The operation Θ is defined by x Θ y = 1/x + 1/y for all

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 erikvm,

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
_________________

Okay well why isnt it then (1/3) + (1/3) / (-2/3)?

I mean, because 1/3 - 1 = -2/3. So it should be 1/3 + (1/3)*(3/-2)?

hi,
it is so because u have taken z as 3 and not 1/3.... replace 3 for z.. and u will get the answer
_________________

I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong?

https://i.imgur.com/ISdtWL7.png

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
_________________

Thanks man. I get it now, appreciate your patience

Hi Bunuel,

I don't know if that is the ideal place to ask my question, but I haven't found any post gathering the broader topic, i.e Functions / Expressions.

In such exercise, can we consider the assumption right in any case if it did work using 1 number (respecting the constraint given by the wording, obviously). I used 2 for that one, but lost time double checking with 8 and 9, probably a bad reflex acquired after thoroughly testing numbers in DS questions!

Thanks in advance,

RCM

The easiest way to answer this is to use substitution... I used 5 to check
so
I) Z Θ (-Z) = \(\frac{1}{5}\) - \(\frac{1}{5}\) = 0

II) Z Θ \(\frac{Z}{Z-1}\) = \(\frac{1}{5}\) + 1/\(\frac{5}{4}\) = \(\frac{1}{5}\) + \(\frac{4}{5}\) = 1

III) \(\frac{2}{Z}\) Θ \(\frac{2}{Z}\) = \(\frac{5}{2}\)+\(\frac{5}{2}\) = \(\frac{10}{2}\) = 5 =Z

Hence I,II,III are all possible


------------------------------------------------------
Kindly press"+1 Kudos" to appreciate

Hi brunel, could you help expand II and III please ? I did not quite get the calculation? Thanks

A hint for anyone looking at this in 2023 and beyond. It's much easier to solve this question if you think about it this way...
The question is essentially asking you the sum of the reciprocal of each term on either side of the unique symbol. You can skip a few steps if you go straight to writing out the equations.
_________________