NetNews Usenet Archive 1992 #30/NN_1992_30.iso/spool/sci/math/17029

Internet Message Format | 1992-12-16 | 2.5 KB

Path: sparky!uunet!pipex!unipalm!uknet!comlab.ox.ac.uk!mbeattie
From: [e-mail] (Malcolm Beattie)
Newsgroups: sci.math
Subject: Re: '-' operation
Message-ID: <[e-mail]>
Date: 16 Dec 92 16:07:54 GMT
References: <[e-mail]>
Organization: Oxford University Computing Service, 13 Banbury Rd, Oxford, U
Lines: 43
Originator: mbeattie@black

In article <[e-mail]> [e-mail] writes:
>Recently my daughter brought back Problem Set II of Wisconsin Mathematics
>Science and Engineering Talent Search, and Question 4 in it is as following:
>         Operation @ satisfies the conditions that
>                X @ (Y @ Z) = X @ Y + Z and X @ X = 0 for any real numbers
>          X, Y, Z. Show that @ must be subtraction.
>
>   Since my daughter does not care much whether she has talent so I tried to
> solve it myself:-). It took me more than an hour but still I am not sure I get
> a right answer. The problem I have is: How is operation '-' defined? How about
> other operations like '+', '*', and '/'? Although I do not have Ph.D
> in math but I consider myself one of the best in learning math in those
> math classes. It is sort of ashameful that I never learned or remember
> how the basic operation is defined. For the above question, is only
> thing I have to prove is X @ 0 = X?, which is not very difficult because
> X @ 0 = X @ (X @ X) = X @ X + X = 0 + X = X.
>
> I hope no one will blame me for bringing 'homework' to the net or trying
> to be 'talent' since the deadline was passed and it is already too late
> to send the answer in for that purpose:-). I just like to know what you
> have to do to say some operation MUST be one of the basic operations.
>
> Duane

Just show that x - y = x @ y for all x, y \in R.
Your conditions are that, for all reals x, y:
(1)   x @ (y @ z) = x @ y + z
(2)   x @ x = 0
Notice that x @ y + z brackets means (x @ y) + z
or else the question is wrong.
Take x,y \in R.
(x @ y) + y = x @ (y @ y)          by (1)
            = x @ (x @ x)          since y @ y = 0 = x @ x by (2)
            = (x @ x) + x          by (1)
            = x                    since x @ x = 0
Subtract y from both sides for your result.

--Malcolm
--
Malcolm Beattie <[e-mail]> | I'm not a kernel hacker
Oxford University Computing Services      | I'm a kernel hacker's mate
13 Banbury Road, Oxford, OX2 6NN (U.K.)   | And I'm only hacking kernels
Tel: [phone] Fax: [phone]   | 'Cos the kernel hacker's late