A problem in algebra: how does $-1=1$? [duplicate]
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This question already has an answer here:
math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = sqrt{(-1)^6}= 1$?
14 answers
I have algebra problem from a friend, that is 1=-1!!! because
$$-1=-1^{3}=-1^{^{frac{6}{2}}}=sqrt{(-1)^6}=sqrt{1}=1$$
I can not see what is wrong with this?
I will appreciate it any help.
exponentiation
edited Jan 10 at 11:00
Ben
329110
asked Jan 10 at 10:22
Ahmed Ahmed
29519
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marked as duplicate by KM101, drhab, Christoph, Masacroso, Lee David Chung Lin Jan 10 at 12:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = sqrt{(-1)^6}= 1$?
14 answers
I have algebra problem from a friend, that is 1=-1!!! because
$$-1=-1^{3}=-1^{^{frac{6}{2}}}=sqrt{(-1)^6}=sqrt{1}=1$$
I can not see what is wrong with this?
I will appreciate it any help.
exponentiation
edited Jan 10 at 11:00
Ben
329110
asked Jan 10 at 10:22
Ahmed Ahmed
29519
$endgroup$
marked as duplicate by KM101, drhab, Christoph, Masacroso, Lee David Chung Lin Jan 10 at 12:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Try a simpler problem $1^2=(-1)^2 notRightarrow 1=-1$.
$endgroup$
– user
Jan 10 at 10:26
$begingroup$
Notice that 1 has two roots: $1$ and $-1$.
$endgroup$
– denklo
Jan 10 at 10:41
add a comment |
$begingroup$
This question already has an answer here:
math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = sqrt{(-1)^6}= 1$?
14 answers
I have algebra problem from a friend, that is 1=-1!!! because
$$-1=-1^{3}=-1^{^{frac{6}{2}}}=sqrt{(-1)^6}=sqrt{1}=1$$
I can not see what is wrong with this?
I will appreciate it any help.
exponentiation
edited Jan 10 at 11:00
Ben
329110
asked Jan 10 at 10:22
Ahmed Ahmed
29519
$endgroup$
This question already has an answer here:
math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = sqrt{(-1)^6}= 1$?
14 answers
I have algebra problem from a friend, that is 1=-1!!! because
$$-1=-1^{3}=-1^{^{frac{6}{2}}}=sqrt{(-1)^6}=sqrt{1}=1$$
I can not see what is wrong with this?
I will appreciate it any help.
This question already has an answer here:
math fallacy problem: $-1= (-1)^3 = (-1)^{6/2} = sqrt{(-1)^6}= 1$?
14 answers
exponentiation
exponentiation
edited Jan 10 at 11:00
Ben
329110
asked Jan 10 at 10:22
Ahmed Ahmed
29519
edited Jan 10 at 11:00
Ben
329110
asked Jan 10 at 10:22
Ahmed Ahmed
29519
edited Jan 10 at 11:00
Ben
329110
edited Jan 10 at 11:00
Ben
329110
edited Jan 10 at 11:00
Ben
329110
329110
asked Jan 10 at 10:22
Ahmed Ahmed
29519
asked Jan 10 at 10:22
Ahmed Ahmed
29519
asked Jan 10 at 10:22
Ahmed Ahmed
29519
29519
marked as duplicate by KM101, drhab, Christoph, Masacroso, Lee David Chung Lin Jan 10 at 12:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by KM101, drhab, Christoph, Masacroso, Lee David Chung Lin Jan 10 at 12:45
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Try a simpler problem $1^2=(-1)^2 notRightarrow 1=-1$.
$endgroup$
– user
Jan 10 at 10:26
$begingroup$
Notice that 1 has two roots: $1$ and $-1$.
$endgroup$
– denklo
Jan 10 at 10:41
add a comment |
1
$begingroup$
Try a simpler problem $1^2=(-1)^2 notRightarrow 1=-1$.
$endgroup$
– user
Jan 10 at 10:26
$begingroup$
Notice that 1 has two roots: $1$ and $-1$.
$endgroup$
– denklo
Jan 10 at 10:41
1
1
$begingroup$
Try a simpler problem $1^2=(-1)^2 notRightarrow 1=-1$.
$endgroup$
– user
Jan 10 at 10:26
$begingroup$
Try a simpler problem $1^2=(-1)^2 notRightarrow 1=-1$.
$endgroup$
– user
Jan 10 at 10:26
$begingroup$
Notice that 1 has two roots: $1$ and $-1$.
$endgroup$
– denklo
Jan 10 at 10:41
$begingroup$
Notice that 1 has two roots: $1$ and $-1$.
$endgroup$
– denklo
Jan 10 at 10:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The usual properties of roots/exponents apply when the basis is (real) positive. In this case, we can't have
$$;-1^3=(-1)^3=(-1)^{6/2}color{red}{stackrel {!!}=}left[(-1)^{1/2}right]^6$$
as $;(-1)^{1/2}=sqrt{-1};$ cannot be done within the real numbers (and this is also another reason why the above mentioned properties don't apply to complex numbers...)
answered Jan 10 at 10:28
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = sqrt[c]{a^b}$ is not always true.
$endgroup$
– Christoph
Jan 10 at 10:33
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@Christoph Agree with you, so I'll add. Thankx
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– DonAntonio
Jan 10 at 10:59
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I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $left[ (-1)^6 right]^{1/2}$, which is $1$.
$endgroup$
– Christoph
Jan 10 at 11:30
$begingroup$
@Christoph Over the complex numbers it is false in general: $$1=sqrt1=sqrt{(-1)(-1)}=sqrt{-1}sqrt{-1}=icdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches.
$endgroup$
– DonAntonio
Jan 10 at 12:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The usual properties of roots/exponents apply when the basis is (real) positive. In this case, we can't have
$$;-1^3=(-1)^3=(-1)^{6/2}color{red}{stackrel {!!}=}left[(-1)^{1/2}right]^6$$
as $;(-1)^{1/2}=sqrt{-1};$ cannot be done within the real numbers (and this is also another reason why the above mentioned properties don't apply to complex numbers...)
answered Jan 10 at 10:28
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = sqrt[c]{a^b}$ is not always true.
$endgroup$
– Christoph
Jan 10 at 10:33
$begingroup$
@Christoph Agree with you, so I'll add. Thankx
$endgroup$
– DonAntonio
Jan 10 at 10:59
$begingroup$
I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $left[ (-1)^6 right]^{1/2}$, which is $1$.
$endgroup$
– Christoph
Jan 10 at 11:30
$begingroup$
@Christoph Over the complex numbers it is false in general: $$1=sqrt1=sqrt{(-1)(-1)}=sqrt{-1}sqrt{-1}=icdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches.
$endgroup$
– DonAntonio
Jan 10 at 12:50
add a comment |
$begingroup$
The usual properties of roots/exponents apply when the basis is (real) positive. In this case, we can't have
$$;-1^3=(-1)^3=(-1)^{6/2}color{red}{stackrel {!!}=}left[(-1)^{1/2}right]^6$$
as $;(-1)^{1/2}=sqrt{-1};$ cannot be done within the real numbers (and this is also another reason why the above mentioned properties don't apply to complex numbers...)
answered Jan 10 at 10:28
DonAntonioDonAntonio
179k1494230
$endgroup$
$begingroup$
All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = sqrt[c]{a^b}$ is not always true.
$endgroup$
– Christoph
Jan 10 at 10:33
$begingroup$
@Christoph Agree with you, so I'll add. Thankx
$endgroup$
– DonAntonio
Jan 10 at 10:59
$begingroup$
I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $left[ (-1)^6 right]^{1/2}$, which is $1$.
$endgroup$
– Christoph
Jan 10 at 11:30
$begingroup$
@Christoph Over the complex numbers it is false in general: $$1=sqrt1=sqrt{(-1)(-1)}=sqrt{-1}sqrt{-1}=icdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches.
$endgroup$
– DonAntonio
Jan 10 at 12:50
add a comment |
$begingroup$
The usual properties of roots/exponents apply when the basis is (real) positive. In this case, we can't have
$$;-1^3=(-1)^3=(-1)^{6/2}color{red}{stackrel {!!}=}left[(-1)^{1/2}right]^6$$
as $;(-1)^{1/2}=sqrt{-1};$ cannot be done within the real numbers (and this is also another reason why the above mentioned properties don't apply to complex numbers...)
answered Jan 10 at 10:28
DonAntonioDonAntonio
179k1494230
$endgroup$
The usual properties of roots/exponents apply when the basis is (real) positive. In this case, we can't have
$$;-1^3=(-1)^3=(-1)^{6/2}color{red}{stackrel {!!}=}left[(-1)^{1/2}right]^6$$
as $;(-1)^{1/2}=sqrt{-1};$ cannot be done within the real numbers (and this is also another reason why the above mentioned properties don't apply to complex numbers...)
answered Jan 10 at 10:28
DonAntonioDonAntonio
179k1494230
edited Jan 10 at 11:01
answered Jan 10 at 10:28
DonAntonioDonAntonio
179k1494230
answered Jan 10 at 10:28
DonAntonioDonAntonio
179k1494230
answered Jan 10 at 10:28
DonAntonioDonAntonio
179k1494230
179k1494230
$begingroup$
All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = sqrt[c]{a^b}$ is not always true.
$endgroup$
– Christoph
Jan 10 at 10:33
$begingroup$
@Christoph Agree with you, so I'll add. Thankx
$endgroup$
– DonAntonio
Jan 10 at 10:59
$begingroup$
I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $left[ (-1)^6 right]^{1/2}$, which is $1$.
$endgroup$
– Christoph
Jan 10 at 11:30
$begingroup$
@Christoph Over the complex numbers it is false in general: $$1=sqrt1=sqrt{(-1)(-1)}=sqrt{-1}sqrt{-1}=icdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches.
$endgroup$
– DonAntonio
Jan 10 at 12:50
add a comment |
$begingroup$
All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = sqrt[c]{a^b}$ is not always true.
$endgroup$
– Christoph
Jan 10 at 10:33
$begingroup$
@Christoph Agree with you, so I'll add. Thankx
$endgroup$
– DonAntonio
Jan 10 at 10:59
$begingroup$
I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $left[ (-1)^6 right]^{1/2}$, which is $1$.
$endgroup$
– Christoph
Jan 10 at 11:30
$begingroup$
@Christoph Over the complex numbers it is false in general: $$1=sqrt1=sqrt{(-1)(-1)}=sqrt{-1}sqrt{-1}=icdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches.
$endgroup$
– DonAntonio
Jan 10 at 12:50
$begingroup$
All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = sqrt[c]{a^b}$ is not always true.
$endgroup$
– Christoph
Jan 10 at 10:33
$begingroup$
All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = sqrt[c]{a^b}$ is not always true.
$endgroup$
– Christoph
Jan 10 at 10:33
$begingroup$
@Christoph Agree with you, so I'll add. Thankx
$endgroup$
– DonAntonio
Jan 10 at 10:59
$begingroup$
@Christoph Agree with you, so I'll add. Thankx
$endgroup$
– DonAntonio
Jan 10 at 10:59
$begingroup$
I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $left[ (-1)^6 right]^{1/2}$, which is $1$.
$endgroup$
– Christoph
Jan 10 at 11:30
$begingroup$
I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $left[ (-1)^6 right]^{1/2}$, which is $1$.
$endgroup$
– Christoph
Jan 10 at 11:30
$begingroup$
@Christoph Over the complex numbers it is false in general: $$1=sqrt1=sqrt{(-1)(-1)}=sqrt{-1}sqrt{-1}=icdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches.
$endgroup$
– DonAntonio
Jan 10 at 12:50
$begingroup$
@Christoph Over the complex numbers it is false in general: $$1=sqrt1=sqrt{(-1)(-1)}=sqrt{-1}sqrt{-1}=icdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches.
$endgroup$
– DonAntonio
Jan 10 at 12:50
add a comment |
1
$begingroup$
Try a simpler problem $1^2=(-1)^2 notRightarrow 1=-1$.
$endgroup$
– user
Jan 10 at 10:26
$begingroup$
Notice that 1 has two roots: $1$ and $-1$.
$endgroup$
– denklo
Jan 10 at 10:41