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Bob has been given the following question during a test:

   x is a number.  
   x is half of 2(x+1).  
   What is x?  

Can you help Bob to solve this?

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    $\begingroup$ Are we allowed cyclic groups? What is your definition of number? $\endgroup$ – Lacklub Mar 18 '16 at 14:53
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    $\begingroup$ My smartass answer to the puzzle: "A number." $\endgroup$ – Khale_Kitha Mar 18 '16 at 14:54
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To answer the question:

Can you help Bob to solve it?

No.


Here's the explanation of why:

Bob has been given the below question during a test:

Assisting a student with a problem during a test is cheating.

But if I were to attempt to solve the system of equations:

x is a number.
x is half of 2(x+1).
What is x?

x is a number that is half of 2(x+1). Without further definition of which operators are used, and what x represents in context no further progress may be made without assumptions.

See other answers for following this system of equations with the standard definitions used for the given operators.

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x is $1$
x is a number (check)
1 is half of 2(which is 1+1) - we all assume it's multiplication, but parentheses can also be used to clarify

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how about

x = 2

because

"2" is half of "23" which is "2" "2+1".

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    $\begingroup$ On the same line of thought: x=8, 2(x+1) = 18, half of which is 8 $\endgroup$ – Lacklub Mar 18 '16 at 15:14
  • $\begingroup$ @Lacklub - yep. I'm sure that some interpretation like this is what Alex is after, but unfortunately, there are too many possibilities that equally match the information given. So it is a shot in the dark what was intended. $\endgroup$ – Paul Sinclair Mar 18 '16 at 15:19
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    $\begingroup$ Or 3, which is half (vertically) of 8. $\endgroup$ – f'' Mar 18 '16 at 16:01
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I think the answer might be

1 or 3

Reasoning

Assuming that "x is a number" implies that x is not $\infty$ then we must subscribe to a different interpretation of the second line. Given that "2(x+1)" is six characters long, it may be that we form x using three characters from that expression.

Valid solutions then for x are "(x)", "(1)" or "2+1" (given that ordering is important and any bracket must have a match). The first solution is too general to be able to determine x directly so we'll take it that it's one of the other two.

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I'm not sure about this, but why not:

0 or 2

because

2(x+1) comes out to 2x+2. Thinking laterally, half of that could be 2x, or half could be 2. For 2x, we have x = 2x, which works for only 0. 2 is self-explanatory.

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x is

Infinity. Because Inf+1 = Inf.

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  • $\begingroup$ infinity is not a number, though $\endgroup$ – Khale_Kitha Mar 18 '16 at 14:52
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    $\begingroup$ @Khale_Kitha - sure infinity is a number. It just isn't a real number (that is, an element of the set of real numbers). It is a element of the set of extended real numbers, or the set of hyperreal numbers or the set of surreal numbers, or a host of other definitions of sets of numbers. $\endgroup$ – Paul Sinclair Mar 18 '16 at 15:15
  • $\begingroup$ @Paul: Those aren't "infinity" though. Infinity is the quality of not being finite. You can have infinite numbers, but you'd need to point to a specific one, such as $\aleph_0$ or $\omega$. $\endgroup$ – Deusovi Mar 25 '16 at 3:08
  • $\begingroup$ @Deusovi - let's not play semantic games, please. My point to Khale_Kitha is that it is completely valid to think of infinity as a number (but not as a Real number). I also pointed out that there are varying definitions of the concept, so you can't just assume a particular one. All of the definitions I mentioned have the quality of not being finite. $\endgroup$ – Paul Sinclair Mar 25 '16 at 12:46
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Well, it's quite simple. x is :

impossible

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It's a computing test. 'is' is the code to assign a string to a variable, so x="half of 2(x+1)" (a string).

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$\frac{2 * (x + 1)}{2} = x$
$2*x + 2 = 2 * x$
$ 0 =2 $ So not possible unless x is infinity or -infinity

Or

$\frac{2 * (x + 1)}{2} = x$
$x + 1 = x$
$ 1 =0 $

Same thing as above.

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