\(f(x) = \sqrt{x+1/x-1} g(x) = \sqrt{x+1} / \sqrt{x-1}\)

explain why f and g are different

Guest Mar 13, 2017

#9**+16 **

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 ?

Guest Mar 15, 2017

#1**+3 **

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.

Pasplox
Mar 13, 2017

#2**+10 **

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

.

Alan
Mar 13, 2017

#3**+10 **

**Alan I have a query??**

\(f(x) = \sqrt{\frac{x+1}{x-1}}\qquad g(x) =\frac{ \sqrt{x+1} }{ \sqrt{x-1}}\\ if\;\;x=-5\\ f(-5) = \sqrt{\frac{-4}{-6}}\\ f(-5) = \sqrt{\frac{4}{6}}\\ f(-5) =\frac{2}{\sqrt{6}}\qquad \text{Definitely real}\\~\\ g(-5) =\frac{ \sqrt{-4} }{ \sqrt{-6}}\\ g(-5) =\frac{ 2i }{ i\sqrt{6}}\\ \text{Can't I cancel out the i's and have }\frac{2}{\sqrt{6}}???\)

Melody
Mar 14, 2017

#6**+5 **

ok thanks Alan

Now I will look at a value of x between -1 and 1

\(f(x) = \sqrt{\frac{x+1}{x-1}}\qquad g(x) =\frac{ \sqrt{x+1} }{ \sqrt{x-1}}\\ if\;\;x=0\\ f(0) = \sqrt{\frac{1}{-1}}\\ f(0) = \sqrt{-1}\\ f(0) =i \\~\\ g(0) =\frac{ \sqrt{1} }{ \sqrt{-1}}\\ g(0) =\frac{ 1}{i}\\ =\frac{ i}{-1}\\ =-i\\ so\\ f(0)\ne g(0)\)

Melody
Mar 14, 2017

#4**0 **

This question is also answered here :

http://web2.0calc.com/questions/is-there-a-difference-if-we-plug-in-any-number-numerically-into

Melody
Mar 14, 2017

#7**0 **

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).

Guest Mar 15, 2017

#9**+16 **

Best Answer

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 ?

Guest Mar 15, 2017

#10**+5 **

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: \(\sqrt{\frac{1}{-1}}\rightarrow\sqrt{(\frac{1}{-1})}\rightarrow\sqrt{-1}=i\)

.

Alan
Mar 16, 2017

#12**0 **

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 \(\frac{1}{-1} \quad \text{does equal } \;\;\frac{-1}{1}\)

Melody
Mar 17, 2017

#13**0 **

It's to do with PEMDAS.

\(\sqrt{\frac{1}{-1}}\rightarrow (\frac{1}{-1})^{1/2}\)

Parentheses first, then exponent.

.

Alan
Mar 17, 2017

#14**0 **

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. ............

Melody
Mar 17, 2017