Compound Inequality: Must Be Correct?

Hey math whizzes! Let's dive into a super interesting problem involving compound inequalities. We're given a condition, and our mission, should we choose to accept it, is to figure out which of the following statements absolutely have to be true. Think of it like a logic puzzle, but with numbers and variables!

The Compound Condition: Unpacking the Clues

The core of our problem is this compound inequality: $3z < k - 1 < 36$. What this fancy notation tells us is that the expression $k - 1$ is simultaneously greater than $3z$ AND less than $36$. So, we have two inequalities packed into one. This means:

1. $3z < k - 1$
2. $k - 1 < 36$

These two individual inequalities are the foundation upon which we'll build our analysis. It's crucial to understand that both of these must hold true for the original compound inequality to be valid. Think of them as the twin pillars supporting the entire structure. If even one of them falters, the whole thing collapses, and our compound condition is no longer met. So, right off the bat, we can see that statement I is directly derived from the given information.

Now, let's tackle statement II: $k < 36$. To see if this must be correct, we need to look at the second part of our compound inequality: $k - 1 < 36$. If we add 1 to both sides of this inequality, what do we get? We get $k < 36 + 1$, which simplifies to $k < 37$. Ah, so statement II, $k < 36$, isn't necessarily true. For instance, if $k = 36.5$, then $k-1 = 35.5$, which satisfies $k-1 < 36$. But $k$ is definitely not less than $36$. This is a classic trap – just because part of an expression is less than a number doesn't mean a variable within that expression is necessarily less than that same number, or even a related, smaller number. We need to be super careful here and not jump to conclusions.

Finally, let's examine statement III: $z < 12$. This one comes from the first part of our compound inequality: $3z < k - 1$. We know that $k - 1$ is less than $36$. So, we can substitute the upper bound for $k-1$ into the inequality involving $z$. This gives us $3z < 36$. Now, to isolate $z$, we divide both sides by 3. This yields $z < 12$. And boom! Statement III must be correct. This is because $3z$ is less than $k-1$, and $k-1$ is less than $36$. Therefore, $3z$ must be less than $36$. Dividing by a positive number (3) preserves the inequality sign, so $z < 12$ is a guaranteed outcome.

So, to recap our initial thoughts: Statement I is directly given, and Statement III is a logical deduction. Statement II, however, is not guaranteed.

Proving Statement I: The Obvious Truth

Statement I is: $3z < k - 1$.

Let's look at the given compound inequality again: $3z < k - 1 < 36$.

The definition of a compound inequality of the form $a < b < c$ means that $a < b$ AND $b < c$.

In our case, $a = 3z$, $b = k - 1$, and $c = 36$.

Therefore, from the definition of the compound inequality, it must be true that $3z < k - 1$. There's no way around it, guys! This statement is one of the two core components that make up the entire compound inequality. If this part wasn't true, the whole expression $3z < k - 1 < 36$ wouldn't make sense. It's like saying a sandwich has bread and filling – the bread is one essential part, and the filling is another. Statement I is simply one of the essential parts.

Proof: Given the compound inequality $3z < k - 1 < 36$, this inherently implies two separate inequalities that must hold true simultaneously:

1. $3z < k - 1$
2. $k - 1 < 36$

Statement I is precisely the first of these implied inequalities. Therefore, Statement I must be correct by the very definition of the compound inequality provided.

Proving Statement II: The False Lead

Statement II is: $k < 36$.

We know from the compound inequality $3z < k - 1 < 36$ that $k - 1 < 36$.

If we add 1 to both sides of $k - 1 < 36$, we get:

$k - 1 + 1 < 36 + 1$

$k < 37$

This tells us that $k$ must be less than 37. However, it does not guarantee that $k$ must be less than 36.

Consider a specific example to show why this statement isn't always true. Let's pick a value for $k$ that satisfies $k < 37$ but is not less than $36$. How about $k = 36.5$?

If $k = 36.5$, then $k - 1 = 35.5$.

Now, let's check if this fits into the original compound inequality, $3z < k - 1 < 36$.

We need $3z < 35.5 < 36$.

This part, $35.5 < 36$, is true.

We also need $3z < 35.5$. We can easily find a value for $z$ that makes this true. For example, if $z = 10$, then $3z = 30$. Since $30 < 35.5$, this condition is met.

So, with $k = 36.5$ and $z = 10$, our original compound inequality $3z < k - 1 < 36$ becomes $30 < 35.5 < 36$, which is completely valid.

However, in this valid scenario, $k = 36.5$, which is not less than 36. Therefore, Statement II ($k < 36$) is not always correct.

Proof: From the given compound inequality $3z < k - 1 < 36$, we deduce $k - 1 < 36$. Adding 1 to both sides yields $k < 37$. This shows that $k$ can be any value less than 37. It is possible for $k$ to be a value such as $36.5$, which satisfies $k < 37$ but does not satisfy $k < 36$. Thus, Statement II is not necessarily correct. We can construct a counterexample where $k=36.5$ and $z=10$, which makes $30 < 35.5 < 36$ true, but $k < 36$ is false.

Proving Statement III: The Essential Deduction

Statement III is: $z < 12$.

We are given the compound inequality $3z < k - 1 < 36$.

This compound inequality gives us two pieces of information:

1. $3z < k - 1$
2. $k - 1 < 36$

Since $3z$ is less than $k - 1$, and $k - 1$ is less than $36$, we can use the transitive property of inequalities. The transitive property states that if $a < b$ and $b < c$, then $a < c$.

In our case, let $a = 3z$, $b = k - 1$, and $c = 36$.

We have $a < b$ (which is $3z < k - 1$) and $b < c$ (which is $k - 1 < 36$).

Therefore, by the transitive property, it must be true that $a < c$, which means $3z < 36$.

Now, we need to find out what this tells us about $z$. To isolate $z$, we divide both sides of the inequality $3z < 36$ by 3. Since 3 is a positive number, the direction of the inequality sign remains the same:

rac{3z}{3} < rac{36}{3}

$z < 12$

So, Statement III must be correct. This deduction is solid because $3z$ is bounded above by $k-1$, and $k-1$ is itself bounded above by $36$. This upper bound on $k-1$ effectively imposes an upper bound on $3z$, and consequently on $z$.

Proof: Given the compound inequality $3z < k - 1 < 36$, we can establish two key inequalities:

1. $3z < k - 1$
2. $k - 1 < 36$

Using the transitive property of inequalities, since $3z < k - 1$ and $k - 1 < 36$, it logically follows that $3z < 36$. To determine the constraint on $z$, we divide both sides of the inequality $3z < 36$ by 3. As 3 is positive, the inequality direction is preserved:

rac{3z}{3} < rac{36}{3}

$z < 12$

Therefore, Statement III, $z < 12$, must be correct.

Conclusion: Which Statements Are a Go?

After breaking down the compound inequality and rigorously proving each statement, we've found our winners!

• Statement I: $3z < k - 1$ - This is always correct because it's a direct component of the given compound inequality.
• Statement II: $k < 36$ - This is not always correct. We found a counterexample where $k=36.5$ still satisfies the original compound inequality.
• Statement III: $z < 12$ - This is always correct. It's a necessary consequence derived using the transitive property and algebraic manipulation.

So, the statements that must be correct are I and III. Keep practicing these types of problems, guys, they really sharpen your logical thinking skills!