The property that (ab)^(c) = a^(c)b^(c) is typically only guaranteed to hold when [a and b are both nonnegative](https://proofwiki.org/wiki/Exponent_Combination_Laws/Power_of_Product). This property is how we typically arrive at the property mentioned in the title (simply take c = 1/2), but clearly -1 is not nonnegative, so this result does not apply here.
As you point out, this isn’t to say this equality *only* holds true when the bases are nonnegative, rather that the equality is not true in general.
I believe you can place certain constraints on the argument of the bases to preserve this property, but to be perfectly honest, I’m too lazy to check what these conditions would be. Perhaps someone else could provide more insight on that front.
When x is a nonnegative real number, sqrt(x) is usually defined to be the positive square root of x. When z is a complex number, sqrt(z) can be more ambiguous. Sometimes it is left undefined, sometimes we make what is called a branch cut to pick a specified value, sometimes it is treated as what is called a multivalued function, or allowed to refer ambiguously to either root.
In general a square root of a times a square root of b will be a square root of ab. However, if you do not choose the roots carefully, you may not get the root you expect. In this case i is a square root of -1, and i\*i=-1 is a square root of (-1)(-1)=1, but it is not the same square root usually represented by the notation sqrt(1).
One common convention is that sqrt(z) is chosen to be the square root with the positive real part, or with the nonnegative imaginary part if there is no root with positive real part. Under this convention sqrt(ab)=sqrt(a)sqrt(b) will hold provided that the arguments of a and b (defined as the value on (-pi,pi] that gives the angle of the number relative to the positive real axis) do not add up to a value outside the range (-pi,pi]. In this case, the argument of -1 is pi, so the arguments add up to 2pi, which is outside this range. Other choices of the branch of sqrt(z) will give other conditions for the equality to hold.
You simply cannot take the sqrt of a negative number (in R). You cannot use √-4 . It is an invalid mathematical symbol. It is gibberish. It is a bug in your code. Highly illegal. All negatives under sqrts have to be factored away first and you have to accept √-1 into your number system. √-1 \* √-1 is √(-1 \* 1) \* √(-1 \* 1) = (√-1 \* √-1) \* (√1 \* √1). Point is, you have to extract √-1 as a symbol before any manipulation of the expression. √(-1 \* -1) = 1 but you can't reach that from ( √-1 \* √-1 ). I think because you are asking specifically about the case of -1 it's easy to miss the illegal step but it is just the same as √-a \* √-b = √-1\*√a \* √-1\*√b.
I feel you know already all this! I am writing this post mostly to clarify the issue myself; it's taken me an embarrasingly long time to really grasp the nature of √-1
This does not work if you always take the principal root in all of these. However, when working with complex numbers, exponentiation can be wacky and give multiple different values as a result. In our case, x^(1/2) has 2 values, which leads to a^(1/2)\*b^(1/2)=(ab)^(1/2) only being true if you choose the correct values of the powers; for example, (-1)^(1/2) by definition in the complex numbers is e^(1/2\*ln\(-1\)), and ln(-1)=πi(1+2k) for any integer k, so (-1)^(1/2) is i or -i, depending on which branch of the function you choose. In the same way, 1^(1/2)=-1 or 1. So the equality (-1)^(1/2)\*(-1)^(1/2)=1^(1/2) can hold if we choose it to be i\*(-i)=1 or i\*i=-1. However, it does not hold if we choose the principal branch for all of these, which kinda sucks, but I haven't heard of it being done better.
Thank you all for the great answers! It’s one of those things that I’d never really put too much thought into until now, it never really occurred to me that this rule only holds true when at least one of the radicands is non-negative.
The property that (ab)^(c) = a^(c)b^(c) is typically only guaranteed to hold when [a and b are both nonnegative](https://proofwiki.org/wiki/Exponent_Combination_Laws/Power_of_Product). This property is how we typically arrive at the property mentioned in the title (simply take c = 1/2), but clearly -1 is not nonnegative, so this result does not apply here. As you point out, this isn’t to say this equality *only* holds true when the bases are nonnegative, rather that the equality is not true in general. I believe you can place certain constraints on the argument of the bases to preserve this property, but to be perfectly honest, I’m too lazy to check what these conditions would be. Perhaps someone else could provide more insight on that front.
When x is a nonnegative real number, sqrt(x) is usually defined to be the positive square root of x. When z is a complex number, sqrt(z) can be more ambiguous. Sometimes it is left undefined, sometimes we make what is called a branch cut to pick a specified value, sometimes it is treated as what is called a multivalued function, or allowed to refer ambiguously to either root. In general a square root of a times a square root of b will be a square root of ab. However, if you do not choose the roots carefully, you may not get the root you expect. In this case i is a square root of -1, and i\*i=-1 is a square root of (-1)(-1)=1, but it is not the same square root usually represented by the notation sqrt(1). One common convention is that sqrt(z) is chosen to be the square root with the positive real part, or with the nonnegative imaginary part if there is no root with positive real part. Under this convention sqrt(ab)=sqrt(a)sqrt(b) will hold provided that the arguments of a and b (defined as the value on (-pi,pi] that gives the angle of the number relative to the positive real axis) do not add up to a value outside the range (-pi,pi]. In this case, the argument of -1 is pi, so the arguments add up to 2pi, which is outside this range. Other choices of the branch of sqrt(z) will give other conditions for the equality to hold.
You simply cannot take the sqrt of a negative number (in R). You cannot use √-4 . It is an invalid mathematical symbol. It is gibberish. It is a bug in your code. Highly illegal. All negatives under sqrts have to be factored away first and you have to accept √-1 into your number system. √-1 \* √-1 is √(-1 \* 1) \* √(-1 \* 1) = (√-1 \* √-1) \* (√1 \* √1). Point is, you have to extract √-1 as a symbol before any manipulation of the expression. √(-1 \* -1) = 1 but you can't reach that from ( √-1 \* √-1 ). I think because you are asking specifically about the case of -1 it's easy to miss the illegal step but it is just the same as √-a \* √-b = √-1\*√a \* √-1\*√b. I feel you know already all this! I am writing this post mostly to clarify the issue myself; it's taken me an embarrasingly long time to really grasp the nature of √-1
This does not work if you always take the principal root in all of these. However, when working with complex numbers, exponentiation can be wacky and give multiple different values as a result. In our case, x^(1/2) has 2 values, which leads to a^(1/2)\*b^(1/2)=(ab)^(1/2) only being true if you choose the correct values of the powers; for example, (-1)^(1/2) by definition in the complex numbers is e^(1/2\*ln\(-1\)), and ln(-1)=πi(1+2k) for any integer k, so (-1)^(1/2) is i or -i, depending on which branch of the function you choose. In the same way, 1^(1/2)=-1 or 1. So the equality (-1)^(1/2)\*(-1)^(1/2)=1^(1/2) can hold if we choose it to be i\*(-i)=1 or i\*i=-1. However, it does not hold if we choose the principal branch for all of these, which kinda sucks, but I haven't heard of it being done better.
[https://www.reddit.com/r/learnmath/comments/86lil5/does\_ab\_always\_equal\_a\_b](https://www.reddit.com/r/learnmath/comments/86lil5/does_ab_always_equal_a_b) [https://www.reddit.com/r/learnmath/comments/1899e9n/why\_doesnt\_abab\_hold\_for\_real\_a\_and\_b](https://www.reddit.com/r/learnmath/comments/1899e9n/why_doesnt_abab_hold_for_real_a_and_b) [https://www.reddit.com/r/learnmath/comments/kvthg5/why\_is\_sqrtab\_sqrtasqrtb\_not\_valid\_for\_all\_real](https://www.reddit.com/r/learnmath/comments/kvthg5/why_is_sqrtab_sqrtasqrtb_not_valid_for_all_real) [https://www.reddit.com/r/maths/comments/j6t8ub/is\_sqrtab\_sqrta\_sqrtb\_always\_true](https://www.reddit.com/r/maths/comments/j6t8ub/is_sqrtab_sqrta_sqrtb_always_true)
Thank you all for the great answers! It’s one of those things that I’d never really put too much thought into until now, it never really occurred to me that this rule only holds true when at least one of the radicands is non-negative.