Problem 19
Question
Mark each statement as true or false. $$x \uparrow y=y \uparrow x$$
Step-by-Step Solution
Verified Answer
The given statement \(x \uparrow y = y \uparrow x\) or \(x^y = y^x\) is false. A counterexample is when x = 3 and y = 2, where \(3^2 \neq 2^3\).
1Step 1: Identify notation
The given problem provides the following equation: \(x \uparrow y = y \uparrow x\). We can recognize that the \(\uparrow\) symbol represents exponentiation operation. So we can rewrite it as \(x^y = y^x\).
2Step 2: Analyze the expression
In order to determine whether the expression \(x^y = y^x\) is true or false, we can evaluate the expression for different values of x and y. If there are counterexamples where the equation doesn't hold true, then we can conclude that it's false.
Example 1: Let's take x = 2 and y = 4:
\[2^4 = 16\]
\[4^2 = 16\]
In this particular case, the expression is true.
Example 2: Let's take x = 3 and y = 2:
\[3^2 = 9\]
\[2^3 = 8\]
In this case, the expression is false.
Since we have found a counterexample where the given expression \(x^y = y^x\) doesn't hold true, we can conclude that the statement is false.
Key Concepts
Commutative PropertyCounterexampleMathematical Notation
Commutative Property
The commutative property is a fundamental principle in mathematics. It states that you can change the order of numbers in an operation without changing the result. The commutative property applies to addition (e.g., \(a + b = b + a\)) and multiplication (e.g., \(a \times b = b \times a\)). However, not all operations are commutative. This means changing order might change the result.
Exponentiation, as discussed in our exercise, is one such operation that isn’t commutative. Comparing the expressions \(x^y\) and \(y^x\), you might immediately realize that swapping the base and the exponent will generally result in different outcomes. While for certain values, such as \(x = 2\) and \(y = 4\), the result remains the same, this isn’t universally true. This lack of universal commutativity is precisely why exponentiation is non-commutative.
Exponentiation, as discussed in our exercise, is one such operation that isn’t commutative. Comparing the expressions \(x^y\) and \(y^x\), you might immediately realize that swapping the base and the exponent will generally result in different outcomes. While for certain values, such as \(x = 2\) and \(y = 4\), the result remains the same, this isn’t universally true. This lack of universal commutativity is precisely why exponentiation is non-commutative.
Counterexample
When exploring whether a statement is universally true in mathematics, providing a counterexample can be a powerful method for demonstrating that it's false. A counterexample is a specific case where the general statement doesn’t hold.
In the context of our exercise about exponentiation, we were testing if \(x^y = y^x\) is always true. By plugging in numbers like \(x = 3\) and \(y = 2\), we found that \(3^2 = 9\) and \(2^3 = 8\), so the statement fails here.
In the context of our exercise about exponentiation, we were testing if \(x^y = y^x\) is always true. By plugging in numbers like \(x = 3\) and \(y = 2\), we found that \(3^2 = 9\) and \(2^3 = 8\), so the statement fails here.
- Once a single counterexample is found, it’s enough to prove the general statement false.
- This method of using specific instances to show a statement doesn't hold is commonly used for disproving hypotheses.
Mathematical Notation
Mathematical notation is a standardized way of communicating complex mathematical ideas succinctly and precisely. It uses symbols and numbers to express operations, expressions, and equations.
- For instance, in our exercise, \(x \uparrow y\) was used to denote exponentiation, which can be more commonly represented as \(x^y\).
- Understanding these symbols is crucial for correctly interpreting and solving problems.
Other exercises in this chapter
Problem 19
Using a logic table, verify each. $$(x+y)^{\prime}=x^{\prime} y^{\prime}$$
View solution Problem 19
The set \(D_{70}=\\{1,2,5,7,10,14,35,70\\}\) of positive factors of 70 is a boolean algebra under the operations \(\oplus, \odot,\) and ' defined by \(x \oplus
View solution Problem 20
Find the boolean expression represented by each Karnaugh map.
View solution Problem 20
Using a logic table, verify each. $$(x y)^{\prime}=x^{\prime}+y^{\prime}$$
View solution