An imaginary error?
The following is my attempt at proving that 1 < − 1 . In which of these steps did I first make a mistake by using flawed logic?
Step 1: Let i = − 1 , then i 2 = − 1 and i 4 = ( i 2 ) 2 = ( − 1 ) 2 = 1 .
Step 2: Hence, we have i 2 < i 4 .
Step 3: We divide both sides by i to get i < i 3 .
Step 4: We divide both sides by i again to get 1 < i 2 or, equivalently, 1 < − 1 .
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4 solutions
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i 3 = i ( i 2 ) = − i , so both i and i 3 are red marked points on the grid above.
As they do not lie on the real number line, they cannot be compared with the standard > and < symbols.
Right. If a < b , then c a < c b is only true if c is a positive number.
In this case, c = i = − 1 is not a positive number, thus we cannot say that i i 2 < i i 4 is true.
Bonus: By Euler's formula , we know that e i x = cos x + i sin x , then e π i = − 1 and e 3 π i = − 1 . So e π i = e 3 π i , but this means π i = 3 π i or 1 = 3 ???? What went wrong?
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I'm no math expert, but i'd say it was when you introduced e^3(pi(i)) = -1. When you multiply a number, example n, by m, and turn it into a power, the answer is never really allowed to be the same as before, whether m be positive, negative, or even surreal. I could probably think a little further and prove it, definitely, but i have a strong belief the mistake is within e^3((pi)i)
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Yup, you're very close! Look up "complex exponentiation" and "multivalued function".
it's like saying cos(pi) = cos(3pi) so pi = 3pi and hence 1 = 3 and that's impossible because pi han goh :D
It cannot be concluded directly for periodic function f(x) like e^(ix) that if f(n1) = f(n2), then n1=n2. This result can only be applied to bijective functions only which is clearly not the case here.
The fact that i is imaginary isn't the problem. One could have divided through by -1 (a real number) straight off, rather than dividing by i twice, and arrive at the same result.
Looking at the graph, the inequality would treat the imaginary component as 0 and zero is smaller than 1 and greater than -1
When you divide an inequality by a positive number, you leave the inequality sign as it is. When you divide by a negative number, you change the direction of the inequality.
But there is no corresponding rule for dividing an inequality by a (non real) complex number, so the error is introduced at step 3, and another error is made at step 4.
Notice that if you combine steps 3 and 4 you can divide by i 2 because this is a real number. But because it is negative you must reverse the inequality sign, producing the perfectly acceptable 1 > − 1 .
My explanation is for the real numbers and for any non-reals this rule doesnot exist. So the fault is made in 3rd step and if we combine both 3rd and 4th step is becomes like dividing by negative integer
Yup, you've indirectly stated this misconception: Does cross multiply always work for inequalities? .
Nicely done!
In inequality ,we can not divide by an imaginary number.
You lost me at step 3:
Although i =✓(-1), i^(3)= i^2 × i = -i ,
The question here is: Is i < i^(3) = -i ?
I would think that -i (or i^(3)) < i
Therefore, dividing both sides by i should yield:
(i^3)/i < i/i =>
(i^2) < 1 =>
-1< 1 (which checks out)
However, another approach would be to assume that Step 3 is correct in the realm of imaginary numbers. This would mean that:
i < i^(3) =>
i < -i ,
However, rather than divide everything by i, let's divide every thing by a negative i. I know that you're not supposed to divide by a negative number in the imaginary world and here's why...This would mean that:
i /-i < -i / -i
=> -1> 1,
Which is incorrect, therefore we can conclude that step 3's assumption is where the error for this proof "officially" begins...
*I believe the moral to the story is :
We need to have a better understanding as to how imaginary numbers are supposed to function within the realm of Mathematical Computations and Logic (i.e. knowing and understanding the rules and limitations to dealing imaginary numbers and why these "limitations" are necessary in Mathematics)
You cannot definitively "compare" or "equate" imaginary numbers that are neither "real" number nor are "equivalent" to each other...
For example:
I can compare i^5 and i^13, because they , produce the same imaginary product(✓-1), However if I attempt to equate i^5 with i^8 , I will almost always run into a logistical error because i^5=✓-1 =/= i^8=+1.
This because all imaginary numbers fall into one of four modules:
i=i^5=i^9=....= i^(n+4),
i^2=i^6=i^10=....= i^(2n+4),
i^3=i^7=i^11=....= i^(3n+4),
i^4=i^8=....= i^(4n+4))
Attempting to compare imaginary numbers who are not apart of the same module... usually leads to some.... rather.... "questionable" logic... not always.... but enough times to where understanding the limitations of Imaginary numbers has become imperative...
I don't think you have properly answered the question of "Why can't we divide an inequality by an imaginary number?"
Yes, we can't do that, but why?
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Imaginary numbers are subset of complex numbers so when dividing complex numbers we must multiply numerator and denominator by the complex conjugate of the denominator
Although we arrive at i < i 3 , it is wrong because we can't compare imaginary numbers(only their modulus can be compared). But the second step is correct because i 2 and i 4 are purely real and can be compared.