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My friend taught me the followings without his memory of the answer:

He said that it is known that there exists an numerical expression which satisfies the following two conditions :

Condition 1 : The expression is represented only by "$x, y, 0, 1, +, -, \times, \div$".(each can be used multiple times.)

Condition 2 : The expression can represent each of $$x+y, x-y, -x+y, -x-y, x\times y, -x\times y, x\div y, -x\div y, y\div x, -y\div x$$ if we use some brackets appropriately.

Suppose that $xy$ does not mean $x\times y$, which means that multiplication always needs "$\times$".

I've been trying to find the answer, but I'm facing difficulty. Can anyone help?

Example : By the way, $-0-x+y$ can represent each of $x+y, x-y, -x+y, -x-y$ if we use some brackets appropriately : $$-(0-x)+y=x+y, -(0-x+y)=x-y,$$$$ (-0-x)+y=-x+y, -0-(x+y)=-x-y.$$

mathlove
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2 Answers2

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$$0 \times 1 + f(x,y)$$ can be written as $$(0 \times 1) + f(x,y) = f(x,y)$$ or as $$0 \times (1 + f(x,y)) = 0$$ so just string a sum of a large number of these together for all the possible $f(x,y)$ and switch on the ones you want.

Henry
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  • Thank you for pointing it out. Your ieda is very nice. But in my opinion, the interesting point of this question seems to be to find a specific example. – mathlove Oct 03 '13 at 07:41
  • $ 0 \times 1 + x+y+ 0 \times 1 + x-y+ 0 \times 1 -x+y+ 0 \times 1 -x-y+ 0 \times 1 + x\times y+ 0 \times 1 -x\times y+ 0 \times 1 + x\div y+ 0 \times 1 -x\div y+ 0 \times 1 + y\div x+ 0 \times 1 -y\div x$ – Henry Oct 03 '13 at 07:43
  • Oh, I see what you mean. Now you solved my question. – mathlove Oct 03 '13 at 07:46
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I've just got the following example :

$$-0+1\div1\times x-0\div 1\times 0+y$$

I think this is an example. You'll know that if you observe it.

mathlove
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