What's wrong with this 1 = -1 proof? [closed]
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Stared at this proof for 10 minutes, perhaps even more. Still a quite stumped, but I'm pretty sure the answer is staring me right in the face.
Okay, so we know that $i^2 = -1$.
Dividing both sides by $i$:
$$i = - frac1i$$
Squaring both sides:
$$i^2 = -frac1^2i^2$$
Obviously $i^2 = -1$, as previously shown, so therefore:
$$-1 = -frac1-1$$
Both negatives become a positive, so we're left with:
$$-1 = frac11$$
Which simplifies to:
$$-1 = 1$$
I'm not quite sure what's wrong here. Unless I'm there's an important step I skipped, I don't really see any problem here.
proof-verification complex-numbers
edited Sep 7 at 11:13
Jendrik Stelzner
7,69121137
asked Sep 7 at 5:02
nothing
41
closed as off-topic by A. Pongrácz, user91500, amWhy, Jendrik Stelzner, Shailesh Sep 7 at 14:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user91500, amWhy, Jendrik Stelzner, Shailesh
add a comment |Â
up vote
-1
down vote
favorite
Stared at this proof for 10 minutes, perhaps even more. Still a quite stumped, but I'm pretty sure the answer is staring me right in the face.
Okay, so we know that $i^2 = -1$.
Dividing both sides by $i$:
$$i = - frac1i$$
Squaring both sides:
$$i^2 = -frac1^2i^2$$
Obviously $i^2 = -1$, as previously shown, so therefore:
$$-1 = -frac1-1$$
Both negatives become a positive, so we're left with:
$$-1 = frac11$$
Which simplifies to:
$$-1 = 1$$
I'm not quite sure what's wrong here. Unless I'm there's an important step I skipped, I don't really see any problem here.
proof-verification complex-numbers
edited Sep 7 at 11:13
Jendrik Stelzner
7,69121137
asked Sep 7 at 5:02
nothing
41
closed as off-topic by A. Pongrácz, user91500, amWhy, Jendrik Stelzner, Shailesh Sep 7 at 14:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user91500, amWhy, Jendrik Stelzner, Shailesh
2
"Squaring both sides"... you should know that if you square $-a$ you get $a^2$. Recognize then that squaring $(-frac1i)$ gives $(-frac1i)^2$ i.e. $frac1^2i^2$.
– JMoravitz
Sep 7 at 5:06
As you play with $-1=1$ "(fake) proofs", you should be careful about several things. One of which is how you use various identities involving roots or exponents which in reality are only guaranteed to work for positive real numbers. See this post for one such example.
– JMoravitz
Sep 7 at 5:08
"Minus times minus equals plus: The reason for this we need not discuss"
– Lord Shark the Unknown
Sep 7 at 5:10
$(-a)^2 ne -(a^2)$ and $(-frac 1i)^2 ne -(frac 1i)^2$.
– fleablood
Sep 7 at 5:45
Possible duplicate of Why $sqrt-1 times -1 neq sqrt-1^2$?
– A. Pongrácz
Sep 7 at 5:56
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Stared at this proof for 10 minutes, perhaps even more. Still a quite stumped, but I'm pretty sure the answer is staring me right in the face.
Okay, so we know that $i^2 = -1$.
Dividing both sides by $i$:
$$i = - frac1i$$
Squaring both sides:
$$i^2 = -frac1^2i^2$$
Obviously $i^2 = -1$, as previously shown, so therefore:
$$-1 = -frac1-1$$
Both negatives become a positive, so we're left with:
$$-1 = frac11$$
Which simplifies to:
$$-1 = 1$$
I'm not quite sure what's wrong here. Unless I'm there's an important step I skipped, I don't really see any problem here.
proof-verification complex-numbers
edited Sep 7 at 11:13
Jendrik Stelzner
7,69121137
asked Sep 7 at 5:02
nothing
41
Stared at this proof for 10 minutes, perhaps even more. Still a quite stumped, but I'm pretty sure the answer is staring me right in the face.
Okay, so we know that $i^2 = -1$.
Dividing both sides by $i$:
$$i = - frac1i$$
Squaring both sides:
$$i^2 = -frac1^2i^2$$
Obviously $i^2 = -1$, as previously shown, so therefore:
$$-1 = -frac1-1$$
Both negatives become a positive, so we're left with:
$$-1 = frac11$$
Which simplifies to:
$$-1 = 1$$
I'm not quite sure what's wrong here. Unless I'm there's an important step I skipped, I don't really see any problem here.
proof-verification complex-numbers
proof-verification complex-numbers
edited Sep 7 at 11:13
Jendrik Stelzner
7,69121137
asked Sep 7 at 5:02
nothing
41
edited Sep 7 at 11:13
Jendrik Stelzner
7,69121137
asked Sep 7 at 5:02
nothing
41
edited Sep 7 at 11:13
Jendrik Stelzner
7,69121137
edited Sep 7 at 11:13
Jendrik Stelzner
7,69121137
edited Sep 7 at 11:13
Jendrik Stelzner
7,69121137
7,69121137
asked Sep 7 at 5:02
nothing
41
asked Sep 7 at 5:02
nothing
41
asked Sep 7 at 5:02
nothing
41
41
closed as off-topic by A. Pongrácz, user91500, amWhy, Jendrik Stelzner, Shailesh Sep 7 at 14:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user91500, amWhy, Jendrik Stelzner, Shailesh
closed as off-topic by A. Pongrácz, user91500, amWhy, Jendrik Stelzner, Shailesh Sep 7 at 14:24
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user91500, amWhy, Jendrik Stelzner, Shailesh
2
"Squaring both sides"... you should know that if you square $-a$ you get $a^2$. Recognize then that squaring $(-frac1i)$ gives $(-frac1i)^2$ i.e. $frac1^2i^2$.
– JMoravitz
Sep 7 at 5:06
As you play with $-1=1$ "(fake) proofs", you should be careful about several things. One of which is how you use various identities involving roots or exponents which in reality are only guaranteed to work for positive real numbers. See this post for one such example.
– JMoravitz
Sep 7 at 5:08
"Minus times minus equals plus: The reason for this we need not discuss"
– Lord Shark the Unknown
Sep 7 at 5:10
$(-a)^2 ne -(a^2)$ and $(-frac 1i)^2 ne -(frac 1i)^2$.
– fleablood
Sep 7 at 5:45
Possible duplicate of Why $sqrt-1 times -1 neq sqrt-1^2$?
– A. Pongrácz
Sep 7 at 5:56
add a comment |Â
2
"Squaring both sides"... you should know that if you square $-a$ you get $a^2$. Recognize then that squaring $(-frac1i)$ gives $(-frac1i)^2$ i.e. $frac1^2i^2$.
– JMoravitz
Sep 7 at 5:06
As you play with $-1=1$ "(fake) proofs", you should be careful about several things. One of which is how you use various identities involving roots or exponents which in reality are only guaranteed to work for positive real numbers. See this post for one such example.
– JMoravitz
Sep 7 at 5:08
"Minus times minus equals plus: The reason for this we need not discuss"
– Lord Shark the Unknown
Sep 7 at 5:10
$(-a)^2 ne -(a^2)$ and $(-frac 1i)^2 ne -(frac 1i)^2$.
– fleablood
Sep 7 at 5:45
Possible duplicate of Why $sqrt-1 times -1 neq sqrt-1^2$?
– A. Pongrácz
Sep 7 at 5:56
2
2
"Squaring both sides"... you should know that if you square $-a$ you get $a^2$. Recognize then that squaring $(-frac1i)$ gives $(-frac1i)^2$ i.e. $frac1^2i^2$.
– JMoravitz
Sep 7 at 5:06
"Squaring both sides"... you should know that if you square $-a$ you get $a^2$. Recognize then that squaring $(-frac1i)$ gives $(-frac1i)^2$ i.e. $frac1^2i^2$.
– JMoravitz
Sep 7 at 5:06
As you play with $-1=1$ "(fake) proofs", you should be careful about several things. One of which is how you use various identities involving roots or exponents which in reality are only guaranteed to work for positive real numbers. See this post for one such example.
– JMoravitz
Sep 7 at 5:08
As you play with $-1=1$ "(fake) proofs", you should be careful about several things. One of which is how you use various identities involving roots or exponents which in reality are only guaranteed to work for positive real numbers. See this post for one such example.
– JMoravitz
Sep 7 at 5:08
"Minus times minus equals plus: The reason for this we need not discuss"
– Lord Shark the Unknown
Sep 7 at 5:10
"Minus times minus equals plus: The reason for this we need not discuss"
– Lord Shark the Unknown
Sep 7 at 5:10
$(-a)^2 ne -(a^2)$ and $(-frac 1i)^2 ne -(frac 1i)^2$.
– fleablood
Sep 7 at 5:45
$(-a)^2 ne -(a^2)$ and $(-frac 1i)^2 ne -(frac 1i)^2$.
– fleablood
Sep 7 at 5:45
Possible duplicate of Why $sqrt-1 times -1 neq sqrt-1^2$?
– A. Pongrácz
Sep 7 at 5:56
Possible duplicate of Why $sqrt-1 times -1 neq sqrt-1^2$?
– A. Pongrácz
Sep 7 at 5:56
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
7
down vote
accepted
The square of $-x$ is $x^2$, not $-(x^2)$. Your error is in the "squaring both sides" step.
answered Sep 7 at 5:05
Hurkyl
110k9114257
add a comment |Â
up vote
1
down vote
You're doing mistake in the "squaring both side" step. You're squaring like this -(1/i)²
It should be done like this (-1/i)²
answered Sep 7 at 5:15
Abhishek VERMA
112
add a comment |Â
up vote
0
down vote
$left(frac-1iright)^2=frac(-1)^2i^2=frac1-1=-1$
answered Sep 7 at 5:09
gandalf61
6,159522
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
The square of $-x$ is $x^2$, not $-(x^2)$. Your error is in the "squaring both sides" step.
answered Sep 7 at 5:05
Hurkyl
110k9114257
add a comment |Â
up vote
7
down vote
accepted
The square of $-x$ is $x^2$, not $-(x^2)$. Your error is in the "squaring both sides" step.
answered Sep 7 at 5:05
Hurkyl
110k9114257
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
The square of $-x$ is $x^2$, not $-(x^2)$. Your error is in the "squaring both sides" step.
answered Sep 7 at 5:05
Hurkyl
110k9114257
The square of $-x$ is $x^2$, not $-(x^2)$. Your error is in the "squaring both sides" step.
answered Sep 7 at 5:05
Hurkyl
110k9114257
answered Sep 7 at 5:05
Hurkyl
110k9114257
answered Sep 7 at 5:05
Hurkyl
110k9114257
answered Sep 7 at 5:05
Hurkyl
110k9114257
110k9114257
add a comment |Â
add a comment |Â
up vote
1
down vote
You're doing mistake in the "squaring both side" step. You're squaring like this -(1/i)²
It should be done like this (-1/i)²
answered Sep 7 at 5:15
Abhishek VERMA
112
add a comment |Â
up vote
1
down vote
You're doing mistake in the "squaring both side" step. You're squaring like this -(1/i)²
It should be done like this (-1/i)²
answered Sep 7 at 5:15
Abhishek VERMA
112
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You're doing mistake in the "squaring both side" step. You're squaring like this -(1/i)²
It should be done like this (-1/i)²
answered Sep 7 at 5:15
Abhishek VERMA
112
You're doing mistake in the "squaring both side" step. You're squaring like this -(1/i)²
It should be done like this (-1/i)²
answered Sep 7 at 5:15
Abhishek VERMA
112
answered Sep 7 at 5:15
Abhishek VERMA
112
answered Sep 7 at 5:15
Abhishek VERMA
112
answered Sep 7 at 5:15
Abhishek VERMA
112
112
add a comment |Â
add a comment |Â
up vote
0
down vote
$left(frac-1iright)^2=frac(-1)^2i^2=frac1-1=-1$
answered Sep 7 at 5:09
gandalf61
6,159522
add a comment |Â
up vote
0
down vote
$left(frac-1iright)^2=frac(-1)^2i^2=frac1-1=-1$
answered Sep 7 at 5:09
gandalf61
6,159522
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$left(frac-1iright)^2=frac(-1)^2i^2=frac1-1=-1$
answered Sep 7 at 5:09
gandalf61
6,159522
$left(frac-1iright)^2=frac(-1)^2i^2=frac1-1=-1$
answered Sep 7 at 5:09
gandalf61
6,159522
answered Sep 7 at 5:09
gandalf61
6,159522
answered Sep 7 at 5:09
gandalf61
6,159522
answered Sep 7 at 5:09
gandalf61
6,159522
6,159522
add a comment |Â
add a comment |Â
2
"Squaring both sides"... you should know that if you square $-a$ you get $a^2$. Recognize then that squaring $(-frac1i)$ gives $(-frac1i)^2$ i.e. $frac1^2i^2$.
– JMoravitz
Sep 7 at 5:06
As you play with $-1=1$ "(fake) proofs", you should be careful about several things. One of which is how you use various identities involving roots or exponents which in reality are only guaranteed to work for positive real numbers. See this post for one such example.
– JMoravitz
Sep 7 at 5:08
"Minus times minus equals plus: The reason for this we need not discuss"
– Lord Shark the Unknown
Sep 7 at 5:10
$(-a)^2 ne -(a^2)$ and $(-frac 1i)^2 ne -(frac 1i)^2$.
– fleablood
Sep 7 at 5:45
Possible duplicate of Why $sqrt-1 times -1 neq sqrt-1^2$?
– A. Pongrácz
Sep 7 at 5:56