Page 24 - Algebra 1
P. 24

Explanation
1. P = {𝐴𝑙𝑙 𝑑𝑒π‘₯π‘‘π‘π‘œπ‘œπ‘˜π‘  π‘œπ‘€π‘›π‘’π‘‘ 𝑏𝑦 π‘†π‘Žπ‘š}
R = {(𝑝, π‘ž): 𝑝 π‘Žπ‘›π‘‘ π‘ž hπ‘Žπ‘£π‘’ 𝑑h𝑒 π‘ π‘Žπ‘šπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿ π‘œπ‘“ π‘π‘Žπ‘”π‘’π‘  }
R is reflexive on A since (p, p) ∈ R
R is symmetric on A since books p and q have the same number of page (p, q) ∈ R.
Also, q and p have the same number of pages. So, (q, p) ∈ R
Therefore, R is symmetric on A
R is transitive on A, since books p, q, and r have the same number of pages (p, q) ∈ R and (q, r) ∈R.So,(p,r) ∈R
R is reflexive, symmetric, and transitive on A. So, it’s equivalent on A.
2. P={(𝑝,π‘ž):π‘β‰€π‘ž4},where(p,q)∈R
Reflexive:
1 ≀ (1)4 doesn’t hold true.
So, R is not reflexive.
Symmetry:
–1 ≀ (3)4 hold true and 3 ≀ (– 1)4 doesn’t holds true. (3, –1) Ι‡ R and (–1, 3) ∈ R
So, R is not symmetric.
Transitive:
4 ≀ (– 5)4 hold true and –5 ≀ (1)4 holds true. 4 ≀ (1)4 doesn’t hold true.
So, R is not transitive.
Therefore, R is neither symmetric, nor reflexive nor transitive.
3. Relation from Q to P
P = {1,2} and Q = {𝑝,π‘ž,π‘Ÿ}
R = {(𝑝, 1), (𝑝, 2), (π‘ž, 1), (π‘ž, 2), (π‘Ÿ, 1), (π‘Ÿ, 2)}
4. R is not reflexive since (3, 3), (4, 4) Ι‡ R
R is symmetric since (3, 4) ∈ R and (4, 3) ∈ R
R is not transitive since (3, 4) ∈ R but (4, 5), (3, 4) Ι‡ R, so R is transitive R is not equivalence since it’s only symmetric
5. Number of relations possible from P to Q = 2rs Number of relations possible = 2(6 Γ— 3) = 218
33
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 ALGEBRA
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