Decoding The Equation: |S₁| + |S₂| + ... + |Sₘ|

Hey guys! Ever stumbled upon a math problem that looks like it's written in another language? Well, today we're going to tackle one of those head-scratchers and turn it into something crystal clear. We're diving into the world of sets, fractional parts, and a cool equation that ties them all together. So, buckle up, put on your thinking caps, and let's get started!

Unraveling the Problem: A Step-by-Step Guide

Our mission, should we choose to accept it (and we do!), is to prove a fascinating identity. It revolves around sets, which are just collections of things (in our case, numbers), and a special function called the fractional part. The identity states that the sum of the number of elements in these sets, up to a certain point, equals a neat expression involving ceiling functions.

Defining Our Players: Sets and Fractional Parts

First, let's break down the key players in our problem. We're dealing with sets labeled as $S_n$, where 'n' is any positive integer (that's what $\mathbb N^*$ means). Each set $S_n$ contains numbers 'x' that live in the interval [0, 1). This interval includes 0 but doesn't include 1. The condition for a number 'x' to be in $S_n$ is a bit quirky: it involves the fractional part of 'nx'.

The fractional part, denoted by curly braces like this {nx}, is simply the decimal part of a number. For example, the fractional part of 3.14 is 0.14, and the fractional part of 7 is 0 (since there's no decimal part). So, the condition $x{nx} = \frac{1}{2}$ means that when you multiply 'x' by 'n' and take the fractional part, then multiply that result by x, you get exactly one-half. This is the core concept to understand.

Now, let's introduce the notation $|S_n|$. The vertical bars around $S_n$ mean "the number of elements in the set $S_n$". So, $|S_n|$ is simply a count of how many numbers 'x' satisfy the condition $x{nx} = \frac{1}{2}$ within the interval [0, 1).

The Grand Equation: Our Target

The equation we're aiming to prove is:

$|S_1| + |S_2| + ... + |S_m| = \left[\frac{m+1}{2}\right] \left[\frac{m+2}{2}\right]$

On the left side, we have the sum of the sizes of the sets $S_1$, $S_2$, up to $S_m$. The right side looks a bit mysterious with those square brackets. Those are called ceiling functions. The ceiling function, denoted by $\lceil x \rceil$ (or in our case, using square brackets), gives you the smallest integer that is greater than or equal to 'x'. For example, $\lceil 3.14 \rceil = 4$, $\lceil 7 \rceil = 7$, and $\lceil 2.99 \rceil = 3$.

So, the right side of the equation involves plugging $\frac{m+1}{2}$ and $\frac{m+2}{2}$ into the ceiling function and then multiplying the results. This equation is the heart of our problem, and we need to show that the left side always equals the right side, no matter what positive integer 'm' we choose.

Why This Matters: The Significance of the Problem

Before we dive into the proof, let's take a moment to appreciate why this problem is interesting. It beautifully connects a few different mathematical ideas: sets, fractional parts, and ceiling functions. It's a great example of how seemingly simple concepts can lead to surprising and elegant results. Problems like these often pop up in areas like number theory and analysis, and mastering them can really sharpen your mathematical intuition.

Cracking the Code: The Proof Unveiled

Alright, enough talk! Let's get our hands dirty and prove this thing. The proof might seem a bit daunting at first, but we'll break it down into manageable chunks. The key idea is to analyze the condition $x{nx} = \frac{1}{2}$ carefully and figure out what it tells us about the possible values of 'x'.

Unpacking the Condition: $x{nx} = \frac{1}{2}$

Let's focus on the core condition: $x{nx} = \frac{1}{2}$. Remember that 'x' is a number between 0 and 1, and 'n' is a positive integer. The fractional part {nx} is also a number between 0 and 1. So, we're multiplying two numbers between 0 and 1 and getting exactly one-half. This gives us a crucial clue about the possible values of 'x' and {nx}.

Let's rewrite the fractional part nx} in a more useful way. We know that any number 'nx' can be written as the sum of its integer part (let's call it 'k') and its fractional part $nx = k + {nx$. Here, 'k' is a non-negative integer (0, 1, 2, ...). Now we can express the fractional part as ${nx} = nx - k$. Substituting this into our condition, we get:

$x(nx - k) = \frac{1}{2}$

This is a quadratic equation in 'x'! Let's rearrange it to make it look more familiar:

$nx^2 - kx - \frac{1}{2} = 0$

Solving the Quadratic: Finding Possible Values of 'x'

Now we have a quadratic equation, and we know how to solve those! We can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, a = n, b = -k, and c = -\frac{1}{2}. Plugging these values into the quadratic formula, we get:

$x = \frac{k \pm \sqrt{k^2 + 2n}}{2n}$

This formula gives us two possible solutions for 'x' for each value of 'n' and 'k'. However, remember that 'x' must be between 0 and 1. This constraint will help us narrow down the valid solutions. Also, since x is positive, we can disregard the solution with the minus sign in the quadratic formula, so:

$x = \frac{k + \sqrt{k^2 + 2n}}{2n}$

The Integer 'k': A Key Insight

The integer 'k' plays a crucial role here. Since 0 ≤ x < 1, we have:

$0 \le \frac{k + \sqrt{k^2 + 2n}}{2n} < 1$

Multiplying by 2n, we get:

$0 \le k + \sqrt{k^2 + 2n} < 2n$

Subtracting k, we have:

$-k \le \sqrt{k^2 + 2n} < 2n - k$

Since the square root is always non-negative, the left inequality is always true as long as k is non-negative, which it is. Squaring the right inequality, we get:

$k^2 + 2n < (2n - k)^2 = 4n^2 - 4nk + k^2$

Simplifying, we obtain:

$2n < 4n^2 - 4nk$

Dividing by 2n (since n is positive), we have:

$1 < 2n - 2k$

Rearranging, we get:

$2k < 2n - 1$

Finally, dividing by 2, we get:

$k < n - \frac{1}{2}$

Since 'k' is an integer, this inequality tells us that k can take values from 0 up to n - 1. This is a critical result because it tells us how many possible values of 'k' we need to consider for each 'n'.

Counting the Solutions: Finding |Sₙ|

For each value of 'n', we have 'n' possible values of 'k' (0, 1, 2, ..., n-1). Each of these values of 'k' gives us a unique solution for 'x' using the formula:

$x = \frac{k + \sqrt{k^2 + 2n}}{2n}$

Therefore, the number of elements in the set $S_n$, denoted by $|S_n|$, is simply 'n'.

Summing It Up: Proving the Identity

Now we're ready to tackle the main equation. We want to show that:

$|S_1| + |S_2| + ... + |S_m| = \left[\frac{m+1}{2}\right] \left[\frac{m+2}{2}\right]$

We know that $|S_n| = n$, so the left side of the equation becomes:

$1 + 2 + 3 + ... + m$

This is just the sum of the first 'm' positive integers, which has a well-known formula:

$\frac{m(m+1)}{2}$

So, we need to show that:

$\frac{m(m+1)}{2} = \left[\frac{m+1}{2}\right] \left[\frac{m+2}{2}\right]$

This is where things get a little tricky with the ceiling functions. We need to consider two cases: when 'm' is even and when 'm' is odd.

Case 1: 'm' is even

If 'm' is even, we can write it as m = 2p, where 'p' is an integer. Then the right side of the equation becomes:

$\left[\frac{2p+1}{2}\right] \left[\frac{2p+2}{2}\right] = \left[p + \frac{1}{2}\right] [p + 1] = (p + 1)(p + 1) = (p+1)^2$

And the left side becomes:

$\frac{2p(2p+1)}{2} = p(2p+1)$ which is incorrect! So our previous logic was flawed and |S_n| is not equal to n. This will require further analysis using number theory principles to correctly calculate |S_n| and prove the identity.

We need to go back to the equation $x = \frac{k + \sqrt{k^2 + 2n}}{2n}$ and condition $k < n - \frac{1}{2}$ more carefully.

The original condition was $x \{nx\} = \frac{1}{2}$. Let $nx = k + f$ where $k$ is an integer and $f = \{nx\}$ is the fractional part. Then $x f = \frac{1}{2}$ and $f = \frac{1}{2x}$. Substituting into $nx = k + f$, we get $nx = k + \frac{1}{2x}$. Multiplying by $2x$, we have $2nx^2 = 2kx + 1$, so $2nx^2 - 2kx - 1 = 0$. Solving for $x$ using the quadratic formula:

$x = \frac{2k \pm \sqrt{4k^2 + 8n}}{4n} = \frac{k \pm \sqrt{k^2 + 2n}}{2n}$.

Since $x$ must be positive, we have $x = \frac{k + \sqrt{k^2 + 2n}}{2n}$. Since $0 \le x < 1$, we must have $0 < k + \sqrt{k^2 + 2n} < 2n$, so $\sqrt{k^2 + 2n} < 2n - k$. Squaring gives $k^2 + 2n < 4n^2 - 4nk + k^2$, so $2n < 4n^2 - 4nk$, or $1 < 2n - 2k$, which means $2k < 2n - 1$, or $k < n - \frac{1}{2}$. Therefore, $k$ can take integer values from 0 to $n - 1$. For each such $k$, we have a potential solution, but we must make sure that the fractional part condition is satisfied. So for a given integer $n$, $k$ ranges from 0 to n-1. So there are floor(n/2) + 1 solutions when n is even and floor(n/2) + 1 when n is odd. This means |S_n| = d(2n), the number of divisors of 2n.

We have $\sum_{n=1}^m |S_n| = \sum_{n=1}^m d(2n)$. The original identity is equivalent to showing $\sum_{n=1}^m d(2n) = [\frac{m+1}{2}][\frac{m+2}{2}]$.

This result requires a more advanced number-theoretic argument to establish. We have shown the initial steps and where the difficulty arises.

Key Takeaways and Next Steps

We've journeyed through a fascinating problem involving sets, fractional parts, and ceiling functions. We learned how to break down a complex condition, solve a quadratic equation, and use inequalities to narrow down possible solutions. We discovered that proving the final identity requires a deeper dive into number theory and a more sophisticated counting argument.

Final Thoughts

Math problems like these are like puzzles. They might seem intimidating at first, but with careful analysis and a bit of creativity, you can unlock their secrets. The journey itself is just as important as the solution, as it helps you develop your problem-solving skills and mathematical intuition.