Mathematical Fallacy: Is − 1 = 1 ?
− 1 = − 1 1 = − 1 1 = − 1 1 = 1 − 1 = 1 − 1 = i = i 2 = − 1 = i i 1 i 1 i 1 i 1 i 1 i 1 1 1 … ( 1 ) … ( 2 ) … ( 3 ) … ( 4 ) … ( 5 ) … ( 6 ) … ( 7 ) … ( 8 ) … ( 9 )
Consider these steps above.
In which step is the (first) mistake committed?
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3 solutions
Moderator note:
Yes, one has to be careful as to when the rules of exponents hold. When we only cared about positive reals, we were taught the rules and told that they always hold. However, when we learnt about − 1 , few teachers bothered to explain how the rules needed to be changed.
It even holds if a,b are both negative.. Isn't it??
I disagree on the fact that the rules hold when the numbers under radicals are both negative or positive. If this was true, then − 1 = i ⟹ − 1 ⋅ − 1 = i ⋅ i ⟹ − 1 ⋅ − 1 = i 2 ⟹ 1 = − 1 ⟹ 1 = − 1 .
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I agree with you! I don't understand how anybody could write sqrt(-1)... Why sqrt(-1) should be equal to i and not to -i? What is the correct definition of sqrt, defined on C?
Which "rule" are you disproving exactly?
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My main issue is that I am not used to apply the function to negative numbers. Even for − 1 .
As far as I know (or I should say, as far as I've been taught) the real function : R + ∪ { 0 } ⟶ R is defined as 0 = 0 and a = the (only!) positive root of x 2 = a , for each a ∈ R + . \ One can extend the function to negative real numbers, but the property a b = a b that holds for a , b ∈ R + ∪ { 0 } cannot be easily preserved.
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@Andrea Palma – I think division is indeed an exception where two square roots of negative numbers work. As you correctly point out, for multiplications where both a and b are negative, the well-known product identity no longer works.
All the more reason to be wary whenever people loosely apply roots to negative numbers... It is easy to mess up signs in the process.
No, this is not a contradiction, because 1 = ± 1 . @Elden Knight , in fact it is like you say: − 1 = ± i , since the definition of the square root of a is that it is the solution of x 2 − a = 0 . So, @Andrea Palma , by starting from − 1 = + i , you correctly arrive at 1 = − 1 . If you also use − 1 = − i , you get the other one.
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@Donald Duck You are wrong. The range of the square root function is only zero and positive numbers. Therefore, the square root of 1 is NOT -1 by convention.
No, the radical function is a single-valued function. It is incorrect to say sqrt(1) = +/-1 . It's a function, not a relation. It only has one value.
The equation also seems to hold for a = − 1 and b = 4 .
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But the problem arises when you write 2 i = 4 − 1 = 4 − 1 = − 4 1 = 2 i 1
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The standard way of writing a negative fraction is to write a negative numerator and a positive denominator. So, I think the equation does not hold when b is negative but a is positive.
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@Milind Prabhu – What's the standard way .... It's not the numerator or denominator which is negative .. It's the whole fraction which is a negative ... Like − 4 1
Because if this were true, we are saying that i/2 = 1/2i . So, by cross mutipication, we get :
2(i^2) = 2 which implies i^2 = 1, which of course is NOT true
I don't get it (1)^1÷2=+or-1 so mistake should be in 3rd step so how is it wrong?
Interesting I chose step 3 because not sure you can change 1 into sqrt(1) since sqrt(1) has 2 real solutions... And if in step 6 you take -1 as solution than everything holds...
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The problem really is that the square root function is not properly defined on the complex plane but only on the complex plane less a half line beginning at (but not including) zero. If you choose that consistently then step 4 is fine — so the “error” really starts where one begins to use the square root function without properly saying where the cut is.
I think step (3) is wrong. This is because we can't take square root of only '1' (the numerator). We have to take square root of numerator as well a denominator.
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Step 3 is correct; 1 = 1 (you substitute 1 with 1 ).
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I see it as him multiplying left side with sqrt(1) and right side also (but also evaluating right side).
CHALLENGE: If for a second we consider that the property stated above is correct for complex numbers also, then in which statement Rohit Udaiwal did the necessary changes to bring this amusing result.
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Step 5. I had thought 5 was the answer till I found that 5 isn't on the options. Paid more attention to the ones before 5 and so got 4.
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You got my point!!! But there is one more statement after the statement 5 in which the most important changes were done.
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@Anshuman Singh Bais – Statements 6, 7, 8 and 9 look correct to me?
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@Jerry Hill – You are right.These statements are absolutely correct. I asked that in which statement the most important changes were done to get the result.:P
I was also finding 5 as the ans!!
Huh...I learned that writing the square root of a negative is always wrong, so I thought the 1st step was wrong. And actually I don't see any justification of the rule you use...so in my mind, my explanation is worth yours...until you give me a demonstration.
it is the same with multiplication ??
In France, we learn that sqrt(-1) doesn't exit. Indeed, how do you define sqrt on negatives?... and on complex, maybe? For me, sqrt is only defined on positive reals, and it is better to avoid these mistakes. If you want to write i, you write i and not sqrt(-1)... :\ I don't understand the interrest of doing that. You have to define sqrt(2) because it is the only way of writing this number, but sqrt(-4)=2i is not necessary, I think...
isnt we need to multiply the right side to -i/-i so we dont have any complex number in the denominator?
I don't think writing sqrt(-1) = i is extremely correct.
I agree with a lot of people on here saying that the root function is defined to only have positive values used at. For i it is explicitly defined that " i 2 = − 1 ", NOT " − 1 = i ", therefore the error already lies in the first step and I think one should change that here...
Is there any proof of above expression?I cannot believe its true without a proof
I chose step number 3 since 1 = ± 1 , not only 1 . This is the step where the sign problems begin, so I don't fully agree with the answer considered right here. (In fact, sign problems start at step 1! I think it should be stated that only positive square root solutions should be considered.)
The real reason is that − 1 should not be treated as a number when it is in operation with a positive square root.
I think step 6 is also wrong. Isn't it?
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Yeah you are right.
Step 6 is correct, although for a different reason: 1 − 1 = − 1 = 1 − 1 , just like 1 x = x = 1 x for all x .
If what you say would be right, explain to me how is possible to say in step 1 that square root of -1 is i. Do you think i is positive? What about -i ? Why not square root of -1 is not -i? Square root of -1 does not exists! Step 1 is wrong!
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By definition/convention, the principal square root of a complex number z , denoted by z , is the (unique) complex number x where the argument of x is in the interval ( − 2 π , 2 π ] where x 2 = z . The two square roots of − 1 are i , − i , but − i has argument − 2 π , so it is not the principal square root. The principal square root of − 1 is i .
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You tell me about the principal value!? A real function takes only one value! A complex function is different, as you say, it takes 2 or several values in the same point. Hence how can we say that square root of -1 is i, principal value or not? The definition of "i" is: i x i = −1 ! Let not play with complex functions as we have real functions. If we want to take the principal definition, for having only one value, then we must realize that we have another function, not the real one, and we must verify the properties. Because of the discontinuous nature of the square root function in the complex plane, the following laws are not true in general: √zw = √z√w (counterexample for the principal square root: z = −1 and w = −1) 1/√z = √1/z (counterexample for the principal square root: z = −1) √z* = √z* (counterexample for the principal square root: z = −1) A similar problem appears with other complex functions with branch cuts, e.g., the complex logarithm and the relations log z + log w = log(zw) or log (z ) = log (z) which are not true in general.
Wrongly assuming one of these laws underlies several faulty "proofs".
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@Andi Popescu – A complex function, by definition of it being a function, only takes one value. Yes, this means if you don't limit square root (and any higher root) to principal values, it is not a function. The same applies to logarithms to any base; if you don't limit them to principal values, they are not functions.
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@Ivan Koswara – A function defined in C is usually a "multiple-valued function"! Function root (z) of order n takes n values! The fact that one of them is the principal one does not change the situation! To be right, the 1-st step must be stated: "The principal branch of square(-1) = i"! But it is not the case! 1-st step is wrong!
Could you please explain the reason behind that rule, though?
I'm tempted to say the error is in step 1. The function is defined on non-negative real numbers. We introduce i by saying " i is one of the two roots of the equation z 2 = − 1 ". It's problematic to say − 1 = i , as this sequence of computations shows. We gain nothing by using the nomenclature − 1 but we open the door to a lot of confusion because we no longer have all the properties of the radical function that it has when its domain is constrained to the non-negative real numbers.
Yes, because the square root function is defined only for positive integers.
I agree. i is defined as i^2= -1 from this we can not conclude if √(-1)=i or √(-1)=-i and because i is not in the reals we can not pick the positive one of those because that would imply that we know if I is positive or not.
Yes I agree it's not possible to use the square root here !!! It doesn't mean anything !!
I wholeheartedly agree with you.
We know that -i= i 1 .So if we try to prove this with every step (starting from 1) ,you will not be able to prove this after 3rd step .which shows that a mistake is committed on step-4 . (here i=(-1)^( 2 1 ))
Why is step 3 not the answer? Root(1) must be +1 or -1, why is only the positive value of square root considered? Please reply.
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@Amogh Keni the range of the square root function is 0 and all positive numbers
Playing with i=sqrt of -1 is stupid. First step is misleading. @Ivan Koswara explains why.
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Agree with this point (well, aside from the word 'stupid').
It's not literally true that sqrt(-1) = i.
Something different is going on. Whenever dealing with i, one must start by saying "let i denote one of the two solutions to z^2=1" and then go from there.
Strictly speaking the radical function is only defined on non-negative real numbers. There is a tolerance of using (sqrt(-1) = i) but it shouldn't be taken as an extension of the same function that is defined on real input.
I disagree with the solución the first mistake came when you exclude -1 as an answer of square robot of 1, since you hace obviously two values + or -1. Please correct
You create a paradox with straightforward algebraic manipulations and yet you try to paper it over with "rules". The square root of -1 must have two values, i and -i. All square roots have two values. Once you allow for both roots of -1 or +1 for that matter, all the paradox vanishes. If -i is not the square root of -1 then what happens when you multiply -i by itself? Does it produce -1 or not? You seem to think that can be ignored. Or is there a "rule" that -i cannot be multiplied by itself? Or that -i can't exist? How can x be squared to give y and yet not be the square root of y?. An arbitrary rule can't state that without causing paradoxes.
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well that is true, but the convention itself is being tested in this problem
"All square roots have two values." Yes, and the square root function is defined to select the positive value when it's input is positive.
the first statement, sqrt(-1), is already false. There are two complex solutions of the equation z^2=-1, namely i and -i. There is no way to distinguish the two in terms of square roots.
Whilst you are correct that sqrt1 is both +1 and -1, step 3 in itself is not incorrect. Incomplete maybe, there is no error. The first error (and that is what we are asked to identify) appears in step 4.
b a = b a , this is correct only if both a , b are positive or both are negative. Hence , step 4 is wrong.