0
$\begingroup$

enter image description here

I am having problems with problem (i) shown in the image above.

Examine which of the following is a binary operation: (i) $a*b=\frac{a+b}{2}; a,b\in\mathbb{N}$

Now since $a$ and $b$ are natural numbers, they belong to the set $\mathbb{N}$. However the operation $a*b=(a+b)/2$ may not be a binary operation because for one (either $a$ or $b$) odd and the other even, the operation $a*b$ will not give a natural number. Hence the operation should not be a binary operation in $\mathbb{N}$. But the book has written that it is. So what is correct?

$\endgroup$
14
  • $\begingroup$ But the site said that i can't upload image because my reputation is below 10. @Joel_Reyes_Noche $\endgroup$ Commented Mar 22, 2018 at 10:24
  • $\begingroup$ I have added it. $\endgroup$ Commented Mar 22, 2018 at 10:34
  • $\begingroup$ are you claiming that 3[odd natural number]*2[even natural number]=6[even natural number] isn't a natural number? $\endgroup$
    – nonchip
    Commented Mar 22, 2018 at 10:44
  • $\begingroup$ @nonchip, I think the OP is claiming that $(3+2)/2=5/2$ is not a natural number. Note that the operation $*$ is not multiplication. $\endgroup$
    – JRN
    Commented Mar 22, 2018 at 10:47
  • $\begingroup$ (3+2)÷2 is not a natural number. @nonchip $\endgroup$ Commented Mar 22, 2018 at 10:47

1 Answer 1

2
$\begingroup$

Your reasoning seems to be correct and the book is in error.

$\endgroup$
1
  • $\begingroup$ note this might also be intentional (as a question), to trick people into proving it to see if they realize their disprove is actually correct. but the solution still is in error then. also the book is missing the actual proof in the solution, which is always a bad thing. $\endgroup$
    – nonchip
    Commented Mar 23, 2018 at 2:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .