What is wrong with the statement 1=√1=√(−1)⋅(−1)=√−1⋅√−1=−1 ?
Answers
The problem is deep, but let's answer it short.
It's playing with the non unique solution in the equality. E.g. we know that 4–√ is 2 and -2, so we could play it as:
2=4–√=−2 --> False!
Or:
−2=4–√=2 --> False!
The correct equality could be found by simply remembering that the square root has two answers in which 2 or -2 is just one of it. Obviously, we cannot take one solution arbitrarily and then claim it as a solution, it's wrong, and it's cheating!
But it did teach us something. Not all of the solutions to some equation are what they're supposed to be. In here we just need 2 or -2, but not both. So we should write the equality honestly as:
2=4–√=2
−2=4–√=−2
Or simply:
±2=4–√=±2 --> True!
Now we could play a fallacy similar to the fallacy in the question as following:
1=1–√=−1 --> False!
But that's too obvious! Nah, in order to make it more intriguing, this guy is so creative, let's bring the square root of imaginary number into play:
1=1–√=−1∗−1−−−−−−−√=i2=−1 --> False!
Bringing in the complex number is such opening a pandora box, so we won't talk much about this interesting complex number algebra here. But in order to understand why this equality is false, remember that the solution to −1−−−√ is too not unique. It's not just i, but -i as well.
Now watch that i2 term in the equality carefully. It could only mean i * i, without considering -i at all. Thus it's not honest because it doesn't show the complete solution. The same case above, to debunk this fallacy, we have to bring up all the solutions completely:
−1∗−1−−−−−−−√=−1−−−√∗−1−−−√=±i∗±i
Which means all the possible combination of it:
i∗i=−1
i∗−i=1
−i∗i=1
−i∗−i=−1
Or simply:
−1∗−1−−−−−−−√=±i∗±i=±1
That is the honest equality, which is 1 is only one of the solution. So the correct version of the equality in the question is:
±1=1–√=−1∗−1−−−−−−−√=±i∗±i=±1
Well, it's not short, but at least, we could finish it here.