Is Ac < Bc Always True? Exploring The 3k-1 Set
Hey everyone, ever stumbled upon a math problem that makes you scratch your head and wonder, "Is this really as simple as it looks?" Well, today, we're diving deep into just such a brain-teaser! We've got this cool set of numbers, A, defined by the rule x = 3k - 1 where k is any integer. Sounds a bit fancy, right? But trust me, it just means these are numbers like -4, -1, 2, 5, and so on. Then, we're asked to check if a specific inequality, ac < bc, is always true whenever we pick three numbers a, b, c from this set A, with the added condition that a is strictly less than b (a < b). This isn't just a random math exercise; it’s a fantastic way to sharpen our critical thinking skills and truly understand how inequalities work, especially when dealing with different types of numbers. We're going to break it all down, explore every nook and cranny of this problem, and by the end, you'll not only know the answer but also understand the why behind it. So, grab your favorite beverage, get comfy, and let's unravel this mathematical mystery together! It's going to be a fun ride through the world of integers and inequalities.
What's Up with Set A? Cracking the x = 3k - 1 Code
Alright, first things first, let's really get to know our main character in this story: Set A. The definition might look a bit intimidating at first glance: A = {x | x = 3k - 1, k ∈ Z}. But don't sweat it, guys! This simply means that A is the collection of all numbers x that can be expressed in the form 3k - 1, where k is any integer. What's an integer, you ask? Think of them as whole numbers, positive, negative, or zero: ..., -3, -2, -1, 0, 1, 2, 3, ... So, to figure out what numbers are actually in Set A, we just need to plug in a few different integer values for k and see what pops out. Let's give it a shot, shall we?
If k = 0, then x = (3 * 0) - 1 = -1. So, -1 is in A. If k = 1, then x = (3 * 1) - 1 = 2. Voila, 2 is in A. If k = 2, then x = (3 * 2) - 1 = 5. Yep, 5 is in A. We can go the other way too! If k = -1, then x = (3 * -1) - 1 = -3 - 1 = -4. Look at that, -4 is in A. And if k = -2, then x = (3 * -2) - 1 = -6 - 1 = -7. You guessed it, -7 is in A. So, our set A looks something like this: ..., -7, -4, -1, 2, 5, 8, ... Notice a pattern? These numbers are all spaced out by 3. More precisely, they are integers that, when divided by 3, leave a remainder of 2 (or equivalently, a remainder of -1, if you're into modular arithmetic). This is a pretty cool and important property of Set A: it contains both positive and negative integers. It also does not contain zero, which will be a significant point later on. Understanding the composition of Set A – that it includes a mix of positive and negative values, and is essentially an arithmetic progression with a common difference of 3 – is absolutely crucial before we even think about tackling the inequality. Without a clear picture of what kind of numbers we're dealing with, any conclusions we draw about ac < bc would just be guesswork. So, take a moment to really internalize this: Set A is full of diverse integers, some negative, some positive, and none of them are zero. This diversity is going to play a starring role in our upcoming investigation.
The Inequality Challenge: ac < bc – What Does It Really Mean?
Now that we're BFFs with Set A, let's turn our attention to the main event: the inequality ac < bc. We're given three elements, a, b, c, all belonging to our special Set A, and we're also told that a is strictly less than b (written as a < b). Our mission, should we choose to accept it, is to figure out if ac < bc holds true every single time under these conditions. This isn't just about plugging in numbers; it's about understanding the fundamental rules of inequalities, especially when multiplication is involved. Many of us remember the basics from school: if you add or subtract the same number from both sides, the inequality sign stays the same. But what about multiplication and division? That's where things get a little trickier, and where this problem really shines a spotlight on common misconceptions.
Let's break down the inequality ac < bc. Our first instinct might be to divide both sides by c. However, this is where we need to pump the brakes! When you divide (or multiply) an inequality by a number, the direction of the inequality sign can change, and it all depends on the sign of that number. Specifically: if you multiply or divide by a positive number, the inequality sign stays the same. But, and this is a big but, if you multiply or divide by a negative number, you must flip the direction of the inequality sign. For example, we know that 2 < 4. If we multiply by 3 (positive), we get 6 < 12, which is true. But if we multiply 2 < 4 by -1 (negative), we get -2 and -4. Now, is -2 < -4? Absolutely not! Instead, -2 > -4. See? The sign flipped! This fundamental rule is the absolute cornerstone of understanding our problem.
To avoid accidentally dividing by zero (which is undefined, remember?) or flipping signs incorrectly, a safer and often clearer approach is to rearrange the inequality. Let's subtract bc from both sides: ac - bc < 0. Now, we can factor out c from the left side, giving us c(a - b) < 0. This form is super powerful because it simplifies the problem. We already know one crucial piece of information: a < b. What does this tell us about (a - b)? If a is less than b, then when you subtract b from a, the result will always be a negative number. For example, if a=2 and b=5, then a-b = 2-5 = -3. If a=-4 and b=-1, then a-b = -4 - (-1) = -3. So, we've established that (a - b) is always negative under our given conditions. This means our inequality boils down to c * (negative number) < 0. For this product to be less than zero (i.e., negative), what must be true about c? Well, if you multiply a negative number by a negative number, you get a positive result (e.g., -2 * -3 = 6). If you multiply a negative number by a positive number, you get a negative result (e.g., -2 * 3 = -6). Therefore, for c * (negative number) to be less than zero, c must be a positive number. This is the ultimate logical deduction from the rearranged inequality. If c is positive, then (positive c) * (negative (a-b)) will indeed be negative, satisfying c(a-b) < 0. But what if c is negative? Then (negative c) * (negative (a-b)) would be positive, making c(a-b) > 0, which contradicts our desired c(a-b) < 0. This is the critical insight we need to explore further using examples from Set A.
Let's Get Real: Testing ac < bc with Examples from Set A
Alright, theory is great, but nothing beats getting our hands dirty with some actual numbers! We've established that the inequality ac < bc (which we simplified to c(a - b) < 0) will only hold true if c is a positive number, since (a - b) is always negative. Now, let's pick some numbers from our Set A (remember, that's ..., -7, -4, -1, 2, 5, 8, ...) and see what happens when c takes on different signs. This practical testing is how we really uncover the truth behind whether ac < bc is always true.
Case 1: When c is Positive (The "Holds True" Scenario)
Let's start with a scenario where c is positive. Set A contains positive numbers like 2 and 5. Let's choose c = 2 for our first test. Now, we need to pick a and b from Set A such that a < b. How about a = -1 and b = 5? These are both in Set A, and -1 < 5 is definitely true.
So, with a = -1, b = 5, and c = 2, let's check ac < bc:
acbecomes(-1) * (2) = -2bcbecomes(5) * (2) = 10
Is -2 < 10? Absolutely! It holds true in this instance. The inequality stands strong because c was a positive number, allowing the direction of the inequality to remain unchanged. Let's try another one. What if a = -4 and b = -1 (both from A, and -4 < -1), and c = 5 (also from A, and positive)?
acbecomes(-4) * (5) = -20bcbecomes(-1) * (5) = -5
Is -20 < -5? You bet it is! Again, the inequality holds. These examples reinforce our understanding: when c is positive, multiplying both sides of a < b by c maintains the inequality direction, resulting in ac < bc. This is exactly what we deduced from our algebraic rearrangement. It seems like everything is going smoothly when c is on the sunny side of the number line. If Set A only contained positive numbers, or if the problem stated c > 0, then the answer would indeed be a straightforward