you can divide both sides by an imaginary value and get an accurate value as this value is a complex number. a + bi where a is real and bi is the imaginary part. Therefore real values are subsets of complex values, so this is not the case. i think the mistake is when you square root 'i' squared (3rd step) as 'i' can be both negative or negative in the next step. if 'i' was written as '-i', then this proof is not valid.
you can divide both sides by an imaginary value and get an accurate value as this value is a complex number. a + bi where a is real and bi is the imaginary part. Therefore real values are subsets of complex values, so this is not the case. i think the mistake is when you square root 'i' squared (3rd step) as 'i' can be both negative or negative in the next step. if 'i' was written as '-i', then this proof is not valid.
http://2.bp.blogspot.com/_i1lmfumuWlA/TCZZKrkDn4I/AAAAAAAAACQ/44KBYmRfzVc/s320/imaginary+notation_1.bmp is error by itself , isnt true, cant have sqrt(-1) by sqrt def, also i=(0,1) pair of real numbers, with add and multiplications rules
you cannot divide both sides by an imaginary value and get an accurate value.
ReplyDeleteyou can divide both sides by an imaginary value and get an accurate value as this value is a complex number. a + bi where a is real and bi is the imaginary part. Therefore real values are subsets of complex values, so this is not the case.
Deletei think the mistake is when you square root 'i' squared (3rd step) as 'i' can be both negative or negative in the next step. if 'i' was written as '-i', then this proof is not valid.
In one or both cases on step 4, i may be negative instead of positive.
ReplyDeleteyou can divide both sides by an imaginary value and get an accurate value as this value is a complex number. a + bi where a is real and bi is the imaginary part. Therefore real values are subsets of complex values, so this is not the case.
ReplyDeletei think the mistake is when you square root 'i' squared (3rd step) as 'i' can be both negative or negative in the next step. if 'i' was written as '-i', then this proof is not valid.
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ReplyDeletehttp://2.bp.blogspot.com/_i1lmfumuWlA/TCZZKrkDn4I/AAAAAAAAACQ/44KBYmRfzVc/s320/imaginary+notation_1.bmp is error by itself , isnt true, cant have sqrt(-1) by sqrt def, also i=(0,1) pair of real numbers, with add and multiplications rules
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