Q.Let $R = \{(a, b) : a, b \in N \text{ and } a + 3b = 12\}$. Find

The range of a relation is the set of all second components ($b$ values) that actually occur in it.

We are given $R=\{(a,b): a=b-2,\ b<6\}$ defined on $N=\{1,2,3,\dots\}$.

Since $b<6$ and $b\in N$, the candidates are $b\in\{1,2,3,4,5\}$.

But $a=b-2$ must also be a natural number, i.e. $a\ge 1$, so $b-2\ge 1\Rightarrow b\ge 3$.

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The domain is the set of all integers $m$ for which some integer $n$ satisfies $m^2+n^2=25$; listing the integer pairs on the circle of radius $5$ gives the answer.

We need integer pairs $(m,n)$ with $m^2+n^2=25$. Since $25 = 0^2+5^2 = 3^2+4^2 = 4^2+3^2 = 5^2+0^2$, the possible values of $m$ (allowing all sign combinations) are:

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The domain of a relation is the set of all first components of its ordered pairs.

$R=\{(x,2),(x,3),(y,1),(y,5),(y,4)\}$.

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The condition $3x - y = 0$ means $y = 3x$; list the ordered pairs with both entries in $A$, then read off the second coordinates.

$A = \{1, 2, 3, \ldots, 14\}$ and $R = \{(x,y) : 3x - y = 0\}$, i.e. $y = 3x$.

For each $x \in A$, the pair $(x, 3x)$ belongs to $R$ only if $3x \in A$ as well:

• $x = 1 \Rightarrow y = 3$ ✓
• $x = 2 \Rightarrow y = 6$ ✓
• $x = 3 \Rightarrow y = 9$ ✓
• $x = 4 \Rightarrow y = 12$ ✓
• $x = 5 \Rightarrow y = 15 \notin A$ ✗ (and larger $x$ fail too)

Hence $R = \{(1,3), (2,6), (3,9), (4,12)\}$. The range is the set of second coordinates.

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The domain of a relation is the set of all first coordinates and the range is the set of all second coordinates of its ordered pairs.

Given $R = \{(1,-1),(2,-2),(3,-1)\}$.

Domain: the set of all first elements of the ordered pairs in $R$:

$\text{Domain}(R) = \{1, 2, 3\}$

Range: the set of all second elements of the ordered pairs in $R$ (repeats counted once):

$\text{Range}(R) = \{-1, -2\}$

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