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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]
01 Nov 2014, 04:00

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The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]
07 Mar 2015, 07:39

Bunuel wrote:

rskanumuri wrote:

Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!.. So either D or E! cheers~!!~

Edited the question.

The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) = 0\)

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}= 1\)

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= z\)

A. I only B. I and II only C. I and III only D. II and III only E. I, II, and III

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) =\frac{1}{z}-\frac{1}{z}=0\) --> true.

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{z}+\frac{z-1}{z}=1\) --> true.

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= \frac{z}{2}+\frac{z}{2}=z\) --> true.

Answer: E.

I don't understand the algebraic form for option "II". I mean, lets say z = 3. Then I'd get \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{3}+\frac{3}{3-1}=/=1\)

Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all [#permalink]
07 Mar 2015, 07:48

erikvm wrote:

Bunuel wrote:

rskanumuri wrote:

Can anyone verify the question? I think that first option should be z Θ (-z) =0 for this to be true!.. So either D or E! cheers~!!~

Edited the question.

The operation Θ is defined by x Θ y = 1/x + 1/y for all nonzero numbers x and y. If z is a number greater than 1, which of the following must be true?

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) = 0\)

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}= 1\)

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= z\)

A. I only B. I and II only C. I and III only D. II and III only E. I, II, and III

I. \(\hspace{10} z \hspace{2} \theta \hspace{2} (-z) =\frac{1}{z}-\frac{1}{z}=0\) --> true.

II. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{z}+\frac{z-1}{z}=1\) --> true.

III. \(\hspace{10} \frac{2}{z} \hspace{2} \theta \hspace{2} \frac{2}{z}= \frac{z}{2}+\frac{z}{2}=z\) --> true.

Answer: E.

I don't understand the algebraic form for option "II". I mean, lets say z = 3. Then I'd get \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z-1)}=\frac{1}{3}+\frac{3}{3-1}=/=1\)

hi option ll says we are adding reciprocals of x and y..... it will not be \(\frac{1}{z}+\frac{z}{(z-1)}\) but \(\frac{1}{z}+\frac{(z-1)}{z}\) =1/3+2/3=1 hope it helped

in the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z-1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2... 1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1 you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong

in the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z-1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2... 1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1 you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong

Im sorry man, I'm sure you are really annoyed, but I really dont understand.. What values am I suppose to take?

in the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z-1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z-1)}\)=3/(3-1)=3/2... 1/x + 1/y=1/3+1/(3/2)=1/3+2/3=1 you are taking z as 3 for x and z as 1/3 for y.. that is why u are going wrong

Im sorry man, I'm sure you are really annoyed, but I really dont understand.. What values am I suppose to take?

does not matter... look dont take any values and work with z itself.. z Θ \(\frac{z}{(z-1)}\).... since x Θ y = 1/x + 1/y... z Θ \(\frac{z}{(z-1)}\)=\(\frac{1}{z}\)+1/\(\frac{z}{(z-1)}\)= \(\frac{1}{z}\)+\(\frac{(z-1)}{z)}\).... \(\frac{(1+z-1)}{z}\)=z/z=1

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