Rational Numbers Proof: Show √((1+a^2)(1+b^2)(1+c^2)) ∈ Q

Hey guys! Let's dive into a cool math problem today. We've got three rational numbers – let's call them a, b, and c. There's a catch, though: they have to satisfy the equation ab + bc + ca = 1. Our mission, should we choose to accept it, is to show that ${\sqrt{(1+a^2)(1+b^2)(1+c^2)}}$ is also a rational number. Sounds like a challenge? Let's break it down together! This problem is a fantastic example of how seemingly complex expressions can be simplified with the right approach. So grab your thinking caps, and let's get started!

Breaking Down the Problem: Initial Steps

Before we get into the nitty-gritty of the proof, let's take a step back and examine what we're working with. The key here is to leverage the given condition, ab + bc + ca = 1, to manipulate the expression under the square root. Our goal is to somehow rewrite (1 + a²), (1 + b²), and (1 + c²) in a way that allows us to eliminate the square root. To make our lives easier, let's start by focusing on individual terms and see how we can incorporate the given condition. This initial exploration is crucial because it sets the stage for the entire proof. Think of it as laying the foundation for a building – a solid foundation ensures a stable structure. We'll be using algebraic manipulation, which is basically like a mathematical puzzle where we rearrange terms to fit our needs. So, let’s start by looking at (1 + a²) and see how we can rewrite it using the fact that ab + bc + ca = 1. This is where the fun begins, guys!

Rewriting (1 + a²)

Okay, let’s tackle the first part: rewriting (1 + a²). This is where our condition ab + bc + ca = 1 comes into play. We can substitute 1 in the expression (1 + a²) with ab + bc + ca. This might seem like a small step, but it's a crucial one. So, (1 + a²) becomes (ab + bc + ca + a²). Now, let's see if we can spot any opportunities for factoring. Factoring is like finding the building blocks of an expression, and it often helps us simplify things. Looking at our new expression, (ab + bc + ca + a²), we can group the terms and factor out common factors. If we group a² and ab together, and bc and ca together, we can factor out an a from the first group and a c from the second group. This gives us a(a + b) + c(a + b). Do you see what's happening? We now have a common factor of (a + b) in both terms. Factoring this out, we get (a + b)(a + c). Voila! We’ve successfully rewritten (1 + a²) as (a + b)(a + c). This is a significant step forward, guys! We've taken a seemingly simple expression and transformed it into something that involves the other variables, b and c. This is exactly what we need to do to link all three terms together. Now, let’s do the same for (1 + b²) and (1 + c²). Are you ready for the next step?

Rewriting (1 + b²) and (1 + c²)

Now that we've conquered (1 + a²), let’s apply the same strategy to (1 + b²) and (1 + c²). Remember, the key is to use the condition ab + bc + ca = 1 to our advantage. Just like before, we'll substitute 1 with ab + bc + ca. So, for (1 + b²), we get (ab + bc + ca + b²). Can you already see the pattern? We're going to group and factor again. Grouping b² with bc and ab with ca, we can factor out a b from the first group and a c from the second group. This gives us b(b + c) + a(b + c). And just like before, we have a common factor: (b + c). Factoring this out, we get (b + c)(b + a), which we can also write as (a + b)(b + c). See? It's all coming together nicely! Now, let’s tackle (1 + c²). Substituting 1 with ab + bc + ca, we get (ab + bc + ca + c²). Grouping c² with bc and ca with ab, we can factor out a c from the first group and an a from the second group. This gives us c(c + b) + a(c + b). And once again, we have a common factor: (c + b), which is the same as (b + c). Factoring this out, we get (b + c)(a + c) or (a + c)(b + c). We've now successfully rewritten (1 + a²), (1 + b²), and (1 + c²) in terms of factored expressions. This is a huge win, guys! We’re one step closer to proving that the entire expression under the square root is a perfect square. Let's recap what we've got so far:

• (1 + a²) = (a + b)(a + c)
• (1 + b²) = (a + b)(b + c)
• (1 + c²) = (a + c)(b + c)

Now, let’s put these pieces together and see what happens when we multiply them.

Putting it All Together: Multiplying the Expressions

Alright, we've done the groundwork, and now it's time for the grand finale of the algebraic manipulation! We've rewritten (1 + a²), (1 + b²), and (1 + c²) as factored expressions. Now, we need to multiply them together and see if we can simplify the result. Remember, we're trying to show that ${\sqrt{(1+a^2)(1+b^2)(1+c^2)}}$ is a rational number. This means we need to prove that (1 + a²)(1 + b²)(1 + c²) is a perfect square. Let's substitute our factored expressions into the product: (1 + a²)(1 + b²)(1 + c²) = (a + b)(a + c) * (a + b)(b + c) * (a + c)(b + c). Now, let's rearrange the terms to make the pattern clearer. We can group the identical factors together: =(a + b)(a + b) * (a + c)(a + c) * (b + c)(b + c). This can be written as: =(a + b)² * (a + c)² * (b + c)². Boom! Look at that, guys! We have a product of squares. This is fantastic news because it means that the entire expression is a perfect square. Remember, our original expression was ${\sqrt{(1+a^2)(1+b^2)(1+c^2)}}$. Now we know that (1 + a²)(1 + b²)(1 + c²) = (a + b)²(a + c)²(b + c)². So, taking the square root of both sides, we get: ${\sqrt{(1+a^2)(1+b^2)(1+c^2)} = \sqrt{(a + b)^2 (a + c)^2 (b + c)^2}}$. This simplifies to: ${\|(a + b)(a + c)(b + c)\|}$. But wait, we're not quite done yet. We need to show that this result is a rational number. Let's tackle that next!

Showing the Result is Rational

Okay, we've shown that ${\sqrt{(1+a^2)(1+b^2)(1+c^2)} = |(a + b)(a + c)(b + c)|}$. The final piece of the puzzle is to demonstrate that this expression is indeed a rational number. This might seem straightforward, but it's crucial to connect all the dots in our proof. Remember, we were given that a, b, and c are rational numbers. What does this tell us about the sum and product of these numbers? Well, the sum of two rational numbers is always rational, and the product of rational numbers is also always rational. This is a fundamental property of rational numbers that we can rely on. So, if a and b are rational, then (a + b) is rational. Similarly, if a and c are rational, then (a + c) is rational, and if b and c are rational, then (b + c) is rational. Now we have three rational numbers: (a + b), (a + c), and (b + c). When we multiply these three rational numbers together, the result is also a rational number. This is because the product of rational numbers is always rational. Therefore, (a + b)(a + c)(b + c) is a rational number. Taking the absolute value, |(a + b)(a + c)(b + c)|, doesn't change the fact that it's rational because the absolute value of a rational number is also a rational number. And there we have it, guys! We've successfully shown that ${\sqrt{(1+a^2)(1+b^2)(1+c^2)}}$ is a rational number. We started with the condition ab + bc + ca = 1, rewrote the expressions under the square root, multiplied them together, and finally showed that the result is rational. This entire journey showcases the power of algebraic manipulation and logical deduction in mathematics.

Conclusion: A Satisfying Proof

Wow, what a journey! We've successfully proven that given rational numbers a, b, and c such that ab + bc + ca = 1, the expression ${\sqrt{(1+a^2)(1+b^2)(1+c^2)}}$ is indeed a rational number. This problem is a beautiful illustration of how seemingly complex mathematical statements can be unraveled with a systematic approach. We leveraged the given condition, applied clever algebraic manipulations, and used the fundamental properties of rational numbers to arrive at our conclusion. This type of problem not only tests our algebraic skills but also our ability to think logically and connect different mathematical concepts. Remember guys, math is not just about memorizing formulas; it's about understanding the underlying principles and using them to solve problems creatively. This proof is a testament to that. So, keep exploring, keep questioning, and keep challenging yourselves with mathematical puzzles. You never know what amazing discoveries you might make! And that’s a wrap for today’s mathematical adventure. Hope you enjoyed the ride!