3 Marks Question

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 us 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
$n^2 > (3)^2$
$\Rightarrow n^2 > 9,$ which is a contradiction, since we have assumed that $n^2 < 9.$
Thus, the given statement is true. That is, if n is a real number with $n > 3$, then $n^2 > 9.$

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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.

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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.

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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.

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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.

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Question 63 Marks

Show that the statement
$p:$ “If $x$ is a real number such that $x^3 + 4x = 0,$ 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.

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Question 73 Marks

Write the component statements of the following compound statements and check whether the compound statement is true or false.
2 is an even number and a prime number.

Answer

It is true statements and its component statements are:
p: 2 is an even number. (True).
q: 2 is a prime number. (True).

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Question 83 Marks

Check the validity of the following statement.
q : 131 is a multiple of 3 or 11.

Answer

Given that:
q: 131 is a multiple of 3 and 11.
Let p: 131 is a multiple of 3 .
and r: 131 is is a multiple of 11 .
Here p is not true and r is also not true i.e., false.
So, $\text{q}\lor\text{r}$ is false.
Here, q is not valid.

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Question 93 Marks

Write the component statements of the following compound statements and check whether the compound statement is true or false.
24 is a multiple of 4 and 6.

Answer

Here, the given statement is of the form p $\land \,$ q which has the truth value T whenever both p and q have the truth value T. Hence, it is a true statement and its component statement are:
p: 24 is a multiple of 4. (True).
q: 24 is a multiple 6. (True).

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Question 103 Marks

Write the component statements of the following compound statements and check whether the compound statement is true or false.
All living things have two eyes and two legs.

Answer

It is a false statement. Since. p $\land \,$ q has truth value F whenever either p or q or both have the truth value F. Its component statements are
p: All living things have two eyes. (False).
q: All living things have two legs. (False).

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Question 113 Marks

Check the validity of the following statement.
p : 125 is divisible by 5 and 7.

Answer

Given that:
p: 125 is divisible by 5 and 7.
Let q: 125 is divisible by 5.
and r: 125 is divisible by 7.
Here q is true and r is false.
⇒ $\text{q}\land \text{r}$ is false.
Here, p is not valid.

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Question 123 Marks

Prove the following statement by contradication method.
p: The sum of an irrational number and a rational number is irrational.

Answer

Let p be false i.e., sum of an irational and a rational number is rational.
Let $\sqrt{\text{m}}$ is irational and a rational number.
$\Rightarrow \sqrt{\text{m}}+\text{n}=\text{r}$ [rational]
$\Rightarrow \sqrt{\text{m}}=\text{r}-\text{n}$
Now, $\sqrt{\text{m}}$ is irrational, whereas (r - n) is rational.
This is contradiction.
Then, our suppposition is wrong.
Hence, p is true.

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