Okay, let me try to figure out this problem. The question is asking for the smallest number that is one less than twice its reciprocal. Hmm, let's break this down step by step.

First, let's understand what the problem is saying. We need to find a number, let's call it x, such that when you take twice its reciprocal and subtract one, you get the number itself. So, in mathematical terms, that would be 2*(1/x) - 1 = x. Right? Let me write that equation down:

2*(1/x) - 1 = x

Now, I need to solve for x. Let's rearrange this equation to make it easier to handle. Maybe first, I can eliminate the fraction by multiplying every term by x. Let's try that:

Multiply both sides by x: 2*(1/x)*x - 1*x = x*x

Simplify each term:

2 - x = x²

Now, let's bring all terms to one side to form a quadratic equation. Subtract 2 from both sides and add x to both sides:

0 = x² + x - 2

Wait, let me check that again. If I have 2 - x = x², then moving everything to the right side gives x² + x - 2 = 0. Yeah, that's correct. So, the quadratic equation is x² + x - 2 = 0.

Alright, now I need to solve this quadratic equation. Let's see if it factors. The quadratic is x² + x - 2. Let's look for two numbers that multiply to -2 and add to 1. Hmm, 2 and -1: 2*(-1) = -2 and 2 + (-1) = 1. Perfect. So, the equation factors as:

(x + 2)(x - 1) = 0

Setting each factor equal to zero gives the solutions:

x + 2 = 0 => x = -2

x - 1 = 0 => x = 1

So, the solutions are x = -2 and x = 1. The problem asks for the smallest number, so between -2 and 1, the smallest is -2. Wait, but let me verify if this is correct because sometimes when dealing with reciprocals, especially negative numbers, things can get tricky.

Let me plug x = -2 back into the original equation to check:

Left side: 2*(1/x) - 1 = 2*(1/(-2)) - 1 = 2*(-1/2) - 1 = -1 - 1 = -2

Right side: x = -2

Yes, that works. Now check x = 1:

Left side: 2*(1/1) - 1 = 2*1 - 1 = 2 - 1 = 1

Right side: x = 1

That also works. So both solutions are valid. The problem asks for the smallest number, which is -2. But wait, let me think again. Is -2 indeed smaller than 1? Yes, because in the number line, -2 is to the left of 1. So, -2 is smaller.

But hold on, the problem says "the smallest number which is one less than twice its reciprocal." Let me just confirm once again with x = -2. Twice the reciprocal of -2 is 2*(1/(-2)) = -1. Then, one less than that is -1 -1 = -2, which is equal to x. So that's correct. Similarly for x = 1: twice reciprocal is 2*1 = 2, then one less is 1, which is x. So both are correct.

Therefore, the smallest number is -2. But wait, the problem says "the smallest number," but sometimes in math problems, especially when dealing with reciprocals, there might be constraints. For example, the reciprocal of x is 1/x, so x cannot be zero. But in this case, both solutions are non-zero, so that's fine. Also, since the problem doesn't specify that the number has to be positive, -2 is a valid solution.

But let me think again: the problem is asking for the smallest number. Since numbers can be negative, and -2 is smaller than 1, that's the answer. However, sometimes problems might expect positive numbers unless specified otherwise. Wait, the original problem doesn't specify that the number has to be positive. It just says "the smallest number," so negative numbers are allowed. Therefore, -2 is indeed the correct answer.

But let me check if there's any other possible approach. For example, maybe setting up the equation differently. Let's say the number is x. Then twice its reciprocal is 2*(1/x). One less than that is 2*(1/x) - 1. And this is equal to x. So, 2/x - 1 = x. Then, same as before, multiply both sides by x: 2 - x = x², leading to x² + x - 2 = 0. Same quadratic. So, same solutions. So, no alternative here.

Alternatively, maybe using substitution. Let me let y = 1/x. Then the equation becomes 2y - 1 = x. But since y = 1/x, substitute back: 2y - 1 = 1/y. Multiply both sides by y: 2y² - y = 1. Then, 2y² - y - 1 = 0. Solving this quadratic: discriminant is 1 + 8 = 9. So, y = [1 ± 3]/4. Thus, y = (1 + 3)/4 = 1 or y = (1 - 3)/4 = -0.5. Then, since y = 1/x, x = 1/y. So, if y = 1, x = 1; if y = -0.5, x = -2. Same solutions. So, same answer.

Therefore, the smallest number is -2. So, the answer should be \boxed{-2}.

Wait, but hold on a second. Let me just confirm once again to make sure I didn't make a mistake in translation. The problem says "one less than twice its reciprocal." So, twice the reciprocal is 2*(1/x), and one less than that is 2*(1/x) - 1. Then, the number is equal to that. So, x = 2*(1/x) - 1. Then, multiplying both sides by x: x² = 2 - x. Then, x² + x - 2 = 0. Yep, that's correct. So, the quadratic is correct, solutions are correct, and the smallest is -2. Therefore, the answer is -2.

But just to be absolutely thorough, let's consider if there's any other interpretation of the problem. For example, maybe "one less than twice its reciprocal" could be interpreted as two times the reciprocal minus one, but that's exactly how I interpreted it. So, 2*(1/x) - 1. If it were something else, like reciprocal of (twice the number minus one), that would be different. But the problem says "one less than twice its reciprocal," which clearly is 2*(1/x) - 1. So, that's correct.

Alternatively, if someone misreads it as "twice the reciprocal of one less than the number," that would be 2*(1/(x - 1)), which is different. But the problem says "twice its reciprocal," so "its" refers to the number itself. Then, "one less than that" would be 2*(1/x) - 1. So, that's correct.

Another possible confusion could be the order of operations, but the phrasing is "one less than twice its reciprocal," which is unambiguous: first take the reciprocal, multiply by two, then subtract one. So, no ambiguity there.

Therefore, after thorough consideration, the answer is indeed -2. But just to recap, let's check the two solutions again:

For x = 1:

Left side: 2*(1/1) - 1 = 2 - 1 = 1. Which equals x. Correct.

For x = -2:

Left side: 2*(1/(-2)) - 1 = -1 -1 = -2. Which equals x. Correct.

So, both solutions are valid, and since -2 is smaller than 1, that's the answer. Therefore, the smallest number is -2.
Therefore, the smallest number which is one less than twice its reciprocal is $\boxed{-2}$.
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