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Calculator's view on the non-permissible values of the expression $frac1fracxx-1+x$
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When I try to find the non-permissible values of the expression $$ cfrac1fracxx-1+x$$ I rewrite it as
$$cfrac1cfracx^2x-1$$
and conclude that the non-permissible values are $x = 0$ and $x=1$ as those values will make the denominators present in the expression undefined.
But this expression can also be written with the $div$ symbol and then, using the "division is multiplying by a reciprocal," we can have $$1 div fracx^2x-1 stackreltextmultiply by recipimplies 1 cdot fracx-1x^2$$ and then one could conclude that $x=0$ is the only non-permissible value.
Three different calculators accept this claim:
(1) A TI-89
(2) Desmos graphing calculator,
(3) Google calculator, with $1 div (1 div (1-1) +1)$:
These three calculators seem to have no problem with the simplification of $$cfrac1fracxx-1+x = fracx-1x^2$$ as they all treated $x=1$ as a permissible value (i.e. no errors).
What is happening here? Is my initial conclusion of the non-permissible values being $x = 0$ and $x=1$ correct?
I have heard that Desmos and other graphers may simplify the original expression to make operations more efficient, but I am not sure how this might apply to the Google calculator.
(Oddly enough, my Windows calculator gives me a divide-by-zero error for $1 div (1 div (1-1) +1)$.)
algebra-precalculus divisibility
asked Jul 16 at 0:05
holo
797
add a comment |Â
up vote
0
down vote
favorite
When I try to find the non-permissible values of the expression $$ cfrac1fracxx-1+x$$ I rewrite it as
$$cfrac1cfracx^2x-1$$
and conclude that the non-permissible values are $x = 0$ and $x=1$ as those values will make the denominators present in the expression undefined.
But this expression can also be written with the $div$ symbol and then, using the "division is multiplying by a reciprocal," we can have $$1 div fracx^2x-1 stackreltextmultiply by recipimplies 1 cdot fracx-1x^2$$ and then one could conclude that $x=0$ is the only non-permissible value.
Three different calculators accept this claim:
(1) A TI-89
(2) Desmos graphing calculator,
(3) Google calculator, with $1 div (1 div (1-1) +1)$:
These three calculators seem to have no problem with the simplification of $$cfrac1fracxx-1+x = fracx-1x^2$$ as they all treated $x=1$ as a permissible value (i.e. no errors).
What is happening here? Is my initial conclusion of the non-permissible values being $x = 0$ and $x=1$ correct?
I have heard that Desmos and other graphers may simplify the original expression to make operations more efficient, but I am not sure how this might apply to the Google calculator.
(Oddly enough, my Windows calculator gives me a divide-by-zero error for $1 div (1 div (1-1) +1)$.)
algebra-precalculus divisibility
asked Jul 16 at 0:05
holo
797
You are right. Both $x=1$ and $x=0$ need to be excluded. I'm not sure what is happening here though
– Rumpelstiltskin
Jul 16 at 0:07
2
"division by $k$ is multiplying by the reciprocal $1/k$" is only true if $k neq 0$.
– Will Sherwood
Jul 16 at 0:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
When I try to find the non-permissible values of the expression $$ cfrac1fracxx-1+x$$ I rewrite it as
$$cfrac1cfracx^2x-1$$
and conclude that the non-permissible values are $x = 0$ and $x=1$ as those values will make the denominators present in the expression undefined.
But this expression can also be written with the $div$ symbol and then, using the "division is multiplying by a reciprocal," we can have $$1 div fracx^2x-1 stackreltextmultiply by recipimplies 1 cdot fracx-1x^2$$ and then one could conclude that $x=0$ is the only non-permissible value.
Three different calculators accept this claim:
(1) A TI-89
(2) Desmos graphing calculator,
(3) Google calculator, with $1 div (1 div (1-1) +1)$:
These three calculators seem to have no problem with the simplification of $$cfrac1fracxx-1+x = fracx-1x^2$$ as they all treated $x=1$ as a permissible value (i.e. no errors).
What is happening here? Is my initial conclusion of the non-permissible values being $x = 0$ and $x=1$ correct?
I have heard that Desmos and other graphers may simplify the original expression to make operations more efficient, but I am not sure how this might apply to the Google calculator.
(Oddly enough, my Windows calculator gives me a divide-by-zero error for $1 div (1 div (1-1) +1)$.)
algebra-precalculus divisibility
asked Jul 16 at 0:05
holo
797
When I try to find the non-permissible values of the expression $$ cfrac1fracxx-1+x$$ I rewrite it as
$$cfrac1cfracx^2x-1$$
and conclude that the non-permissible values are $x = 0$ and $x=1$ as those values will make the denominators present in the expression undefined.
But this expression can also be written with the $div$ symbol and then, using the "division is multiplying by a reciprocal," we can have $$1 div fracx^2x-1 stackreltextmultiply by recipimplies 1 cdot fracx-1x^2$$ and then one could conclude that $x=0$ is the only non-permissible value.
Three different calculators accept this claim:
(1) A TI-89
(2) Desmos graphing calculator,
(3) Google calculator, with $1 div (1 div (1-1) +1)$:
These three calculators seem to have no problem with the simplification of $$cfrac1fracxx-1+x = fracx-1x^2$$ as they all treated $x=1$ as a permissible value (i.e. no errors).
What is happening here? Is my initial conclusion of the non-permissible values being $x = 0$ and $x=1$ correct?
I have heard that Desmos and other graphers may simplify the original expression to make operations more efficient, but I am not sure how this might apply to the Google calculator.
(Oddly enough, my Windows calculator gives me a divide-by-zero error for $1 div (1 div (1-1) +1)$.)
algebra-precalculus divisibility
asked Jul 16 at 0:05
holo
797
edited Jul 16 at 0:08
asked Jul 16 at 0:05
holo
797
asked Jul 16 at 0:05
holo
797
asked Jul 16 at 0:05
holo
797
797
You are right. Both $x=1$ and $x=0$ need to be excluded. I'm not sure what is happening here though
– Rumpelstiltskin
Jul 16 at 0:07
2
"division by $k$ is multiplying by the reciprocal $1/k$" is only true if $k neq 0$.
– Will Sherwood
Jul 16 at 0:08
add a comment |Â
You are right. Both $x=1$ and $x=0$ need to be excluded. I'm not sure what is happening here though
– Rumpelstiltskin
Jul 16 at 0:07
2
"division by $k$ is multiplying by the reciprocal $1/k$" is only true if $k neq 0$.
– Will Sherwood
Jul 16 at 0:08
You are right. Both $x=1$ and $x=0$ need to be excluded. I'm not sure what is happening here though
– Rumpelstiltskin
Jul 16 at 0:07
You are right. Both $x=1$ and $x=0$ need to be excluded. I'm not sure what is happening here though
– Rumpelstiltskin
Jul 16 at 0:07
2
2
"division by $k$ is multiplying by the reciprocal $1/k$" is only true if $k neq 0$.
– Will Sherwood
Jul 16 at 0:08
"division by $k$ is multiplying by the reciprocal $1/k$" is only true if $k neq 0$.
– Will Sherwood
Jul 16 at 0:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
They are different functions since they have different domains. But, they agree on whatever numbers are common to their domain. By simplifying to
$$
fracx^2x - 1
$$
you are changing to a new function. For example, if I say your function is to take a number, $x$, divide $1$ by it, then divide $1$ by that result, you will get:
$$
x mapsto frac1frac1x.
$$
I could then define a different function that just repeats the input:
$$
x mapsto x.
$$
For $x neq 0$, they agree. But, the first cannot be computed at $x = 0$.
answered Jul 16 at 0:13
Joe Johnson 126
13.6k32567
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
They are different functions since they have different domains. But, they agree on whatever numbers are common to their domain. By simplifying to
$$
fracx^2x - 1
$$
you are changing to a new function. For example, if I say your function is to take a number, $x$, divide $1$ by it, then divide $1$ by that result, you will get:
$$
x mapsto frac1frac1x.
$$
I could then define a different function that just repeats the input:
$$
x mapsto x.
$$
For $x neq 0$, they agree. But, the first cannot be computed at $x = 0$.
answered Jul 16 at 0:13
Joe Johnson 126
13.6k32567
add a comment |Â
up vote
1
down vote
They are different functions since they have different domains. But, they agree on whatever numbers are common to their domain. By simplifying to
$$
fracx^2x - 1
$$
you are changing to a new function. For example, if I say your function is to take a number, $x$, divide $1$ by it, then divide $1$ by that result, you will get:
$$
x mapsto frac1frac1x.
$$
I could then define a different function that just repeats the input:
$$
x mapsto x.
$$
For $x neq 0$, they agree. But, the first cannot be computed at $x = 0$.
answered Jul 16 at 0:13
Joe Johnson 126
13.6k32567
add a comment |Â
up vote
1
down vote
up vote
1
down vote
They are different functions since they have different domains. But, they agree on whatever numbers are common to their domain. By simplifying to
$$
fracx^2x - 1
$$
you are changing to a new function. For example, if I say your function is to take a number, $x$, divide $1$ by it, then divide $1$ by that result, you will get:
$$
x mapsto frac1frac1x.
$$
I could then define a different function that just repeats the input:
$$
x mapsto x.
$$
For $x neq 0$, they agree. But, the first cannot be computed at $x = 0$.
answered Jul 16 at 0:13
Joe Johnson 126
13.6k32567
They are different functions since they have different domains. But, they agree on whatever numbers are common to their domain. By simplifying to
$$
fracx^2x - 1
$$
you are changing to a new function. For example, if I say your function is to take a number, $x$, divide $1$ by it, then divide $1$ by that result, you will get:
$$
x mapsto frac1frac1x.
$$
I could then define a different function that just repeats the input:
$$
x mapsto x.
$$
For $x neq 0$, they agree. But, the first cannot be computed at $x = 0$.
answered Jul 16 at 0:13
Joe Johnson 126
13.6k32567
answered Jul 16 at 0:13
Joe Johnson 126
13.6k32567
answered Jul 16 at 0:13
Joe Johnson 126
13.6k32567
answered Jul 16 at 0:13
Joe Johnson 126
13.6k32567
13.6k32567
add a comment |Â
add a comment |Â
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You are right. Both $x=1$ and $x=0$ need to be excluded. I'm not sure what is happening here though
– Rumpelstiltskin
Jul 16 at 0:07
2
"division by $k$ is multiplying by the reciprocal $1/k$" is only true if $k neq 0$.
– Will Sherwood
Jul 16 at 0:08