(Each question 3 marks)

Question 13 Marks

Check the validity of the statements given below by the method given against it. q: If n is a real number with n > 3, then n2 > 9 (by contradiction method).

Answer

The given statement, $q$, is as follows. If $n$ is a real number with $n>3$, then $n^2>9$. Let $u s$ assume that $n$ is a real number with $n>3$, but $n^2>9$ is not true. That is, $n^2<9$ Then, $n>3$ and $n$ is a real number. Squaring both the sides, we obtain $\mathrm{n}^2>(3)^2 \Rightarrow \mathrm{n}^2>9$, which is a contradiction, since we have assumed that $\mathrm{n}^2<9$. Thus, the given statement is true. That is, if $n$ is a real number with $n>3$, then $n^2>9$.

View full question & answer→

Question 23 Marks

Give three examples of sentences which are not statements. Give reasons for the answers.

Answer

The three examples of sentences, which are not statements, are as follows.

He is a doctor.

It is not evident from the sentence as to whom ‘he’ is referred to. Therefore, it is not a statement.

Geometry is difficult.

This is not a statement because for some people, geometry can be easy and for some others, it can be difficult.

Where is she going?

This is a question, which also contains ‘she’, and it is not evident as to who ‘she’ is. Hence, it is not a statement.

View full question & answer→

Question 33 Marks

Show that the following statement is true by the method of contrapositive. p: If x is an integer and $x^2$ is even, then x is also even.

Answer

$p$ : If $x$ is an integer and $x^2$ is even, then $x$ is also even. Let $q: x$ is an integer and $x^2$ is even. $r: x$ is even. To prove that $p$ is true by contra positive method, we assume that $r$ is false, and prove that $q$ is also false. Let $x$ is not even. To prove that $q$ is false, it has to be proved that $x$ is not an integer or $x^2$ is not even. $x$ is not even implies that $x^2$ is also not even. Therefore, statement q is false. Thus, the given statement p is true.

View full question & answer→

Question 43 Marks

Check the validity of the statements given below by the method given against it. p: The sum of an irrational number and a rational number is irrational (by contradiction method).

Answer

The given statement is as follows. p: the sum of an irrational number and a rational number is irrational. Let us assume that the given statement, p, is false. That is, we assume that the sum of an irrational number and a rational number is rational. Therefore, $\sqrt{\text{a}}+\frac{\text{b}}{\text{c}}=\frac{\text{d}}{\text{e}},$ where $\sqrt{\text{a}}$ is irrational and and b, c, d, e are integers. ⇒ de − bc = a But here, $\frac{\text{d}}{\text{e}}-\frac{\text{b}}{\text{c}}$ is a rational number and $\sqrt{\text{a}}$ is an irrational number. This is a contradiction. Therefore, our assumption is wrong. Therefore, the sum of an irrational number and a rational number is rational. Thus, the given statement is true.

View full question & answer→

Question 53 Marks

Show that the statement “For any real numbers a and b, $\text{a}^2=\text{b}^2$ implies that $\text{a = b}$” is not true by giving a counter-example.

Answer

The given statement can be written in the form of “if-then” as follows.If a and b are real numbers such that $\text{a}^2=\text{b}^2,$ then $\text{a = b}.$
Let p: a and b are real numbers such that $\text{a}^2=\text{b}^2.$
$\text{q}:\text{a = b}$
The given statement has to be proved false. For this purpose, it has to be proved that if $\text{p},$ then $\sim\text{q}.$ To show this, two real numbers, a and b, with $\text{a}^2=\text{b}^2$ are required such that $\text{a}\neq \text{b}.$
Let $\text{a}=1$ and $\text{b}=-1$
$\text{a}^2=(1)^2=1$ and $\text{b}^2=(-1)^2=1$
$\therefore\text{a}^2=\text{b}^2$
However, $\text{a}\neq \text{b}$
Thus, it can be concluded that the given statement is false.

View full question & answer→

Question 63 Marks

Show that the statement p: “If x is a real number such that x3 + 4x = 0, then x is 0” is true by

direct method,
method of contradiction,
method of contrapositive

Answer

p: “If x is a real number such that $\text{x}^3+4\text{x}=0,$ then x is 0”. Let q: x is a real number such that $\text{x}^3+4\text{x}=0$ r: x is 0.

To show that statement p is true, we assume that q is true and then show that r is true.

Therefore, let statement q be true.
$\therefore\text{x}^3+4\text{x}=0$
$\text{x}(\text{x}^2+4)=0$
$\Rightarrow\text{x}=0$ or $\text{x}^2+4=0$
However, since x is real, it is 0.
Thus, statement r is true.
Therefore, the given statement is true.

To show statement p to be true by contradiction, we assume that p is not true.

Let x be a real number such that $\text{x}^3+4\text{x}=0$ and let x is not 0.
Therefore, $\text{x}^3+4\text{x}=0$
$\text{x}(\text{x}^2+4)=0$
$\text{x}=0$ or $\text{x}^2+4=0$
$\text{x}=0$ or $\text{x}^2=-4$
However, x is real. Therefore, $\text{x}=0,$ which is a contradiction since we have assumed that x is not 0.
Thus, the given statement p is true.

To prove statement p to be true by contra positive method, we assume that r is false and prove that q must be false.

Here, r is false implies that it is required to consider the negation of statement r. This obtains the following statement.
$\sim\text{r . x}$ is not 0.
It can be seen that $(\text{x}^2+4)$ will always be positive.
$\text{x}\neq0$ implies that the product of any positive real number with x is not zero.
Let us consider the product of x with $(\text{x}^2+4)$.
$\therefore\text{x}(\text{x}^2+4)\neq0$
$\Rightarrow\text{x}^3+4\text{x}\neq0$
This shows that statement q is not true.
Thus, it has been proved that
$\sim\text{r}\Rightarrow\sim\text{q}$
Therefore, the given statement p is true.

View full question & answer→

MATHS (Each question 3 marks)

Take a timed test