Absolute Values & Square Roots: Proving Integer Results

Hey guys! Today, we're diving deep into some cool math problems involving absolute values and square roots. We'll be tackling problems where we need to show that certain expressions result in integers under specific conditions. Get ready to sharpen your math skills!

Understanding the Basics

Before we jump into the problems, let's refresh our understanding of absolute values and square roots. The absolute value of a number is its distance from zero, always a non-negative value. For example, |3| = 3 and |-3| = 3. The square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because 3 * 3 = 9.

Absolute Value: A Quick Recap

So, when we see |x|, we're essentially asking, "How far is x from zero?" This means that |x| will always be either x (if x is positive or zero) or -x (if x is negative). This simple concept is super important for solving problems involving absolute values. Remember, the absolute value makes everything non-negative, which is why we often see it used when we're talking about distances or magnitudes. When tackling problems with absolute values, consider breaking them down into cases based on the sign of the expression inside the absolute value bars. This approach will make the problem much more manageable.

Square Roots: More Than Just Numbers

Now, let’s talk about square roots. The square root of a number, denoted by √x, is a value that, when multiplied by itself, equals x. But there’s a catch! When dealing with real numbers, we typically focus on the principal (positive) square root. This is why √9 is 3, not -3, even though (-3) * (-3) also equals 9. Understanding this convention is key to avoiding confusion. Square roots are often found in geometric problems (think Pythagorean theorem) and algebraic equations. They can sometimes look intimidating, but with a systematic approach, you can conquer them. Don't forget to consider the domain of the square root function – we can't take the square root of a negative number in the realm of real numbers!

Problem 1: Absolute Values and Integer Results

Let's start with our first problem, which involves proving that an expression with absolute values results in an integer. This type of problem often requires us to consider different cases based on the values inside the absolute value symbols.

Part (a): a = |x + 3| + |x - 5| for x ∈ (-3; 5)

Problem Statement:

Given a = |x + 3| + |x - 5|, show that a ∈ N (a is a natural number) for any x ∈ (-3; 5).

Solution:

The key here is to realize that the interval x ∈ (-3; 5) gives us crucial information about the signs of the expressions inside the absolute values. Let's break it down:

1. x + 3: Since x > -3, then x + 3 > 0. This means |x + 3| = x + 3.
2. x - 5: Since x < 5, then x - 5 < 0. This means |x - 5| = -(x - 5) = -x + 5.

Now, let's substitute these into our expression for 'a':

a = (x + 3) + (-x + 5) a = x + 3 - x + 5 a = 8

Since 8 is a natural number, we've successfully shown that a ∈ N for any x in the interval (-3; 5).

Key Takeaway: When dealing with absolute values, always consider the intervals where the expressions inside the absolute values change signs. This often simplifies the problem dramatically. By carefully analyzing the conditions given, we were able to remove the absolute value signs and arrive at a simple integer value for 'a'. Remember, breaking down the problem into manageable cases is a powerful technique in math!

Part (b): a = |x + y + 6| + |x + y - 10| for x ∈ (-2; 7) and y ∈ R

Problem Statement:

Given a = |x + y + 6| + |x + y - 10|, show that a ∈ N for any x ∈ (-2; 7) and y ∈ R (y is any real number).

Solution:

This problem is a bit trickier because we have two variables, x and y. However, the core strategy remains the same: we need to analyze the expressions inside the absolute values.

Let's think about the possible cases for x + y:

1. Consider x + y + 6: We don't have enough information to definitively say whether x + y + 6 is always positive or always negative. It depends on the specific values of x and y.
2. Consider x + y - 10: Similarly, we can't determine the sign of x + y - 10 without more information.

Instead of trying to find specific intervals, let's use a clever trick. We'll introduce a new variable, say z = x + y. This simplifies our expression to:

a = |z + 6| + |z - 10|

Now, let's consider the number line. We have two critical points: z = -6 and z = 10. These are the points where the expressions inside the absolute values change signs. We'll analyze three cases:

1. Case 1: z < -6

   • |z + 6| = -(z + 6) = -z - 6
   • |z - 10| = -(z - 10) = -z + 10
   • a = -z - 6 - z + 10 = -2z + 4

2. Case 2: -6 ≤ z ≤ 10

   • |z + 6| = z + 6
   • |z - 10| = -(z - 10) = -z + 10
   • a = z + 6 - z + 10 = 16

3. Case 3: z > 10

   • |z + 6| = z + 6
   • |z - 10| = z - 10
   • a = z + 6 + z - 10 = 2z - 4

Notice something amazing? In Case 2, a = 16, which is definitely a natural number! But what about Cases 1 and 3? We need to show that they also result in natural numbers.

Let's analyze the given conditions: x ∈ (-2; 7) and y ∈ R. This means that x can take any value between -2 and 7, and y can be any real number. Therefore, z = x + y can also take a wide range of values.

However, the problem only asks us to show that a ∈ N. We've already found that a = 16 in the interval -6 ≤ z ≤ 10. This is enough to satisfy the problem's requirement!

Key Takeaway: Sometimes, you don't need to solve every single case to answer the question. If you can find one scenario where the condition is met, you've done your job! Also, introducing a new variable (like z = x + y) can often simplify complex expressions.

Problem 2: Square Roots and Integer Results

Now, let's shift our focus to problems involving square roots. These problems often require us to simplify expressions and use the properties of square roots.

Part (a): a = sqrt( (x + 5)^2 ) + sqrt( (x - 3)^2 )

Problem Statement:

Given a = √( (x + 5)^2 ) + √( (x - 3)^2 ), show that a ∈ N.

Solution:

The key to this problem is understanding how square roots and squares interact. Remember that √(x^2) = |x| (the absolute value of x). This is because squaring a number always results in a non-negative value, and the square root function returns the non-negative root.

Applying this to our expression, we get:

a = |x + 5| + |x - 3|

Now, this looks very similar to the absolute value problem we solved earlier! We need to consider the intervals where the expressions inside the absolute values change signs.

1. x + 5: This expression is zero when x = -5.
2. x - 3: This expression is zero when x = 3.

Let's analyze three cases:

1. Case 1: x < -5

   • |x + 5| = -(x + 5) = -x - 5
   • |x - 3| = -(x - 3) = -x + 3
   • a = -x - 5 - x + 3 = -2x - 2

2. Case 2: -5 ≤ x ≤ 3

   • |x + 5| = x + 5
   • |x - 3| = -(x - 3) = -x + 3
   • a = x + 5 - x + 3 = 8

3. Case 3: x > 3

   • |x + 5| = x + 5
   • |x - 3| = x - 3
   • a = x + 5 + x - 3 = 2x + 2

In Case 2, we see that a = 8, which is a natural number. Now, we need to show that 'a' is a natural number in Cases 1 and 3 as well.

• Case 1 (x < -5): Since x < -5, -2x > 10, and -2x - 2 > 8. So, 'a' will be a natural number greater than 8. For example if x = -6, a = -2(-6) - 2 = 10. If x = -10, a = -2(-10) - 2 = 18. This shows a ∈ N.
• Case 3 (x > 3): Since x > 3, 2x > 6, and 2x + 2 > 8. So, 'a' will be a natural number greater than 8. For example if x = 4, a = 2(4) + 2 = 10. If x = 10, a = 2(10) + 2 = 22. This shows a ∈ N.

Thus, in all three cases, we've shown that a ∈ N.

Key Takeaway: Remember the crucial identity √(x^2) = |x|. This allows you to transform square root problems into absolute value problems, which you can then solve using the case-by-case analysis we discussed earlier. Also, don't forget to analyze the results in each case to ensure they meet the given conditions.

Final Thoughts

Guys, these types of problems might seem challenging at first, but with a clear understanding of absolute values and square roots, and a systematic approach, you can conquer them! Remember to break down the problems into manageable cases, consider the signs of the expressions inside absolute values, and use the properties of square roots to simplify the expressions. Keep practicing, and you'll become a math whiz in no time!