\(f(x) = \sqrt{x+1/x-1} g(x) = \sqrt{x+1} / \sqrt{x-1}\)
explain why f and g are different
So how would you resolve this problem ?
\(\displaystyle \sqrt{-1}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{\imath} \\ \displaystyle\sqrt{-1}=\sqrt{\frac{-1}{1}}=\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\imath}{1}\)
Equating and cross multiplying,
\(\displaystyle \imath^{2}=1.\)
At what point has a mistake been made ?
+44 f is different from g because in f, the square root of the result of the expression is taken, where as in g, the root of x+ 1 and x-1 are divided by each other rather than the result.
+33661 When x>1 f(x) = g(x) and both are real
When -1 < x < 1 both f(x) and g(x) are imaginary
When x< -1 f(x) is real but g(x) is imaginary
.
+118687 Alan I have a query??
\(f(x) = \sqrt{\frac{x+1}{x-1}}\qquad g(x) =\frac{ \sqrt{x+1} }{ \sqrt{x-1}}\\ if\;\;x=-5\\ f(-5) = \sqrt{\frac{-4}{-6}}\\ f(-5) = \sqrt{\frac{4}{6}}\\ f(-5) =\frac{2}{\sqrt{6}}\qquad \text{Definitely real}\\~\\ g(-5) =\frac{ \sqrt{-4} }{ \sqrt{-6}}\\ g(-5) =\frac{ 2i }{ i\sqrt{6}}\\ \text{Can't I cancel out the i's and have }\frac{2}{\sqrt{6}}???\)
+118687 ok thanks Alan
Now I will look at a value of x between -1 and 1 
\(f(x) = \sqrt{\frac{x+1}{x-1}}\qquad g(x) =\frac{ \sqrt{x+1} }{ \sqrt{x-1}}\\ if\;\;x=0\\ f(0) = \sqrt{\frac{1}{-1}}\\ f(0) = \sqrt{-1}\\ f(0) =i \\~\\ g(0) =\frac{ \sqrt{1} }{ \sqrt{-1}}\\ g(0) =\frac{ 1}{i}\\ =\frac{ i}{-1}\\ =-i\\ so\\ f(0)\ne g(0)\)
+118687 This question is also answered here :
http://web2.0calc.com/questions/is-there-a-difference-if-we-plug-in-any-number-numerically-into
Every number has two square roots.
\(\displaystyle \sqrt{-1}= \pm\imath\)
If you choose to say that f(0) = i,
then you are making a choice.
Ditto with g(x).
So how would you resolve this problem ?
\(\displaystyle \sqrt{-1}=\sqrt{\frac{1}{-1}}=\frac{\sqrt{1}}{\sqrt{-1}}=\frac{1}{\imath} \\ \displaystyle\sqrt{-1}=\sqrt{\frac{-1}{1}}=\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\imath}{1}\)
Equating and cross multiplying,
\(\displaystyle \imath^{2}=1.\)
At what point has a mistake been made ?
+33661 The mistake occurs in the first line: \(\sqrt{\frac{1}{-1}}\ne\frac{\sqrt1}{\sqrt{-1}}\)
Your subsequent development shows that equality here leads to an inconsistency.
We need to look at this as follows: \(\sqrt{\frac{1}{-1}}\rightarrow\sqrt{(\frac{1}{-1})}\rightarrow\sqrt{-1}=i\)
.
+118687 Can I have some clarification on that please Alan?
Are you saying that you cannot take the squareroot of a negative number unless the negative sign is in the numerator?
Because obviously \(\frac{1}{-1} \quad \text{does equal } \;\;\frac{-1}{1}\)
+33661 It's to do with PEMDAS.
\(\sqrt{\frac{1}{-1}}\rightarrow (\frac{1}{-1})^{1/2}\)
Parentheses first, then exponent.
.
+118687 ok thanks Alan but that still means that
\(\sqrt{-\frac{1}{2}} \;\;\;\text{Must be interpreted as }\;\;\sqrt{\frac{-1}{2}}\)
I suppose if you think of it as \(\sqrt{-0.5}\)
the problem is fixed, but that still means the - sign must be in the numerator.
Mmmm still thinking. ............
