9th Class Math Chapter 8 Notes

Q1(i) — For $\sim p\to q$

Write the converse, inverse and contrapositive.

1. Converse swaps the two sides: $q\to \sim p$.
2. Inverse negates both sides of the original: $\sim (\sim p)\to \sim q$, i.e. $p\to \sim q$.
3. Contrapositive negates and swaps: $\sim q\to \sim (\sim p)$, i.e. $\sim q\to p$.

Answer: Converse: $q\to \sim p$; Inverse: $p\to \sim q$; Contrapositive: $\sim q\to p$

Q1(ii) — For $q\to p$

Write the converse, inverse and contrapositive.

1. Converse: $p\to q$.
2. Inverse: $\sim q\to \sim p$.
3. Contrapositive: $\sim p\to \sim q$.

Answer: Converse: $p\to q$; Inverse: $\sim q\to \sim p$; Contrapositive: $\sim p\to \sim q$

Q1(iii) — For $\sim p\to \sim q$

Write the converse, inverse and contrapositive.

1. Converse: $\sim q\to \sim p$.
2. Inverse: $\sim (\sim p)\to \sim (\sim q)$, i.e. $p\to q$.
3. Contrapositive: $\sim (\sim q)\to \sim (\sim p)$, i.e. $q\to p$.

Answer: Converse: $\sim q\to \sim p$; Inverse: $p\to q$; Contrapositive: $q\to p$

Q1(iv) — For $\sim q\to \sim p$

Write the converse, inverse and contrapositive.

1. Converse: $\sim p\to \sim q$.
2. Inverse: $\sim (\sim q)\to \sim (\sim p)$, i.e. $q\to p$.
3. Contrapositive: $\sim (\sim p)\to \sim (\sim q)$, i.e. $p\to q$.

Answer: Converse: $\sim p\to \sim q$; Inverse: $q\to p$; Contrapositive: $p\to q$

Q2(i) — Truth table: $\sim (p\vee q)\vee (\sim q)$

Construct the truth table.

1. List all four $p,q$ combinations, compute $p\vee q$, then $\sim (p\vee q)$, then $\sim q$, then OR them together.
2. $T,T\to F$; $T,F\to T$; $F,T\to F$; $F,F\to T$.

Answer: Result column: $F,T,F,T$ (for $p,q=$ TT,TF,FT,FF)

Q2(ii) — Truth table: $\sim (\sim q\vee \sim p)$

Construct the truth table.

1. By De Morgan's law this simplifies to $p\wedge q$.
2. $T,T\to T$; $T,F\to F$; $F,T\to F$; $F,F\to F$.

Answer: Result column: $T,F,F,F$ — identical to $p\wedge q$

Q2(iii) — Truth table: $(p\vee q)\leftrightarrow (p\wedge q)$

Construct the truth table.

1. $p\vee q$: $T,T,T,F$; $p\wedge q$: $T,F,F,F$.
2. Biconditional is true exactly when both columns match.

Answer: Result column: $T,F,F,T$ (for $p,q=$ TT,TF,FT,FF)

Q3(i) — Prove: $(x-4{)}^2+9=x^2-8x+25$

Prove the L.H.S. equals the R.H.S.

1. $(x-4{)}^2=x^2-8x+16$.
2. L.H.S. $=x^2-8x+16+9=x^2-8x+25=$ R.H.S.

Answer: Proved

Q3(ii) — Prove: $(x+1{)}^2-(x-1{)}^2=4x$

Prove the identity.

1. Using difference of squares: $[(x+1)+(x-1)][(x+1)-(x-1)]=(2x)(2)=4x$.

Answer: Proved

Q3(iii) — Prove: $(x+5{)}^2-(x-5{)}^2=20x$

Prove the identity.

1. $[(x+5)+(x-5)][(x+5)-(x-5)]=(2x)(10)=20x$.

Answer: Proved

Q4(i) — Prove: $\frac{4+16x}{4}=1+4x$

Prove, justifying each step.

1. Split the fraction: $\frac{4}{4}+\frac{16x}{4}=1+4x$.

Answer: Proved

Q4(ii) — Prove: $\frac{6x^2+18x}{3x^2-27}=\frac{2x}{x-3}$

Prove, justifying each step.

1. Numerator $=6x(x+3)$; denominator $=3(x-3)(x+3)$.
2. Cancel the common factor $(x+3)$ (valid since $x\ne -3$): $\frac{6x(x+3)}{3(x-3)(x+3)}=\frac{6x}{3(x-3)}=\frac{2x}{x-3}$.

Answer: Proved

Q4(iii) — Prove: $\frac{x^2+7x+10}{x^2-3x-10}=\frac{x+5}{x-5}$

Prove, justifying each step.

1. Numerator $=(x+2)(x+5)$; denominator $=(x-5)(x+2)$.
2. Cancel $(x+2)$ (valid since $x\ne -2$): $\frac{x+5}{x-5}$.

Answer: Proved

Q5(i) — If $x$ is odd, prove $9x+4$ is odd

Prove the statement.

1. Let $x=2k+1$ for some integer $k$.
2. $9x+4=9(2k+1)+4=18k+13=2(9k+6)+1$ — of the form $2m+1$, hence odd.

Answer: Proved: $9x+4$ is odd

Q5(ii) — If $x$ is odd, prove $7x+5$ is even

Prove the statement.

1. $x=2k+1$: $7x+5=7(2k+1)+5=14k+12=2(7k+6)$ — even.

Answer: Proved: $7x+5$ is even

Q6(i) — If $x$ is odd, prove $x^2-4x+6$ is odd

Prove the statement.

1. $x=2k+1$: $x^2=4k^2+4k+1$.
2. $x^2-4x+6=4k^2+4k+1-8k-4+6=4k^2-4k+3=2(2k^2-2k+1)+1$ — odd.

Answer: Proved: $x^2-4x+6$ is odd

Q6(ii) — If $x$ is even, prove $x^2+2x+4$ is even

Prove the statement.

1. $x=2k$: $x^2+2x+4=4k^2+4k+4=2(2k^2+2k+2)$ — even.

Answer: Proved: $x^2+2x+4$ is even

Q7 — Prove $(A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }$

Prove De Morgan's law for any two non-empty sets $A,B$.

1. If $x\in (A\cap B{)}^{\prime }$, then $x\notin (A\cap B)$, i.e. NOT($x\in A$ AND $x\in B$), which is equivalent to $x\notin A$ OR $x\notin B$, i.e. $x\in A^{\prime }\cup B^{\prime }$.
2. Conversely, if $x\in A^{\prime }\cup B^{\prime }$, then $x\notin A$ or $x\notin B$, so $x\notin (A\cap B)$, i.e. $x\in (A\cap B{)}^{\prime }$.
3. Each side implies the other, so the sets are equal.

Answer: Proved: $(A\cap B{)}^{\prime }=A^{\prime }\cup B^{\prime }$

Q8 — If $x,y>0$ and $x^2<y^2$, prove $x<y$

Prove the statement.

1. $x^2<y^2\Rightarrow y^2-x^2>0\Rightarrow (y-x)(y+x)>0$.
2. Since $x,y>0$, $y+x>0$; dividing both sides by this positive quantity preserves the inequality: $y-x>0$.

Answer: Proved: $x<y$

Q9 — Prove the sum of a triangle's interior angles is $180°$

Prove the statement.

1. Through one vertex, draw a line parallel to the opposite side.
2. The two angles formed on this line equal the triangle's other two interior angles (alternate interior angles with the two transversal sides).
3. These two angles plus the vertex's own angle lie on a straight line, summing to $180°$ — hence all three interior angles sum to $180°$.

Answer: Proved: interior angles sum to $180°$

Q10(i) — If $a,b,c,d\ne 0$, prove $\frac{a}{b}=\frac{c}{d}\iff ad=bc$

Prove both directions.

1. ($\Rightarrow$) Multiply both sides by $bd$: $\frac{a}{b}(bd)=\frac{c}{d}(bd)\Rightarrow ad=bc$.
2. ($\Leftarrow$) Divide both sides by $bd$ (nonzero): $\frac{ad}{bd}=\frac{bc}{bd}\Rightarrow \frac{a}{b}=\frac{c}{d}$.

Answer: Proved both directions

Q10(ii) — Prove $\frac{a}{b}\cdot \frac{c}{d}=\frac{ac}{bd}$

Prove the identity.

1. By the definition of fraction multiplication, multiplying numerators and denominators gives $\frac{a\times c}{b\times d}=\frac{ac}{bd}$.

Answer: Proved

Q10(iii) — Prove $\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}$

Prove the identity.

1. $\frac{a}{b}+\frac{c}{b}=a\cdot \frac{1}{b}+c\cdot \frac{1}{b}=(a+c)\cdot \frac{1}{b}$ by the distributive property, $=\frac{a+c}{b}$.

Answer: Proved