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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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closed as too broad by Nautilus, Tyler Seacrest, JonTheMon, Khale_Kitha, Justin Mar 18 '16 at 15:31

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ 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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10
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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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  • 1
    $\begingroup$ 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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1
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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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