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The operation Θ is defined by x Θ y = 1/x + 1/y for all
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Updated on: 23 Jul 2018, 22:47
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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. \(z \theta (z) = 0\) II. \(z \theta \frac{z}{(z1)}= 1\) III. \(\frac{2}{z} \theta \frac{2}{z}= z\) A. I only B. I and II only C. I and III only D. II and III only E. I, II, and III
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Originally posted by TOUGH GUY on 28 Sep 2006, 03:17.
Last edited by Bunuel on 23 Jul 2018, 22:47, edited 4 times in total.
Edited the question.



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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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08 Sep 2013, 04:19
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~!!~



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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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08 Sep 2013, 06:48
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. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}= 1\) 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. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}=\frac{1}{z}+\frac{z1}{z}=1\) > 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.
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 08:39
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. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}= 1\) 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. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}=\frac{1}{z}+\frac{z1}{z}=1\) > 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 \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}=\frac{1}{3}+\frac{3}{31}=/=1\)



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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 08:48
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. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}= 1\) 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. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}=\frac{1}{z}+\frac{z1}{z}=1\) > 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 \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}=\frac{1}{3}+\frac{3}{31}=/=1\) hi erikvm, hi option ll says we are adding reciprocals of x and y..... it will not be \(\frac{1}{z}+\frac{z}{(z1)}\) but \(\frac{1}{z}+\frac{(z1)}{z}\) =1/3+2/3=1 hope it helped
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 08:57
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)?



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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 09:04
erikvm wrote: 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
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 09:10
I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong? http://i.imgur.com/ISdtWL7.png



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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 09:17
erikvm wrote: I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong? http://i.imgur.com/ISdtWL7.pngin the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z1)}\)=3/(31)=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
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 09:34
chetan2u wrote: erikvm wrote: I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong? http://i.imgur.com/ISdtWL7.pngin the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z1)}\)=3/(31)=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?



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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 09:47
erikvm wrote: chetan2u wrote: erikvm wrote: I never get the fractions to work so I just uploaded a picture of it instead, where do I go wrong? http://i.imgur.com/ISdtWL7.pngin the attached fig...x Θ y = 1/x + 1/y here x is z and y is\(\frac{z}{(z1)}\)... you have taken z as 3... so x=z=3..and y=\(\frac{z}{(z1)}\)=3/(31)=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 \(\frac{z}{(z1)}\).... since x Θ y = 1/x + 1/y... z Θ \(\frac{z}{(z1)}\)=\(\frac{1}{z}\)+1/\(\frac{z}{(z1)}\)= \(\frac{1}{z}\)+\(\frac{(z1)}{z)}\).... \(\frac{(1+z1)}{z}\)=z/z=1
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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07 Mar 2015, 09:54
Thanks man. I get it now, appreciate your patience



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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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04 Jul 2015, 18:18
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. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}= 1\) 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. \(\hspace{10} z \hspace{2} \theta \hspace{2} \frac{z}{(z1)}=\frac{1}{z}+\frac{z1}{z}=1\) > 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. 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



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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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28 Aug 2015, 05:25
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 \(\frac{Z}{Z1}\) = \(\frac{1}{5}\) + 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
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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01 Nov 2018, 07:56
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Re: The operation Θ is defined by x Θ y = 1/x + 1/y for all
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