Sets, Relations and Functions
Let $$A$$ be a non-empty set such that $$A \times A$$ has $$9 $$ elements among which are found $$(-1, 0)$$ and $$(0, 1)$$, then
If $$\displaystyle A = \{2, 3, 5\}$$ and $$\displaystyle B = \{5, 7\}$$, find: $$B \times A$$
Given:
A={2,3,5}A={2,3,5} and B={5,7}B={5,7}
therefore B×AB×A can be written as,
B×A={(5,2),(5,3),(5,5),(7,2),(7,3),(7,5)}B×A={(5,2),(5,3),(5,5),(7,2),(7,3),(7,5)}
Let $$A$$ be a non-empty set such that $$A \times A$$ has $$9 $$ elements among which are found $$(-1, 0)$$ and $$(0, 1)$$, then
Given,
$$A=\{b,c,d\}$$ and $$B=\{x,y\}$$ : find element of $$A\times B$$ .
Suppose $$S=\{1,2\}$$ and $$T=\{a,b\}$$ then $$T \times S$$
If $$G = \{7, 8\}$$ and $$H = \{5, 4, 2\}$$, find $$\displaystyle G\times H $$ and $$\displaystyle H\times G $$.
State whether each of the following statements are true or false. If the statement is false rewrite the given statement correctly (i) If $$P = \{m, n\}$$ and $$Q = \{n, m\}$$ then $$\displaystyle P\times Q = \{(m, n), (n, m)\}$$ (ii) If $$A$$ and $$B$$ are non-empty sets then $$\displaystyle A\times B$$ is a non-empty set of ordered pairs $$(x, y)$$ such that $$\displaystyle x\in A$$ and $$\displaystyle y\in B$$ (iii) If $$A = \{1, 2\}, B = \{3, 4\}$$ then $$\displaystyle A\times \left ( B\cap \phi \right )=\phi $$
Let $$A$$ and $$B$$ be two sets such that $$n(A) = 3$$ and $$n(B) = 2$$. If $$(x, 1), (y, 2), (z, 1)$$ are in $$\displaystyle A\times B$$ find $$A$$ and $$B$$ where $$x, y$$ and $$z$$ are distinct elements
The cartesian product $$\displaystyle A\times A$$ has $$9$$ elements among which are found $$(-1, 0)$$ and $$(0, 1)$$. Find the set $$A$$ and the remaining elements of $$\displaystyle A\times A$$
If $$P = \{a , b\}$$ and $$Q = \{x, y, z\}$$, show that $$P \times Q \neq Q \times P.$$
For any sets $$\displaystyle A, B$$ and $$\displaystyle C$$, prove that: $$\displaystyle A \times (B \cup C) = (A \times B)\cup (A \times C)$$
Enter 1 if it is true else enter 0. If $$\displaystyle A = \{2, 3, 5\}$$ and $$\displaystyle B = \{5, 7\}$$, then $$\displaystyle A \times B = \{(2, 5), (2, 7), (3, 5), (3, 7), (5, 5), (5, 7)\}$$