Let R be a relation from N to N defined by R = {(x, y): x, y ∈ N and x 2 = y 3}. Which of the following are not correct? 1. (x, x) ∈ R for all x ∈ N 2. (x, y) ∈ R ⇒ (y, x) ∈ R 3. (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R Select the correct answer using the code given below :
1, 2 and 3
Understanding the Relation R on Natural Numbers
The question asks us to examine a specific relation R defined on the set of natural numbers, denoted by N. The relation R is given by R = {(x, y): x, y \(\in\) N and x2 = y3}. We need to determine which of the given properties are not correct for this relation R.
Let's analyze each property one by one.
Checking for Reflexivity Property
A relation R on a set A is called reflexive if (x, x) \(\in\) R for every element x \(\in\) A.
For our relation R on N, reflexivity means that for all x \(\in\) N, the pair (x, x) must be in R. According to the definition of R, this means \(x^2\) must be equal to \(x^3\) for all x \(\in\) N.
We need to check if \(x^2 = x^3\) holds for all x \(\in\) N.
Consider the equation \(x^2 = x^3\).
We can rewrite this as \(x^3 - x^2 = 0\), which factors as \(x^2(x - 1) = 0\).
This equation is true if and only if \(x^2 = 0\) or $x - 1 = 0$. This gives $x = 0$ or $x = 1$.
The set of natural numbers N is typically considered to be \(\{1, 2, 3, \dots\}\). If \(N = \{1, 2, 3, \dots\}\), then only for $x = 1$ is \(x^2 = x^3\) true (\(1^2 = 1\) and \(1^3 = 1\)). For any other natural number $x > 1$, such as $x = 2$, we have \(2^2 = 4\) and \(2^3 = 8\), and \(4 \ne 8\). Thus, \((2, 2) \notin R\).
Therefore, the statement "(x, x) \(\in\) R for all x \(\in\) N" is not correct.
Checking for Symmetry Property
A relation R on a set A is called symmetric if whenever (x, y) \(\in\) R, it follows that (y, x) \(\in\) R for all x, y \(\in\) A.
For our relation R on N, symmetry means that if \((x, y) \in R\), which is \(x^2 = y^3\) for x, y \(\in\) N, then it must be true that \((y, x) \in R\), which is \(y^2 = x^3\).
Let's try to find a pair (x, y) that is in R where x \(\ne\) y.
We need \(x^2 = y^3\). Let's try some natural numbers for y.
- If $y = 1$, \(y^3 = 1\). \(x^2 = 1\), which means $x = 1$ (since x \(\in\) N). This gives the pair (1, 1). For this pair, symmetry holds as (1, 1) \(\in\) R implies (1, 1) \(\in\) R.
- If $y = 2$, \(y^3 = 8\). \(x^2 = 8\). There is no natural number x such that \(x^2 = 8\).
- If $y = 3$, \(y^3 = 27\). \(x^2 = 27\). There is no natural number x such that \(x^2 = 27\).
- If $y = 4$, \(y^3 = 64\). \(x^2 = 64\), which means $x = 8$ (since x \(\in\) N). This gives the pair (8, 4). Let's check if (8, 4) \(\in\) R. Yes, \(8^2 = 64\) and \(4^3 = 64\).
Now, according to the symmetry property, if (8, 4) \(\in\) R, then (4, 8) must also be in R.
Let's check if (4, 8) \(\in\) R. This requires \(4^2 = 8^3\).
\(4^2 = 16\)
\(8^3 = 512\)
Since \(16 \ne 512\), the pair (4, 8) is not in R.
We found a pair (8, 4) \(\in\) R such that (4, 8) \(\notin\) R. Therefore, the relation R is not symmetric.
The statement "(x, y) \(\in\) R \(\rArr\) (y, x) \(\in\) R" is not correct.
Checking for Transitivity Property
A relation R on a set A is called transitive if whenever (x, y) \(\in\) R and (y, z) \(\in\) R, it follows that (x, z) \(\in\) R for all x, y, z \(\in\) A.
For our relation R on N, transitivity means that if \((x, y) \in R\) and \((y, z) \in R\), it follows that \((x, z) \in R\) for all x, y, z \(\in\) N.
Let's start with the given conditions:
- \((x, y) \in R\) means \(x^2 = y^3\)
- \((y, z) \in R\) means \(y^2 = z^3\)
From condition 2, \(y^2 = z^3\). Taking the square root (and considering only natural numbers), we get \(y = z^{3/2}\).
Substitute this expression for y into condition 1:
\(x^2 = (z^{3/2})^3\)
\(x^2 = z^{(3/2) \times 3}\)
\(x^2 = z^{9/2}\)
For transitivity to hold, we need \((x, z) \in R\), which means \(x^2 = z^3\).
So, we would need \(z^{9/2} = z^3\) for all valid x, y, z in N that satisfy the first two conditions.
Let's check if \(z^{9/2} = z^3\) holds true for all such z \(\in\) N.
\(z^{9/2} = z^3\)
\(z^{9/2} - z^3 = 0\)
\(z^3(z^{3/2} - 1) = 0\)
This equation is true if \(z^3 = 0\) (which means $z = 0$, but 0 is usually not in N) or \(z^{3/2} - 1 = 0\).
\(z^{3/2} = 1\) implies \(z = 1^{2/3} = 1\).
This equation \(z^{9/2} = z^3\) only holds when $z = 1$ (assuming z \(\in\) N). It does not hold for all z that can appear in such pairs.
Let's find an example. We need x, y, z \(\in\) N such that \(x^2 = y^3\) and \(y^2 = z^3\).
As found earlier, pairs (x, y) in R are of the form \((k^3, k^2)\) for \(k \in N\) (e.g., (1,1), (8,4), (27,9), ...).
Pairs (y, z) in R are also of the form \((m^3, m^2)\) for \(m \in N\) (e.g., (1,1), (8,4), (27,9), ...).
For a chain (x, y) and (y, z) to exist, the second element of the first pair must be the first element of the second pair. So, we need \(y = k^2 = m^3\).
The solutions in natural numbers for \(k^2 = m^3\) are when \(k = n^3\) and \(m = n^2\) for some \(n \in N\).
So, \(y = k^2 = (n^3)^2 = n^6\). Also \(y = m^3 = (n^2)^3 = n^6\). This is consistent.
If \(y = n^6\), then from \(x^2 = y^3\), we have \(x^2 = (n^6)^3 = n^{18}\). So \(x = n^9\).
From \(y^2 = z^3\), we have \((n^6)^2 = z^3\), so \(n^{12} = z^3\). So \(z = n^4\).
Thus, a chain \((x, y) \in R\) and \((y, z) \in R\) exists when \(x = n^9\), \(y = n^6\), \(z = n^4\) for some \(n \in N\).
Now let's check if \((x, z) \in R\) for these values. This requires \(x^2 = z^3\).
\(x^2 = (n^9)^2 = n^{18}\).
\(z^3 = (n^4)^3 = n^{12}\).
We need to check if \(n^{18} = n^{12}\) for all \(n \in N\).
\(n^{18} = n^{12}\) implies \(n^{18} - n^{12} = 0\), or \(n^{12}(n^6 - 1) = 0\).
This is true only for $n = 1$ (since \(n \in N\)). For $n > 1$, \(n^{18} \ne n^{12}\).
For example, let $n = 2$.
\(x = 2^9 = 512\)
\(y = 2^6 = 64\)
\(z = 2^4 = 16\)
Check (x, y): (512, 64). \(512^2 = 262144\), \(64^3 = 262144\). So \((512, 64) \in R\).
Check (y, z): (64, 16). \(64^2 = 4096\), \(16^3 = 4096\). So \((64, 16) \in R\).
Check (x, z): (512, 16). \(512^2 = 262144\), \(16^3 = 4096\).
Since \(262144 \ne 4096\), \((512, 16) \notin R\).
We found a case where \((x, y) \in R\) and \((y, z) \in R\), but \((x, z) \notin R\). Therefore, the relation R is not transitive.
The statement "(x, y) \(\in\) R and (y, z) \(\in\) R \(\rArr\) (x, z) \(\in\) R" is not correct.
Summary of Findings on Relation Properties
Based on our analysis of the relation R = {(x, y): x, y \(\in\) N and \(x^2 = y^3\)}:
- Statement 1 (Reflexivity): (x, x) \(\in\) R for all x \(\in\) N. This is not correct.
- Statement 2 (Symmetry): (x, y) \(\in\) R \(\rArr\) (y, x) \(\in\) R. This is not correct.
- Statement 3 (Transitivity): (x, y) \(\in\) R and (y, z) \(\in\) R \(\rArr\) (x, z) \(\in\) R. This is not correct.
The question asks which of the given statements are not correct.
All three statements (1, 2, and 3) are not correct.
Conclusion: Identifying Incorrect Statements about Relation R
We have shown that Statement 1 (Reflexivity), Statement 2 (Symmetry), and Statement 3 (Transitivity) do not hold for the relation R defined on natural numbers as \(x^2 = y^3\). Therefore, all three statements are not correct.
Revision Table: Properties of Relation R
| Property | Statement | Correct? | Reason/Counterexample |
|---|---|---|---|
| Reflexivity | (x, x) \(\in\) R \(\forall\) x \(\in\) N | No | Requires \(x^2 = x^3\) for all x \(\in\) N. Only true for x = 1 (in N). Fails for x = 2 (\(2^2 \ne 2^3\)). |
| Symmetry | (x, y) \(\in\) R \(\rArr\) (y, x) \(\in\) R | No | Requires if \(x^2 = y^3\), then \(y^2 = x^3\). (8, 4) \(\in\) R (\(8^2=4^3\)), but (4, 8) \(\notin\) R (\(4^2 \ne 8^3\)). |
| Transitivity | (x, y) \(\in\) R & (y, z) \(\in\) R \(\rArr\) (x, z) \(\in\) R | No | Requires if \(x^2 = y^3\) and \(y^2 = z^3\), then \(x^2 = z^3\). (512, 64) \(\in\) R and (64, 16) \(\in\) R, but (512, 16) \(\notin\) R. |
Additional Information on Relation Properties
A relation is a way of describing a relationship between elements of sets. For a relation R from set A to set B, it is a subset of the Cartesian product A \(\times\) B. If R is a relation on a single set A, it is a subset of A \(\times\) A.
Key properties of relations include:
- Reflexive: Every element is related to itself.
- Symmetric: If a is related to b, then b is related to a.
- Antisymmetric: If a is related to b and b is related to a, then a must be equal to b.
- Transitive: If a is related to b and b is related to c, then a is related to c.
Relations that are reflexive, symmetric, and transitive are called equivalence relations. Relations that are reflexive, antisymmetric, and transitive are called partial order relations.
Understanding these properties helps classify relations and analyze their structure.
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