Exploring Sum Of Set Cardinalities Proving The Floor Function Formula

Hey guys! Ever stumbled upon a math problem that just makes you scratch your head? Well, today we're diving deep into one of those intriguing problems that involves sets, fractional parts, and floor functions. Specifically, we're going to explore the sum of the cardinalities of some special sets, and trust me, it's a fascinating journey. The problem we're tackling involves natural numbers, sets defined by fractional parts, and a cool formula involving floor functions. So, buckle up, grab your thinking caps, and let's get started!

In this article, we will break down the problem step by step, making sure everyone, from math enthusiasts to those just curious about the topic, can follow along. We'll start by understanding the basic definitions and notations, then we'll gradually build our way up to the solution. Along the way, we'll throw in some examples and intuitive explanations to make things crystal clear. By the end of this exploration, you'll not only understand the solution but also appreciate the beauty and elegance of mathematical reasoning. So, let’s jump right into it and unravel this mathematical gem together!

Let’s get the problem statement crystal clear right off the bat. For every natural number n (denoted as n ∈ ℕ*), we define a set S_n. This set S_n consists of all real numbers x within the interval [0, 1) that satisfy a particular condition. This condition involves the fractional part of nx, which we denote as {nx}. Specifically, x belongs to S_n if and only if the product of x and its fractional part of nx equals 1/2. Mathematically, we can write this as:

S_n = {x ∈ [0, 1) | x{nx} = 1/2}

Now, we let |S_n| represent the number of elements in the set S_n, which is also known as the cardinality of S_n. Our main goal is to show that the sum of these cardinalities, from |S₁| up to |S_m|, can be expressed by a formula involving floor functions. The formula we aim to prove is:

|S₁| + |S₂| + ... + |S_m| = ⌊(m + 1) / 2⌋ * ⌊(m + 2) / 2⌋

where ⌊x⌋ denotes the floor function, which gives the greatest integer less than or equal to x. This is quite the equation, isn't it? But don’t worry, we’ll break it down and understand each part thoroughly. We'll need to dive into the properties of fractional parts, floor functions, and set theory to fully appreciate and solve this problem. So, let’s get started by exploring some of the fundamental concepts we'll need along the way.

Before we dive into the solution, let’s make sure we’re all on the same page with some key concepts. Understanding these definitions will make the problem much easier to tackle. Think of it as gathering the right tools for the job – you wouldn’t try to build a house without a hammer and nails, right? Similarly, we need these concepts to solve our mathematical puzzle.

First up, let's talk about the fractional part of a number. The fractional part of a real number x, denoted as {x}, is the part of the number that remains after you subtract its integer part. In other words, if you write x as the sum of an integer and a fraction between 0 and 1 (including 0 but not including 1), that fraction is the fractional part. Mathematically, {x} = x - ⌊x⌋, where ⌊x⌋ is the floor function. For example, if x = 3.14, then {x} = 0.14, and if x = -2.7, then {x} = 0.3. The fractional part is always a non-negative number less than 1.

Next, let's clarify the floor function. The floor function of a real number x, denoted as ⌊x⌋, gives the greatest integer that is less than or equal to x. Essentially, it rounds the number down to the nearest integer. For instance, ⌊3.14⌋ = 3, ⌊-2.7⌋ = -3, and ⌊5⌋ = 5. The floor function is crucial in many areas of mathematics, including number theory and computer science, and it plays a starring role in our problem today.

Lastly, let's touch on set cardinality. The cardinality of a set is simply the number of elements it contains. If a set S has a finite number of elements, we denote its cardinality as |S|. For example, if S = {1, 2, 3}, then |S| = 3. In our problem, |S_n| represents the number of elements in the set S_n, which is the set of real numbers x in the interval [0, 1) that satisfy the condition x{nx} = 1/2.

With these concepts in hand, we're well-equipped to start digging into the problem. Understanding fractional parts, floor functions, and set cardinality is the first step towards unraveling this mathematical challenge. So, let's move on and see how we can apply these ideas to find a solution.

Alright, guys, now that we have a solid grasp of the key concepts, let's start breaking down the problem into manageable chunks. Think of it like tackling a big jigsaw puzzle – you wouldn’t try to assemble it all at once, right? You'd sort the pieces, find the edges, and work on smaller sections first. Similarly, we’re going to dissect our problem and identify the key steps we need to take to reach the solution.

The main goal is to show that the sum of the cardinalities |S₁| + |S₂| + ... + |S_m| is equal to ⌊(m + 1) / 2⌋ * ⌊(m + 2) / 2⌋. To get there, we need to figure out how to find |S_n| for a given n. Remember, S_n is the set of x in [0, 1) such that x{nx} = 1/2. This equation involving the fractional part is the heart of the matter.

Step 1: Understanding the Equation x{nx} = 1/2

This equation tells us that the product of x and the fractional part of nx is exactly 1/2. Since x is in the interval [0, 1), this means that {nx} must be greater than or equal to 1/2 (because if {nx} were less than 1/2, their product would be less than 1/2). This gives us a crucial constraint on the possible values of {nx}.

Step 2: Expressing {nx} in terms of nx

Recall that the fractional part of a number nx is given by {nx} = nx - ⌊nx⌋. Let's substitute this into our equation:

x(nx - ⌊nx⌋) = 1/2

This equation now involves both x and ⌊nx⌋, which is an integer. Let's denote ⌊nx⌋ by k, where k is a non-negative integer. This substitution transforms our equation into a more manageable form.

Step 3: Solving for x in terms of k

Now we have x(nx - k) = 1/2. We can rearrange this equation to solve for x:

nx² - kx = 1/2

nx² - kx - 1/2 = 0

This is a quadratic equation in x. We can use the quadratic formula to find the solutions for x in terms of k and n. This will give us a way to express x explicitly.

Step 4: Finding the Range of k

Since x is in the interval [0, 1), we need to make sure that the solutions we find for x also fall within this interval. Additionally, since {nx} must be greater than or equal to 1/2, we can use this condition to determine the possible range of values for k. This is a critical step because it will tell us how many possible solutions for x there are for each n.

Step 5: Counting the Solutions |S_n|

For each n, we will have a range of possible values for k. Each value of k will give us a potential solution for x. However, we need to make sure that these solutions are valid, meaning they satisfy the condition x ∈ [0, 1) and {nx} ≥ 1/2. By counting the valid solutions, we can find |S_n|.

Step 6: Summing the Cardinalities

Finally, once we know how to find |S_n| for each n, we can sum these cardinalities from n = 1 to m. This will give us the sum |S₁| + |S₂| + ... + |S_m|. Our goal is to show that this sum is equal to ⌊(m + 1) / 2⌋ * ⌊(m + 2) / 2⌋. This might involve some clever algebraic manipulation and the use of properties of the floor function.

So, there you have it – our roadmap to solving the problem! We've broken it down into six key steps, each building on the previous one. Now, let's roll up our sleeves and start working through these steps one by one. Next up, we'll dive into the quadratic equation and start solving for x. Let's do this!

Okay, guys, let's tackle that quadratic equation we arrived at in the previous section. Remember, we transformed our original equation x{nx} = 1/2 into a quadratic equation by letting k = ⌊nx⌋. The quadratic equation we need to solve is:

nx² - kx - 1/2 = 0

This looks like a job for the quadratic formula! The quadratic formula is our trusty tool for finding the roots of any quadratic equation in the form ax² + bx + c = 0. The formula is:

x = (-b ± √(b² - 4ac)) / (2a)

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

x = (k ± √((k)² - 4n(-1/2))) / (2n)

Simplifying this expression, we have:

x = (k ± √(k² + 2n)) / (2n)

So, we have two potential solutions for x: one with the plus sign and one with the minus sign. Let's denote them as x₁ and x₂:

x₁ = (k + √(k² + 2n)) / (2n)

x₂ = (k - √(k² + 2n)) / (2n)

Now, we need to analyze these solutions and determine which ones are valid. Remember, we are looking for solutions x that lie in the interval [0, 1). Also, we need to consider the condition that {nx} ≥ 1/2. Let's start by looking at x₂.

Notice that since √(k² + 2n) is always greater than k, the solution x₂ will always be negative. To see this, consider that √(k² + 2n) > √(k²) = |k| ≥ k. Therefore, k - √(k² + 2n) is negative, and so is x₂. Since we are only interested in x values in the interval [0, 1), we can discard the solution x₂.

This leaves us with only one candidate solution:

x₁ = (k + √(k² + 2n)) / (2n)

We now need to determine the range of values for k that will give us valid solutions for x₁ in the interval [0, 1). This is our next step, and it will help us count the number of elements in the set S_n. So, let's move on and figure out the possible values of k.

Alright, let's zoom in on finding the valid range for k. Remember, k is an integer, specifically k = ⌊nx⌋, and we have our solution for x:

x₁ = (k + √(k² + 2n)) / (2n)

We know that x must be in the interval [0, 1), so we have the condition:

0 ≤ (k + √(k² + 2n)) / (2n) < 1

Let's tackle this inequality. First, since √(k² + 2n) is always positive, (k + √(k² + 2n)) is also positive for non-negative k. Thus, the left-hand side of the inequality, 0 ≤ (k + √(k² + 2n)) / (2n), is automatically satisfied for k ≥ 0. So, we just need to focus on the right-hand side:

(k + √(k² + 2n)) / (2n) < 1

Multiplying both sides by 2n, we get:

k + √(k² + 2n) < 2n

Now, let's isolate the square root:

√(k² + 2n) < 2n - k

Since both sides are positive (we'll address this in a moment), we can square both sides without changing the inequality:

k² + 2n < (2n - k)²

k² + 2n < 4n² - 4nk + k²

Now, we can simplify and rearrange the terms:

2n < 4n² - 4nk

4nk < 4n² - 2n

Dividing both sides by 4n (since n is a positive integer), we get:

k < n - 1/2

Since k is an integer, this inequality means that k can take values from 0 up to ⌊n - 1/2⌋, which is the same as n - 1. So, we have an upper bound for k.

But we also need to consider the condition that {nx} ≥ 1/2. Let's write this condition in terms of x and k:

{nx} = nx - ⌊nx⌋ = nx - k ≥ 1/2

Substituting our solution for x:

n((k + √(k² + 2n)) / (2n)) - k ≥ 1/2

(k + √(k² + 2n)) / 2 - k ≥ 1/2

Multiplying both sides by 2:

k + √(k² + 2n) - 2k ≥ 1

√(k² + 2n) ≥ k + 1

Squaring both sides (again, both sides are positive):

k² + 2n ≥ (k + 1)²

k² + 2n ≥ k² + 2k + 1

2n ≥ 2k + 1

k ≤ n - 1/2

Since k is an integer, this means k ≤ ⌊n - 1/2⌋ = n - 1. This confirms our previous upper bound for k.

From the inequality √(k² + 2n) < 2n - k, we implicitly assumed that 2n - k > 0. If 2n - k ≤ 0, then the inequality would not hold. The condition 2n - k > 0 is equivalent to k < 2n, which is always true since we already found that k ≤ n - 1. So, this condition doesn't add any further restrictions.

However, from √(k² + 2n) ≥ k + 1, we derived k ≤ n - 1/2. For the square root to be greater than or equal to k + 1, we need k + 1 to be non-negative, which means k ≥ -1. But since k represents ⌊nx⌋ and x is in [0, 1), k must be a non-negative integer. Thus, k ≥ 0.

Combining these results, we find that the possible values for k are integers in the range 0 ≤ k ≤ n - 1. This means there are n possible values for k for each n. Great! We're getting closer to finding |S_n|. Now, let's count the solutions.

Okay, let's get down to counting the solutions, which means finding |S_n|. We've already done a lot of the groundwork. We know that for each n, the possible values of k are the integers in the range 0 ≤ k ≤ n - 1. That gives us n possible values for k. For each value of k, we have a potential solution for x given by:

x = (k + √(k² + 2n)) / (2n)

Now, we need to check whether each of these values of k actually gives us a valid solution. Remember, we need to satisfy two conditions:

1. x ∈ [0, 1), which we already ensured when we derived the range for k.
2. {nx} ≥ 1/2, which we also used to find the range for k.

So, it seems like every value of k in the range 0 ≤ k ≤ n - 1 should give us a valid solution for x. However, we need to be a bit careful here. We need to make sure that for each k, the corresponding x actually satisfies the condition k = ⌊nx⌋. This is the crucial step that ties everything together.

Let's substitute our expression for x into ⌊nx⌋ and see what happens:

⌊nx⌋ = ⌊n((k + √(k² + 2n)) / (2n))⌋

⌊nx⌋ = ⌊(k + √(k² + 2n)) / 2⌋

We want to show that this is equal to k. In other words, we want to show:

⌊(k + √(k² + 2n)) / 2⌋ = k

This is equivalent to the inequality:

k ≤ (k + √(k² + 2n)) / 2 < k + 1

Let's multiply all parts of the inequality by 2:

2k ≤ k + √(k² + 2n) < 2k + 2

Subtract k from all parts:

k ≤ √(k² + 2n) < k + 2

The left-hand side, k ≤ √(k² + 2n), is always true since n is a positive integer. So, we just need to focus on the right-hand side:

√(k² + 2n) < k + 2

Square both sides:

k² + 2n < k² + 4k + 4

2n < 4k + 4

n < 2k + 2

n - 2 < 2k

k > (n - 2) / 2

So, we have a new condition on k: k > (n - 2) / 2. Since k is an integer, this is equivalent to k ≥ ⌊(n - 2) / 2⌋ + 1. This means that not all values of k in the range 0 ≤ k ≤ n - 1 will give us a valid solution. We need to count the values of k that satisfy both 0 ≤ k ≤ n - 1 and k ≥ ⌊(n - 2) / 2⌋ + 1.

The number of such k values is given by:

(n - 1) - (⌊(n - 2) / 2⌋) = n - 1 - ⌊(n - 2) / 2⌋

This is the number of valid solutions for x, which is the cardinality of S_n. Therefore,

|S_n| = n - 1 - ⌊(n - 2) / 2⌋

Now we have an explicit formula for |S_n|. Let's simplify this further. We can rewrite ⌊(n - 2) / 2⌋ as ⌊n/2 - 1⌋. If n is even, say n = 2p, then ⌊n/2 - 1⌋ = p - 1. If n is odd, say n = 2p + 1, then ⌊n/2 - 1⌋ = ⌊p + 1/2 - 1⌋ = p - 1. So, in both cases, we have:

|S_n| = n - 1 - ⌊n/2 - 1⌋

Now, let's substitute the two cases for n.

If n = 2p, then |S_n| = 2p - 1 - (p - 1) = p. Since n = 2p, we have p = n/2, so |S_n| = n/2.

If n = 2p + 1, then |S_n| = 2p + 1 - 1 - (p - 1) = p + 1. Since n = 2p + 1, we have p = (n - 1) / 2, so |S_n| = (n - 1) / 2 + 1 = (n + 1) / 2.

In summary:

|S_n| = n/2 if n is even

|S_n| = (n + 1) / 2 if n is odd

Notice that |S_n| = ⌊(n + 1) / 2⌋. This is a much simpler form, and it gives us the number of elements in S_n directly. We're one step closer to our final result! Now, let's sum these cardinalities and see if we can arrive at the desired formula.

Okay, folks, we've reached the final leg of our journey! We've found a nice, neat formula for |S_n|:

|S_n| = ⌊(n + 1) / 2⌋

Now, our mission is to show that the sum of these cardinalities, from n = 1 to m, is equal to ⌊(m + 1) / 2⌋ * ⌊(m + 2) / 2⌋. Let's write out the sum we need to evaluate:

∑_n=1^m |S_n| = ∑_n=1^m ⌊(n + 1) / 2⌋

This looks like a daunting sum, but let's break it down by considering two cases: when m is even and when m is odd. This approach often helps with sums involving floor functions.

Case 1: m is even

Let m = 2p, where p is an integer. Then, our sum becomes:

∑_n=1^2p ⌊(n + 1) / 2⌋ = ⌊2/2⌋ + ⌊3/2⌋ + ⌊4/2⌋ + ⌊5/2⌋ + ... + ⌊(2p + 1) / 2⌋

Let's group the terms in pairs:

= (⌊2/2⌋ + ⌊3/2⌋) + (⌊4/2⌋ + ⌊5/2⌋) + ... + (⌊(2p)/2⌋ + ⌊(2p + 1) / 2⌋)

= (1 + 1) + (2 + 2) + ... + (p + p)

= 2 * (1 + 2 + ... + p)

The sum of the first p integers is p(p + 1) / 2, so:

= 2 * (p(p + 1) / 2)

= p(p + 1)

Since m = 2p, we have p = m/2. Substituting this back into our expression, we get:

= (m/2) * (m/2 + 1)

= (m/2) * ((m + 2) / 2)

= m(m + 2) / 4

Now, let's look at the right-hand side of what we want to prove:

⌊(m + 1) / 2⌋ * ⌊(m + 2) / 2⌋ = ⌊(2p + 1) / 2⌋ * ⌊(2p + 2) / 2⌋

= ⌊p + 1/2⌋ * ⌊p + 1⌋

= p * (p + 1)

= (m/2) * (m/2 + 1)

= m(m + 2) / 4

So, in the case where m is even, the formula holds true!

Case 2: m is odd

Let m = 2p + 1, where p is an integer. Then, our sum becomes:

∑_n=1^{2p + 1} ⌊(n + 1) / 2⌋ = ⌊2/2⌋ + ⌊3/2⌋ + ⌊4/2⌋ + ... + ⌊(2p + 2) / 2⌋

This is the same sum as in the even case, but with one extra term: ⌊(2p + 2) / 2⌋ = p + 1. So, we can write:

= ∑_n=1^2p ⌊(n + 1) / 2⌋ + ⌊(2p + 2) / 2⌋

= p(p + 1) + (p + 1) (from our result in the even case)

= (p + 1)(p + 1)

= (p + 1)²

Since m = 2p + 1, we have p = (m - 1) / 2. Substituting this back into our expression, we get:

= (((m - 1) / 2) + 1)²

= (((m + 1) / 2))²

Now, let's look at the right-hand side of what we want to prove:

⌊(m + 1) / 2⌋ * ⌊(m + 2) / 2⌋ = ⌊((2p + 1) + 1) / 2⌋ * ⌊((2p + 1) + 2) / 2⌋

= ⌊p + 1⌋ * ⌊p + 3/2⌋

= (p + 1) * (p + 1)

= (p + 1)²

= (((m - 1) / 2) + 1)²

= ((m + 1) / 2)²

So, in the case where m is odd, the formula also holds true!

We've shown that the formula holds for both even and odd values of m. Therefore, we have proven that:

∑_n=1^m |S_n| = ⌊(m + 1) / 2⌋ * ⌊(m + 2) / 2⌋

for any natural number m. 🎉

And there you have it, guys! We've successfully navigated through this intricate mathematical problem, exploring the fascinating interplay between sets, fractional parts, and floor functions. We started with a seemingly complex equation and, step by step, broke it down into manageable pieces. From understanding the key concepts to solving the quadratic equation, determining the range of k, counting the solutions, and finally summing the cardinalities, we've seen how each step builds upon the previous one to lead us to the final proof.

This journey highlights the beauty and power of mathematical reasoning. By carefully analyzing the problem, breaking it into smaller parts, and applying the right tools and techniques, we were able to unravel the mystery and arrive at a satisfying conclusion. The formula we proved, |S₁| + |S₂| + ... + |S_m| = ⌊(m + 1) / 2⌋ * ⌊(m + 2) / 2⌋, not only answers the specific problem but also gives us a deeper appreciation for the elegance and interconnectedness of mathematical concepts.

Whether you're a seasoned math enthusiast or just starting to explore the wonders of mathematics, problems like this remind us that with patience, persistence, and a dash of curiosity, even the most challenging puzzles can be solved. So, keep exploring, keep questioning, and keep pushing the boundaries of your mathematical understanding. Who knows what amazing discoveries you'll make along the way? Keep up the great work, and happy problem-solving!