of $p$ instead.
So in that sense, we say that the percentage change starting from $0$ is nonsensical. Either you have a contradictory answer or just a plain true statement, regardless of your $p$ - $p$ becomes irrelevant in both cases. So it only makes sense to say that the percentage change $p$ is undefined.
answered Dec 6 at 8:28
Eevee Trainer
2,639221
add a comment |
up vote
4
down vote
While the other answer by Siong Thye Goh is right, I want to present a slightly more intuitive approach.
The "percent change" of $x$ to $y$ is essentially saying, for some number $p$,
$$x + (p% text{ of } x) = y$$
For example, the percentage change of $2$ to $5$ can be calculated via the method Siong mentioned:
$$frac{5-2}{2} times 100% = frac{3}{2} times 100% = 150%$$
We can equivalently state what this means by
$$2 + (150% text{ of } 2) = 5$$
Since $150%$ of $2$ is given by $150% cdot 2 = 1.5 cdot 2 = 3$, this idea becomes somewhat clear, this idea being that
$$x + (p% text{ of } x) = y ;;; Leftrightarrow ;;; frac{y-x}{x} = p$$
i.e. this intuitive approach is analogous to the formula you usually get told a la what Siong noted.
So we ask: what if $x = 0$, and $y neq 0$? Well, we have
$$0 + (p% text{ of } 0) = y neq 0$$
Okay, but $0$ plus anything is just that thing, and any percentage of $0$ is $0$, and $0+0=0$. Thus, we get
$$0 = y neq 0$$
i.e. a nonsensical answer, and our $p$ is irrelevant here. No matter what $p$ is, there is always this contradiction. Similarly, if $y=0$, then we get a true statement instead but it's true for all possible values of $p$ instead.
So in that sense, we say that the percentage change starting from $0$ is nonsensical. Either you have a contradictory answer or just a plain true statement, regardless of your $p$ - $p$ becomes irrelevant in both cases. So it only makes sense to say that the percentage change $p$ is undefined.
answered Dec 6 at 8:28
Eevee Trainer
2,639221
add a comment |
up vote
4
down vote
up vote
4
down vote
While the other answer by Siong Thye Goh is right, I want to present a slightly more intuitive approach.
The "percent change" of $x$ to $y$ is essentially saying, for some number $p$,
$$x + (p% text{ of } x) = y$$
For example, the percentage change of $2$ to $5$ can be calculated via the method Siong mentioned:
$$frac{5-2}{2} times 100% = frac{3}{2} times 100% = 150%$$
We can equivalently state what this means by
$$2 + (150% text{ of } 2) = 5$$
Since $150%$ of $2$ is given by $150% cdot 2 = 1.5 cdot 2 = 3$, this idea becomes somewhat clear, this idea being that
$$x + (p% text{ of } x) = y ;;; Leftrightarrow ;;; frac{y-x}{x} = p$$
i.e. this intuitive approach is analogous to the formula you usually get told a la what Siong noted.
So we ask: what if $x = 0$, and $y neq 0$? Well, we have
$$0 + (p% text{ of } 0) = y neq 0$$
Okay, but $0$ plus anything is just that thing, and any percentage of $0$ is $0$, and $0+0=0$. Thus, we get
$$0 = y neq 0$$
i.e. a nonsensical answer, and our $p$ is irrelevant here. No matter what $p$ is, there is always this contradiction. Similarly, if $y=0$, then we get a true statement instead but it's true for all possible values of $p$ instead.
So in that sense, we say that the percentage change starting from $0$ is nonsensical. Either you have a contradictory answer or just a plain true statement, regardless of your $p$ - $p$ becomes irrelevant in both cases. So it only makes sense to say that the percentage change $p$ is undefined.
answered Dec 6 at 8:28
Eevee Trainer
2,639221
While the other answer by Siong Thye Goh is right, I want to present a slightly more intuitive approach.
The "percent change" of $x$ to $y$ is essentially saying, for some number $p$,
$$x + (p% text{ of } x) = y$$
For example, the percentage change of $2$ to $5$ can be calculated via the method Siong mentioned:
$$frac{5-2}{2} times 100% = frac{3}{2} times 100% = 150%$$
We can equivalently state what this means by
$$2 + (150% text{ of } 2) = 5$$
Since $150%$ of $2$ is given by $150% cdot 2 = 1.5 cdot 2 = 3$, this idea becomes somewhat clear, this idea being that
$$x + (p% text{ of } x) = y ;;; Leftrightarrow ;;; frac{y-x}{x} = p$$
i.e. this intuitive approach is analogous to the formula you usually get told a la what Siong noted.
So we ask: what if $x = 0$, and $y neq 0$? Well, we have
$$0 + (p% text{ of } 0) = y neq 0$$
Okay, but $0$ plus anything is just that thing, and any percentage of $0$ is $0$, and $0+0=0$. Thus, we get
$$0 = y neq 0$$
i.e. a nonsensical answer, and our $p$ is irrelevant here. No matter what $p$ is, there is always this contradiction. Similarly, if $y=0$, then we get a true statement instead but it's true for all possible values of $p$ instead.
So in that sense, we say that the percentage change starting from $0$ is nonsensical. Either you have a contradictory answer or just a plain true statement, regardless of your $p$ - $p$ becomes irrelevant in both cases. So it only makes sense to say that the percentage change $p$ is undefined.
answered Dec 6 at 8:28
Eevee Trainer
2,639221
answered Dec 6 at 8:28
Eevee Trainer
2,639221
answered Dec 6 at 8:28
Eevee Trainer
2,639221
answered Dec 6 at 8:28
Eevee Trainer
2,639221
2,639221
add a comment |
add a comment |
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Isn't it always $100$% since the previous number is $0$? Edit: Oh, too late
– NotEinstein
Dec 6 at 7:50
1
It's undefined. There can be no relative change from $0$. The absolute change is $10$ ($10-0 = 10$).
– Eff
Dec 6 at 8:31
(I forgot about the initial value in my previous comment, so here’s the correction.) $frac{v_2-v_1}{v_1}$ is undefined for $v_1 = 0$. If you have $v_1 = 2$ and $v_2 = 3$, for instance, you get a percentage change of $50$%. In other words, $2$ added to $50$% of itself gives $3$. Can $0$ added to any percentage of itself give a non-zero number?
– KM101
Dec 6 at 8:41
The "percentage change" number isn't that useful when your number is hovering around zero anyway.
– James
Dec 6 at 13:44