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Define a binary relation \(P\) from \(\mathbf{R}\) to \(\mathbf{R}\) as follows: For all real numbers \(x\) and \(y_{+}\) $$ (x, y) \in P \quad \Leftrightarrow \quad x=y^{2} . $$ Is \(P\) a function? Explain.

Short Answer

Expert verified
The relation P is not a function because for the domain of 鈩, there are elements (negative real numbers) which do not have a corresponding value in the codomain. In order for a relation to be a function, each element in the domain must have exactly one corresponding element in the codomain, which is not the case here.

Step by step solution

01

The relation P is given as: For all real numbers x and y, (x, y) 鈭 P 鈬 x = y虏 #Step 2: Examine the domain of P#

In this case, the domain of P is the set of all real numbers 鈩. Therefore, we need to determine if every real number x has a unique real number y such that x = y虏. #Step 3: Determine if each element in the domain has exactly one corresponding element in the codomain#
02

Let us take an example to check if each element in the domain has exactly one corresponding element in the codomain: Case 1: x 鈮 0 If x is a non-negative real number, then there exists a unique non-negative real number y such that x = y虏. In this case, the relation P has a unique corresponding real number y for x. Case 2: x < 0 If x is a negative real number, there is no real number y such that x = y虏. This is because the square of a real number is always non-negative. Thus, for negative real numbers x, there is no corresponding real number y in the codomain. #Step 4: Conclude if P is a function or not#

The relation P is not a function because for the domain of 鈩, there are elements (negative real numbers) which do not have a corresponding value in the codomain. In order for a relation to be a function, each element in the domain must have exactly one corresponding element in the codomain, which is not the case here.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Real Numbers
When we talk about real numbers, we're referring to the set of all the numbers that can be found on the number line. This includes every single whole number, fraction, decimal, and irrational number鈥攍ike \( \pi \) or the square root of 2鈥攚hich do not have precise decimal representations.

Real numbers have two main categories: rational numbers, which can be expressed as a fraction of two integers, and irrational numbers, which cannot be expressed so simply. For your setups involving binary relations as in the given exercise, understanding real numbers is critical because they form the universe within which domain and codomain are defined.

A common misunderstanding occurs when dealing with negative numbers and their squares. It's essential to remember that while the square of any real number results in a non-negative real number, the reverse is not always true. Negative numbers do not have real square roots, hence why they're not included in the codomain for the binary relation in the provided exercise, which leads to the conclusion that the relation is not a function.
Function Definition
A function is a special type of binary relation that relates elements of one set, called the domain, to a second set, the codomain, in a very, very particular way.

Specifically, for a relation to be a function, every single element in the domain must be paired with exactly one element in the codomain. It's like every x in the domain has one鈥攁nd only one鈥攂est friend in the codomain. \( f(x) = y \) is the typical way to express it.

If any element in the domain corresponds to two or more elements in the codomain, or none at all, breaking this 'one-to-one' rule, then you don't have a function. In the context of the exercise provided, since negative real numbers can't be paired with real numbers through squaring (they'd need imaginary friends for that!), we can see that our relation P doesn't meet the criteria for a function.
Domain and Codomain
The domain of a binary relation or function is the set of all possible inputs, or x-values, we can use. In simpler terms, it's where the function 'lives.' In the example in the exercise, the domain is all real numbers, denoted by \( \mathbf{R} \).

On the flip side, the codomain is the set of all possible outputs, or y-values, that we can get from the domain after applying the function's rule. Even if some elements of the codomain are never reached by the function, they're still part of it.

Within the context of the exercise, while any real number can square to be another real number (the codomain), not every real number has a corresponding real square root within the real numbers (the domain in our case), particularly negative ones. This disconnect between domain and codomain illustrates why the given relation cannot function鈥攊t's missing crucial connections.

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Most popular questions from this chapter

For each of the algorithm segments in \(6-14\), assume that \(n\) is a positive integer. (a) Compute the actual number of additions, subtractions, multiplications, divisions, and comparisons that must be performed when the algorithm segment is executed. For simplicity, however, count only comparisons thatFor each of the algorithm segments in \(6-14\), assume that \(n\) is a positive integer. (a) Compute the actual number of additions, subtractions, multiplications, divisions, and comparisons that must be performed when the algorithm segment is executed. For simplicity, however, count only comparisons that occur within if-then statements; ignore those implied by fornext loops. (b) Find an order for the algorithm segment from among the set of power functions. for \(i:=1\) to \(n-1\) for \(j:=i+1\) to \(n\) if \(a[j]>a[i]\) then do temp : \(=a[i]\) \(a[i]:=a[j]\) \(a[j]:=\operatorname{tem} p\) end do next \(j\) next \(i\)

Define a binary relation \(S\) on \(B=\\{a, b, c, d\\}\) by \(S=\) \(\\{(a, b),(a, c),(b, c),(d, d)\\}\).

for \(i:=1\) to \(n\) $$ \begin{aligned} \text { for } j &:=1 \text { to }\lfloor(i+1) / 2\rfloor \\ a &:=(n-i) \cdot(n-j) \\ \text { next } j & \end{aligned} $$

Exercises 4-5 explore the fact that for relatively small values of \(n\), algorithms with larger orders can be more efficient than algorithms with smaller orders. Suppose that when run with an input of size \(n\), algorithm \(A\) requires \(2 n^{2}\) operations and algorithm \(B\) requires \(80 n^{3 / 2}\) operations. a. What are orders for algorithms \(A\) and \(B\) from among the set of power functions? b. For what values of \(n\) is algorithm \(A\) more efficient than algorithm \(B\) ? c. For what values of \(n\) is algorithm \(B\) at least 100 times more efficient than algorithm \(A\) ?

Let \(A=\\{3,4,5\\}\) and \(B=\\{4,5,6\\}\) and let \(S\) be the "divides" relation. That is, For all \((x, y) \in A \times B, \quad x S y \Leftrightarrow x \mid y .\) State explicitly which ordered pairs are in \(S\) and \(S^{-1}\).

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