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Wrong implication in Complex number
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Actually, I am facing a problem with a wrong implication/fallacy in complex number, which is
$frac1i=frac1times iitimes
i=fracii^2=fraci-1=-iimplies i=-i$
I know I should not ask this kind of question here, but I want to learn the reason behind this type of unwanted fallacies.
Can anybody help me? Thanks for your assistance in advance.
complex-numbers
asked Jul 23 at 18:52
Arnab Roy
134
 |Â
show 3 more comments
up vote
1
down vote
favorite
Actually, I am facing a problem with a wrong implication/fallacy in complex number, which is
$frac1i=frac1times iitimes
i=fracii^2=fraci-1=-iimplies i=-i$
I know I should not ask this kind of question here, but I want to learn the reason behind this type of unwanted fallacies.
Can anybody help me? Thanks for your assistance in advance.
complex-numbers
asked Jul 23 at 18:52
Arnab Roy
134
2
Fallacy? what fallacy?
– Lord Shark the Unknown
Jul 23 at 18:52
@LordSharktheUnknown $i=-i$
– Arnab Roy
Jul 23 at 18:56
No, it doesn't!
– Lord Shark the Unknown
Jul 23 at 18:56
Now you on earth do you go from $1/i=-i$ (correct) to $i=-i$ (obviously false)?
– Lord Shark the Unknown
Jul 23 at 18:58
2
No, it doesn't: it shows $1/i=-i$.
– Lord Shark the Unknown
Jul 23 at 18:59
 |Â
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Actually, I am facing a problem with a wrong implication/fallacy in complex number, which is
$frac1i=frac1times iitimes
i=fracii^2=fraci-1=-iimplies i=-i$
I know I should not ask this kind of question here, but I want to learn the reason behind this type of unwanted fallacies.
Can anybody help me? Thanks for your assistance in advance.
complex-numbers
asked Jul 23 at 18:52
Arnab Roy
134
Actually, I am facing a problem with a wrong implication/fallacy in complex number, which is
$frac1i=frac1times iitimes
i=fracii^2=fraci-1=-iimplies i=-i$
I know I should not ask this kind of question here, but I want to learn the reason behind this type of unwanted fallacies.
Can anybody help me? Thanks for your assistance in advance.
complex-numbers
asked Jul 23 at 18:52
Arnab Roy
134
edited Jul 23 at 18:56
asked Jul 23 at 18:52
Arnab Roy
134
asked Jul 23 at 18:52
Arnab Roy
134
asked Jul 23 at 18:52
Arnab Roy
134
134
2
Fallacy? what fallacy?
– Lord Shark the Unknown
Jul 23 at 18:52
@LordSharktheUnknown $i=-i$
– Arnab Roy
Jul 23 at 18:56
No, it doesn't!
– Lord Shark the Unknown
Jul 23 at 18:56
Now you on earth do you go from $1/i=-i$ (correct) to $i=-i$ (obviously false)?
– Lord Shark the Unknown
Jul 23 at 18:58
2
No, it doesn't: it shows $1/i=-i$.
– Lord Shark the Unknown
Jul 23 at 18:59
 |Â
show 3 more comments
2
Fallacy? what fallacy?
– Lord Shark the Unknown
Jul 23 at 18:52
@LordSharktheUnknown $i=-i$
– Arnab Roy
Jul 23 at 18:56
No, it doesn't!
– Lord Shark the Unknown
Jul 23 at 18:56
Now you on earth do you go from $1/i=-i$ (correct) to $i=-i$ (obviously false)?
– Lord Shark the Unknown
Jul 23 at 18:58
2
No, it doesn't: it shows $1/i=-i$.
– Lord Shark the Unknown
Jul 23 at 18:59
2
2
Fallacy? what fallacy?
– Lord Shark the Unknown
Jul 23 at 18:52
Fallacy? what fallacy?
– Lord Shark the Unknown
Jul 23 at 18:52
@LordSharktheUnknown $i=-i$
– Arnab Roy
Jul 23 at 18:56
@LordSharktheUnknown $i=-i$
– Arnab Roy
Jul 23 at 18:56
No, it doesn't!
– Lord Shark the Unknown
Jul 23 at 18:56
No, it doesn't!
– Lord Shark the Unknown
Jul 23 at 18:56
Now you on earth do you go from $1/i=-i$ (correct) to $i=-i$ (obviously false)?
– Lord Shark the Unknown
Jul 23 at 18:58
Now you on earth do you go from $1/i=-i$ (correct) to $i=-i$ (obviously false)?
– Lord Shark the Unknown
Jul 23 at 18:58
2
2
No, it doesn't: it shows $1/i=-i$.
– Lord Shark the Unknown
Jul 23 at 18:59
No, it doesn't: it shows $1/i=-i$.
– Lord Shark the Unknown
Jul 23 at 18:59
 |Â
show 3 more comments
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
Your argument is not a fallacy: it just proves that $i^-1=-i$, which is also clear from the fact that
$$
(-i)i=-i^2=-(-1)=1
$$
However it's easy to make a true fallacy out of it:
$$
sqrt-1=sqrtfrac1-1colorred=
fracsqrt1sqrt-1=frac1sqrt-1tag⚡
$$
Together with your argument this seems to prove that $i=-i$.
However, the step marked in red is fallacious. And it's very bad practice using the ambiguous symbols $sqrt-1$: in the complex numbers there is $i$, such that $i^2=-1$, but writing it the other way may lead to false arguments. The problem is that it is not possible to define a function $zmapstosqrtz$ such that, for all complex numbers $z_1$ and $z_2$,
$$
sqrtz_1z_2=sqrtz_1sqrtz_2
$$
and the fallacious ⚡ uses such a (non existent) function.
answered Jul 23 at 19:08
egreg
164k1180187
add a comment |Â
up vote
1
down vote
From $frac1i=-i$ we obtain $1=icdot -i$, and in no way that $i=-i$. So your last step is incorrect, and you also give no computation to show it.
For any complex number with $z=-z$ we obtain $2z=0$, and since $2neq 0$ this implies $z=0$. Again, $ineq 0$.
answered Jul 23 at 18:58
Dietrich Burde
74.6k64184
add a comment |Â
up vote
1
down vote
What makes you think it's incorrect? :)
It happens to be true that
$$frac1i = -i$$
All your first computations are justified and correct. However, we have that $1/ineq i$. I think you're confusing the two. Remember that in general we don't have $1/x = x$, either.
More generally for a complex number $z = a+bi$, we have that
$$frac1z = frac1a+bi = fraca-ib(a+ib)(a-ib) = fraca-iba^2+b^2. $$
Here the denominator is real, and you can see when $a = 0$ and $b =1$ (which is your example):
$$frac1a+ib = frac10+icdot 1 = frac1i = frac0-icdot 10^2+1^2 = -i. $$
answered Jul 23 at 18:53
Eff
11k21537
It becomes $i=-i$, Why?
– Arnab Roy
Jul 23 at 18:57
1
It does not become this. Why should it?
– Dietrich Burde
Jul 23 at 18:58
@ArnabRoy Ehmm.. No, it doesn't. You wrote $1/i$ not just $i$, they are not the same thing.
– Eff
Jul 23 at 18:59
@Eff, Understood.
– Arnab Roy
Jul 23 at 19:00
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your argument is not a fallacy: it just proves that $i^-1=-i$, which is also clear from the fact that
$$
(-i)i=-i^2=-(-1)=1
$$
However it's easy to make a true fallacy out of it:
$$
sqrt-1=sqrtfrac1-1colorred=
fracsqrt1sqrt-1=frac1sqrt-1tag⚡
$$
Together with your argument this seems to prove that $i=-i$.
However, the step marked in red is fallacious. And it's very bad practice using the ambiguous symbols $sqrt-1$: in the complex numbers there is $i$, such that $i^2=-1$, but writing it the other way may lead to false arguments. The problem is that it is not possible to define a function $zmapstosqrtz$ such that, for all complex numbers $z_1$ and $z_2$,
$$
sqrtz_1z_2=sqrtz_1sqrtz_2
$$
and the fallacious ⚡ uses such a (non existent) function.
answered Jul 23 at 19:08
egreg
164k1180187
add a comment |Â
up vote
3
down vote
accepted
Your argument is not a fallacy: it just proves that $i^-1=-i$, which is also clear from the fact that
$$
(-i)i=-i^2=-(-1)=1
$$
However it's easy to make a true fallacy out of it:
$$
sqrt-1=sqrtfrac1-1colorred=
fracsqrt1sqrt-1=frac1sqrt-1tag⚡
$$
Together with your argument this seems to prove that $i=-i$.
However, the step marked in red is fallacious. And it's very bad practice using the ambiguous symbols $sqrt-1$: in the complex numbers there is $i$, such that $i^2=-1$, but writing it the other way may lead to false arguments. The problem is that it is not possible to define a function $zmapstosqrtz$ such that, for all complex numbers $z_1$ and $z_2$,
$$
sqrtz_1z_2=sqrtz_1sqrtz_2
$$
and the fallacious ⚡ uses such a (non existent) function.
answered Jul 23 at 19:08
egreg
164k1180187
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your argument is not a fallacy: it just proves that $i^-1=-i$, which is also clear from the fact that
$$
(-i)i=-i^2=-(-1)=1
$$
However it's easy to make a true fallacy out of it:
$$
sqrt-1=sqrtfrac1-1colorred=
fracsqrt1sqrt-1=frac1sqrt-1tag⚡
$$
Together with your argument this seems to prove that $i=-i$.
However, the step marked in red is fallacious. And it's very bad practice using the ambiguous symbols $sqrt-1$: in the complex numbers there is $i$, such that $i^2=-1$, but writing it the other way may lead to false arguments. The problem is that it is not possible to define a function $zmapstosqrtz$ such that, for all complex numbers $z_1$ and $z_2$,
$$
sqrtz_1z_2=sqrtz_1sqrtz_2
$$
and the fallacious ⚡ uses such a (non existent) function.
answered Jul 23 at 19:08
egreg
164k1180187
Your argument is not a fallacy: it just proves that $i^-1=-i$, which is also clear from the fact that
$$
(-i)i=-i^2=-(-1)=1
$$
However it's easy to make a true fallacy out of it:
$$
sqrt-1=sqrtfrac1-1colorred=
fracsqrt1sqrt-1=frac1sqrt-1tag⚡
$$
Together with your argument this seems to prove that $i=-i$.
However, the step marked in red is fallacious. And it's very bad practice using the ambiguous symbols $sqrt-1$: in the complex numbers there is $i$, such that $i^2=-1$, but writing it the other way may lead to false arguments. The problem is that it is not possible to define a function $zmapstosqrtz$ such that, for all complex numbers $z_1$ and $z_2$,
$$
sqrtz_1z_2=sqrtz_1sqrtz_2
$$
and the fallacious ⚡ uses such a (non existent) function.
answered Jul 23 at 19:08
egreg
164k1180187
answered Jul 23 at 19:08
egreg
164k1180187
answered Jul 23 at 19:08
egreg
164k1180187
answered Jul 23 at 19:08
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
up vote
1
down vote
From $frac1i=-i$ we obtain $1=icdot -i$, and in no way that $i=-i$. So your last step is incorrect, and you also give no computation to show it.
For any complex number with $z=-z$ we obtain $2z=0$, and since $2neq 0$ this implies $z=0$. Again, $ineq 0$.
answered Jul 23 at 18:58
Dietrich Burde
74.6k64184
add a comment |Â
up vote
1
down vote
From $frac1i=-i$ we obtain $1=icdot -i$, and in no way that $i=-i$. So your last step is incorrect, and you also give no computation to show it.
For any complex number with $z=-z$ we obtain $2z=0$, and since $2neq 0$ this implies $z=0$. Again, $ineq 0$.
answered Jul 23 at 18:58
Dietrich Burde
74.6k64184
add a comment |Â
up vote
1
down vote
up vote
1
down vote
From $frac1i=-i$ we obtain $1=icdot -i$, and in no way that $i=-i$. So your last step is incorrect, and you also give no computation to show it.
For any complex number with $z=-z$ we obtain $2z=0$, and since $2neq 0$ this implies $z=0$. Again, $ineq 0$.
answered Jul 23 at 18:58
Dietrich Burde
74.6k64184
From $frac1i=-i$ we obtain $1=icdot -i$, and in no way that $i=-i$. So your last step is incorrect, and you also give no computation to show it.
For any complex number with $z=-z$ we obtain $2z=0$, and since $2neq 0$ this implies $z=0$. Again, $ineq 0$.
answered Jul 23 at 18:58
Dietrich Burde
74.6k64184
answered Jul 23 at 18:58
Dietrich Burde
74.6k64184
answered Jul 23 at 18:58
Dietrich Burde
74.6k64184
answered Jul 23 at 18:58
Dietrich Burde
74.6k64184
74.6k64184
add a comment |Â
add a comment |Â
up vote
1
down vote
What makes you think it's incorrect? :)
It happens to be true that
$$frac1i = -i$$
All your first computations are justified and correct. However, we have that $1/ineq i$. I think you're confusing the two. Remember that in general we don't have $1/x = x$, either.
More generally for a complex number $z = a+bi$, we have that
$$frac1z = frac1a+bi = fraca-ib(a+ib)(a-ib) = fraca-iba^2+b^2. $$
Here the denominator is real, and you can see when $a = 0$ and $b =1$ (which is your example):
$$frac1a+ib = frac10+icdot 1 = frac1i = frac0-icdot 10^2+1^2 = -i. $$
answered Jul 23 at 18:53
Eff
11k21537
It becomes $i=-i$, Why?
– Arnab Roy
Jul 23 at 18:57
1
It does not become this. Why should it?
– Dietrich Burde
Jul 23 at 18:58
@ArnabRoy Ehmm.. No, it doesn't. You wrote $1/i$ not just $i$, they are not the same thing.
– Eff
Jul 23 at 18:59
@Eff, Understood.
– Arnab Roy
Jul 23 at 19:00
add a comment |Â
up vote
1
down vote
What makes you think it's incorrect? :)
It happens to be true that
$$frac1i = -i$$
All your first computations are justified and correct. However, we have that $1/ineq i$. I think you're confusing the two. Remember that in general we don't have $1/x = x$, either.
More generally for a complex number $z = a+bi$, we have that
$$frac1z = frac1a+bi = fraca-ib(a+ib)(a-ib) = fraca-iba^2+b^2. $$
Here the denominator is real, and you can see when $a = 0$ and $b =1$ (which is your example):
$$frac1a+ib = frac10+icdot 1 = frac1i = frac0-icdot 10^2+1^2 = -i. $$
answered Jul 23 at 18:53
Eff
11k21537
It becomes $i=-i$, Why?
– Arnab Roy
Jul 23 at 18:57
1
It does not become this. Why should it?
– Dietrich Burde
Jul 23 at 18:58
@ArnabRoy Ehmm.. No, it doesn't. You wrote $1/i$ not just $i$, they are not the same thing.
– Eff
Jul 23 at 18:59
@Eff, Understood.
– Arnab Roy
Jul 23 at 19:00
add a comment |Â
up vote
1
down vote
up vote
1
down vote
What makes you think it's incorrect? :)
It happens to be true that
$$frac1i = -i$$
All your first computations are justified and correct. However, we have that $1/ineq i$. I think you're confusing the two. Remember that in general we don't have $1/x = x$, either.
More generally for a complex number $z = a+bi$, we have that
$$frac1z = frac1a+bi = fraca-ib(a+ib)(a-ib) = fraca-iba^2+b^2. $$
Here the denominator is real, and you can see when $a = 0$ and $b =1$ (which is your example):
$$frac1a+ib = frac10+icdot 1 = frac1i = frac0-icdot 10^2+1^2 = -i. $$
answered Jul 23 at 18:53
Eff
11k21537
What makes you think it's incorrect? :)
It happens to be true that
$$frac1i = -i$$
All your first computations are justified and correct. However, we have that $1/ineq i$. I think you're confusing the two. Remember that in general we don't have $1/x = x$, either.
More generally for a complex number $z = a+bi$, we have that
$$frac1z = frac1a+bi = fraca-ib(a+ib)(a-ib) = fraca-iba^2+b^2. $$
Here the denominator is real, and you can see when $a = 0$ and $b =1$ (which is your example):
$$frac1a+ib = frac10+icdot 1 = frac1i = frac0-icdot 10^2+1^2 = -i. $$
answered Jul 23 at 18:53
Eff
11k21537
edited Jul 23 at 19:02
answered Jul 23 at 18:53
Eff
11k21537
answered Jul 23 at 18:53
Eff
11k21537
answered Jul 23 at 18:53
Eff
11k21537
11k21537
It becomes $i=-i$, Why?
– Arnab Roy
Jul 23 at 18:57
1
It does not become this. Why should it?
– Dietrich Burde
Jul 23 at 18:58
@ArnabRoy Ehmm.. No, it doesn't. You wrote $1/i$ not just $i$, they are not the same thing.
– Eff
Jul 23 at 18:59
@Eff, Understood.
– Arnab Roy
Jul 23 at 19:00
add a comment |Â
It becomes $i=-i$, Why?
– Arnab Roy
Jul 23 at 18:57
1
It does not become this. Why should it?
– Dietrich Burde
Jul 23 at 18:58
@ArnabRoy Ehmm.. No, it doesn't. You wrote $1/i$ not just $i$, they are not the same thing.
– Eff
Jul 23 at 18:59
@Eff, Understood.
– Arnab Roy
Jul 23 at 19:00
It becomes $i=-i$, Why?
– Arnab Roy
Jul 23 at 18:57
It becomes $i=-i$, Why?
– Arnab Roy
Jul 23 at 18:57
1
1
It does not become this. Why should it?
– Dietrich Burde
Jul 23 at 18:58
It does not become this. Why should it?
– Dietrich Burde
Jul 23 at 18:58
@ArnabRoy Ehmm.. No, it doesn't. You wrote $1/i$ not just $i$, they are not the same thing.
– Eff
Jul 23 at 18:59
@ArnabRoy Ehmm.. No, it doesn't. You wrote $1/i$ not just $i$, they are not the same thing.
– Eff
Jul 23 at 18:59
@Eff, Understood.
– Arnab Roy
Jul 23 at 19:00
@Eff, Understood.
– Arnab Roy
Jul 23 at 19:00
add a comment |Â
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2
Fallacy? what fallacy?
– Lord Shark the Unknown
Jul 23 at 18:52
@LordSharktheUnknown $i=-i$
– Arnab Roy
Jul 23 at 18:56
No, it doesn't!
– Lord Shark the Unknown
Jul 23 at 18:56
Now you on earth do you go from $1/i=-i$ (correct) to $i=-i$ (obviously false)?
– Lord Shark the Unknown
Jul 23 at 18:58
2
No, it doesn't: it shows $1/i=-i$.
– Lord Shark the Unknown
Jul 23 at 18:59