Let A = {x∈N : x²-5x+6=0}, B = {x∈W : 0≤x

Question

Let A = {x∈N : x²-5x+6=0}, B = {x∈W : 0≤x<2} and C = {x∈N : x<3}, then verify that A×(B∪C) = (A×B)∪(A×C).

Accepted Answer

> > [!TLDR] > Explicitly list $A,B,C$, compute both sides of the distributive law for Cartesian products, and confirm they match. > > **Find the sets.** $A=\{x\in\mathbb N: x^2-5x+6=0\}$: solving $x^2-5x+6=0\Rightarrow(x-2)(x-3)=0\Rightarrow x=2,3$. So $A=\{2,3\}$. > > $B=\{x\in W: 0\le x<2\}$, where $W$ is the set of whole numbers $\{0,1,2,\dots\}$: $B=\{0,1\}$. > > $C=\{x\in\mathbb N: x<3\}$: $C=\{1,2\}$. > > **Compute $B\cup C=\{0,1,2\}$.** > > **LHS:** $A imes(B\cup C)=\{2,3\} imes\{0,1,2\}=\{(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)\}$. > > **RHS:** $A imes B=\{2,3\} imes\{0,1\}=\{(2,0),(2,1),(3,0),(3,1)\}$ …

Q.Let A = {x∈N : x²-5x+6=0}, B = {x∈W : 0≤x<2} and C = {x∈N : x<3}, then verify that A×(B∪C) = (A×B)∪(A×C).

Concept understanding — Cartesian Product of Sets

Step 1 — Intersection $B\cap C$

Step 2 — Cartesian product $A\times(B\cap C)$

Step 1 — Form the union

Step 2 — Note the size of $B$

Step 3 — Cardinality of the Cartesian product

Step 1 — Set difference

Step 2 — Intersection

Step 3 — Cartesian product

Q.Let A = {x∈N : x²-5x+6=0}, B = {x∈W : 0≤x<2} and C = {x∈N : x<3}, then verify that A×(B∪C) = (A×B)∪(A×C).

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