Ahsome
  • Ahsome
A simple * question
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
whats the question
Ahsome
  • Ahsome
Develop a proof to show that \[a ° b =\frac{a+b}{ab}\]Has no identity (i.e. \(a ° b = a\)).
Ahsome
  • Ahsome
Any clue?

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Ahsome
  • Ahsome
Here is what I got
wolf1728
  • wolf1728
Is that ever true? Let's use some real numbers: 7 * 3 = (7 + 3) / 7*3 7*3 =? 10/21 21 =? 10/21
anonymous
  • anonymous
Assume that \[a*b=a\] then, \[\frac{ab}{a+b}=a\]\[\frac{1}{a}+\frac{1}{b}=a\]\[\frac{1}{b}=a-\frac{1}{a}\]\[\frac{1}{b}=\frac{a^2-1}{a}\]\[b=\frac{a}{a^2-1}\] Here b would be the identity element, for which all a*b=a however this is not true as at a=1 b is not defined, therefore there is no identity element
wolf1728
  • wolf1728
Why do we assume that a*b = a?
anonymous
  • anonymous
If our assumption is correct, then we'd get an expression for b which is defined for all a, but since b is not defined at a=1, our assumption is incorrect and there's no identity element
anonymous
  • anonymous
for all a belonging to the required set
Ahsome
  • Ahsome
Ahh, that make's sense @Nishant_Garg. Thank you! :D. Also, from my textbook, it says that there can only be one identity element. Since in this case, the identity element would change (the squared would affect it), is that another reason aswell?
Ahsome
  • Ahsome
@Nishant_Garg, the \(ab\) should be at the bottom, not the top
anonymous
  • anonymous
oh wow my bad haha
Ahsome
  • Ahsome
Yeah, that changes a bit ;)
anonymous
  • anonymous
but it should be something like that if u just solve it
Ahsome
  • Ahsome
I don't know how to rearrange to get something similar, can you help @Nishant_Garg?
anonymous
  • anonymous
ill try
Ahsome
  • Ahsome
Thanks so much @Nishant_Garg :D
anonymous
  • anonymous
ok there's only 1 mistake, i've written \[\frac{ab}{a+b}\] instead of \[\frac{a+b}{ab}\] but the rest of the calculation is ok As for the identity element, yes there's only one identity element...so about that I'm also confused O.o
Ahsome
  • Ahsome
Maybe, what we are meant to do is to sub in some value for \(a\), and find the corresponding \(b\) vaule. Then, substitute anoter \(a\) value. If the \(b\) values don't match, then there is no identity element, @Nishant_Garg?
anonymous
  • anonymous
yep, and clearly they won't match! and it's not even defined for some numbers like +-1, so there r definitely many factors to it...
Ahsome
  • Ahsome
Yeah. Thanks so much @Nishant_Garg :D

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