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# help plz

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

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
Sort:

#1
+44
+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
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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
+91229
+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
#5
+26357
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Oops!  Yes you can.

Alan  Mar 14, 2017
#6
+91229
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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
#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
#8
+26357
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No.  If $$x^2=-1$$ then $$x=\pm \sqrt{-1}$$

However $$\sqrt{-1} = i$$

Alan  Mar 15, 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
#10
+26357
+5

The mistake occurs in the first line:  $$\sqrt{\frac{1}{-1}}\ne\frac{\sqrt1}{\sqrt{-1}}$$

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
+91229
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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
+26357
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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
+91229
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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
#11
+26357
+5

f and g differ between -1<x<1:

.

Alan  Mar 16, 2017

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