a) If 10 < 5x + 10 < 5 , prove that 4 < x < 1 . b) Let A = \1, 2, 3\ and B = \1, 3, 5\ . Find A B and also find the relation from A to B satisfy the condition x + y 5 where x A, y B .

a) Solution to the Inequality $-10<5x+10<5$

Step 1: Understand the given inequality We are given the compound inequality: $$-10<5x+10<5$$ This means that $5x+10$ is greater than $-10$ and less than $5$.

Step 2: Subtract 10 from all parts To isolate the term with $x$, subtract 10 from all parts of the inequality: $$-10-10<5x+10-10<5-10$$ $$-20<5x<-5$$

Step 3: Divide all parts by 5 Now, divide every part of the inequality by 5 to solve for $x$: $$\frac{-20}{5}<\frac{5x}{5}<\frac{-5}{5}$$ $$-4<x<-1$$

Conclusion Thus, we have proven that if $-10<5x+10<5$, then $-4<x<-1$.

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b) Cartesian Product $A\times B$ and Relation $R$ from $A$ to $B$ with $x+y\le 5$

Part 1: Find $A\times B$

Given: $$A=\{1,2,3\},B=\{1,3,5\}$$

The Cartesian product $A\times B$ is defined as the set of all ordered pairs $(x,y)$ where $x\in A$ and $y\in B$. We list all possible pairs:

$$A\times B=\{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5)\}$$

Part 2: Find the Relation $R$ from $A$ to $B$ satisfying $x+y\le 5$

We need to find all ordered pairs $(x,y)\in A\times B$ such that $x+y\le 5$.

Let's evaluate each pair:

1. $(1,1)$: $1+1=2\le 5$ → Included
2. $(1,3)$: $1+3=4\le 5$ → Included
3. $(1,5)$: $1+5=6>5$ → Excluded
4. $(2,1)$: $2+1=3\le 5$ → Included
5. $(2,3)$: $2+3=5\le 5$ → Included
6. $(2,5)$: $2+5=7>5$ → Excluded
7. $(3,1)$: $3+1=4\le 5$ → Included
8. $(3,3)$: $3+3=6>5$ → Excluded
9. $(3,5)$: $3+5=8>5$ → Excluded

Thus, the relation $R$ is: $$R=\{(1,1),(1,3),(2,1),(2,3),(3,1)\}$$

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Final Answer

a) Proof:

$$-10<5x+10<5⟹-4<x<-1$$

b)

• Cartesian Product $A\times B$: $$A\times B=\{(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5)\}$$

• Relation $R$ satisfying $x+y\le 5$: $$R=\{(1,1),(1,3),(2,1),(2,3),(3,1)\}$$