Mathematical Reasoning - Notes, Overview, Books, FAQs

Mathematical reasoning is the logical thought process used to analyze, deduce, and infer properties, structures, and patterns within mathematical frameworks. Generally, Logic is the study of valid reasoning. But mathematical logic allows us to represent knowledge in a precise mathematical way and it also allows us to make valid inferences using a set of precise rules. It is regarded as a powerful tool for computer science because it is mainly used to verify the correctness of programs.

The main advantage of studying mathematical reasoning is that its results serve as very good tools for improving reasoning and problem-solving capabilities. Some of the branches of discrete mathematics are combinatorics, mathematical logic, Boolean algebra, graph theory, coding theory etc.

This article is about the concept of Mathematical Reasoning Class 11. Mathematical reasoning chapter is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE, BCECE, and more.

What is Mathematical Reasoning?

Mathematical reasoning is the logical thought process used to analyze, deduce, and infer properties, structures, and patterns within mathematical frameworks.

A mathematical statement is the basic unit of any mathematical reasoning. A sentence is called a mathematical statement if it is either true or false but not both. A true statement is called a valid statement and a false statement is called an invalid statement.

For example,

1. ‘Mumbai is in India’ is a statement because it is true.

2. Mount Everest is the highest mountain in the world.

3. $3+4=8$.

4. $(10-x)=7$.

5. '$x$ is a prime number' is NOT a mathematical statement as it can either be true or false depending on the value that the variable $x$ takes. Such statements are also called Open statements.

A sentence containing varying factors like today, tomorrow, here, there, she, he, it, etc is not a mathematical statement as the validity of the statement may differ with respect to the varying factors. Similarly, types of sentences like imperative sentences (command, order, request), exclamatory sentences (emotions, excitement) and interrogative sentences (question) are not mathematical statements as these sentences may not be true or false. The validity of the statement depends on the people.

Logical Connectives

The words which combine or change simple statements to form new statements or compound statements are called Connectives. The basic connectives (logical) conjunction corresponds to the English word ‘and’, disjunction corresponds to the word ‘or’, and negation corresponds to the word ‘not’. Logical connectives in mathematical reasoning are,

Negation of a Statement

The negation of the statement gives the opposite of the given statement. The negation of a statement $p$ in symbolic form is written as " $\sim$ $p$" and read as "not $p$".

$\begin{array}{|c|c|}\hline \mathrm{\;\;\;\;\;}p\mathrm{\;\;\;\;\;}&\mathrm{\;\;\;}\sim p\mathrm{\;\;\;\;\;} \\ \hline \mathrm{T}& \mathrm{F} \\ \hline \mathrm{F}&\mathrm{T} \\ \hline\end{array}$

Example:

$p$: New Delhi is the capital of India.
The negation of this statement is
$\sim$ $p$: New Delhi is not the capital of India.

Conjunction

Let $p$ and $q$ be any two simple statements. The conjunction of $p$ and $q$ is obtained by connecting $p$ and $q$ by the word and. It is denoted by $p \wedge q$, read as ' $p$ conjunction $q$ or ' $p$ hat $q$ '. The truth value of $p \wedge q$ is $T$, whenever both $p$ and $q$ are $T$ and it is $F$ otherwise.

$
\begin{array}{|c|c|c|}
\hline
\mathrm{\;\;\;\;} p \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} q \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} p \lor q \mathrm{\;\;\;\;} \\
\hline
\mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline
\mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}
$

For example,

${p}:$ Delhi is in India and $2+3=5$.
The statement can be broken into two component statements as
$q$: Delhi is in India.
$r: 2+3=5$

Disjunction

The disjunction of any two simple statements $p$ and $q$ is the compound statement obtained by connecting $p$ and $q$ by the word 'or'. It is denoted by $p \vee q$, read as ' $p$ disjunction $q$ ' or ' $p$ cup $q$ '. The truth value of $p \vee q$ is $F$, whenever both $p$ and $q$ are $F$ and it is $T$ otherwise.

$
\begin{array}{|c|c|c|}
\hline
\mathrm{\;\;\;\;} p \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} q \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} p \lor q \mathrm{\;\;\;\;} \\
\hline
\mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline
\mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}
$

Example:

$p$: Delhi is in India or $2+3=5$.
The statement can be broken into two component statements as
$q:$ Delhi is in India
$
r: 2+3=5
$

The compound statement with 'or' is true if any of its component statements are true.The component statement with 'or' is false if all of its component statements are false.

Conditional Statement

The conditional statement of any two statements $p$ and $q$ is the statement, "If $p$, then $q$ " and it is denoted by $p \rightarrow q$. $p \rightarrow q$ is false only if $p$ is true and $q$ is false. Otherwise it is true.

$
\begin{array}{|c|c|c|}
\hline
\mathrm{\;\;\;\;} p \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} q \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} p \rightarrow q \mathrm{\;\;\;\;} \\
\hline
\mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline
\mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}
$

The conditional statement $p \Rightarrow q$ can be expressed in several different ways. Some of the common expressions are
1. $p$ implies $q$
2. $p$ is a sufficient condition for $q$
3. $p$ only if $q$
4. $q$ is necessary condition for $p$.
5. $\sim q$ implies $\sim p$

Example:

$r$: If you are born in some country, then you are a citizen of that country
$p$ : you are born in some country.
$q$: you are a citizen of that country.
Then the sentence "if $p$ then $q$" says that in the event if $p$ is true, then $q$ must be true.

Biconditional Statement

If two statements $p$ and $q$ are connected by the connective "if and only if then the resulting compound statement " $p$ if and only if $q$ " is called a biconditional of $p$ and $q$ and is written in symbolic form as $p \leftrightarrow q$ or $p \Leftrightarrow q$.

A biconditional is true if and only if both the statements $p \rightarrow q$ and $q \rightarrow p$ are true. Also a biconditional is true if $p$ and $q$ both are true or when both $p$ and $q$ are false.

$
\begin{array}{|c|c|c|}
\hline
\mathrm{\;\;\;\;} p \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} q \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} p \leftrightarrow q \mathrm{\;\;\;\;} \\
\hline
\mathrm{T} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline
\mathrm{F} & \mathrm{T} & \mathrm{F} \\
\hline
\mathrm{F} & \mathrm{F} & \mathrm{T} \\
\hline
\end{array}
$

Exclusive OR

Let $p$ and $q$ be any two statements. Then $p$ EOR $q$ is such a compound statement that its truth value is decided by either $p$ or $q$ but not both. It is denoted by $p \oplus q$. The truth value of $p \oplus q$ is $T$ whenever either $p$ or $q$ is $T$, otherwise it is $F$. The truth table of $p \oplus q$ is given below.

$
\begin{array}{|c|c|c|}
\hline
\mathrm{\;\;\;\;} p \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} q \mathrm{\;\;\;\;} & \mathrm{\;\;\;\;} p \oplus q \mathrm{\;\;\;\;} \\
\hline
\mathrm{T} & \mathrm{T} & \mathrm{F} \\
\hline
\mathrm{T} & \mathrm{F} & \mathrm{T} \\
\hline
\mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline
\mathrm{F} & \mathrm{F} & \mathrm{F} \\
\hline
\end{array}
$

Converse

To form the converse of the conditional statement, interchange $p$ and $q$.

The converse of “If you are born in some country, then you are a citizen of that country” is “If you are a citizen of some country, then you are born in that country.”

Inverse

To form the inverse of the conditional statement, take the negation of both the $p$ and $q$.

The inverse of “If you are born in some country, then you are a citizen of that country” is “If you are not born in some country, then you are not a citizen of that country.”

Contrapositive

To form the contrapositive of the conditional statement, interchange the $p$ and $q$ and take the negation of both.

The Contrapositive of “If you are born in some country, then you are a citizen of that country” is “If you are not a citizen of that country, then you are not born in some country.”

Tautology

A compound statement is called tautology if it is always true for all possible truth values of its component statement.

For example, $p ∨ ~p , ( p ⇒ q ) ∨ ( q ⇒ p ) $

It is denoted by the letter ' $t$ '

Contradiction (fallacy)

A compound statement is called a contradiction if it is always false for all possible truth values of its component statement.

For example, $p ∧ ~p, ∼(( p ⇒ q ) ∨ ( q ⇒ p )) $

It is denoted by letter ' $c$ ' or ' $f$ '

$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \mathrm{\;\;\;\;\;}p\mathrm{\;\;\;\;\;}&\mathrm{\;\;\;}q\mathrm{\;\;\;}&\mathrm{\;\;\;\;\;}p\rightarrow q\mathrm{\;\;\;\;\;}&\mathrm{\;\;\;} q\rightarrow p\mathrm{\;\;\;} &\mathrm{\;\;\;}\left ( p\rightarrow q \right )\vee\left ( q\rightarrow p \right )\mathrm{\;\;}&\mathrm{\;\;\;}\sim\left ( \left ( p\rightarrow q \right )\vee\left ( q\rightarrow p \right ) \right ) \mathrm{\;\;} \\\hline \hline \mathrm{T}&\mathrm{T} & \mathrm{T} &\mathrm{T}&\mathrm{T}&\mathrm{F} \\ \hline \mathrm{T}&\mathrm{F} & \mathrm{F} &\mathrm{T}&\mathrm{T}&\mathrm{F} \\ \hline \mathrm{F}&\mathrm{T} & \mathrm{T} &\mathrm{F}&\mathrm{T}&\mathrm{F} \\ \hline \mathrm{F}&\mathrm{F} & \mathrm{T} &\mathrm{T}&\mathrm{T}&\mathrm{F} \\ \hline\end{array}$

Mathematical Reasoning Symbols

Mathematical reasoning symbols are,

Mathematical Reasoning Formulas

Mathematical reasoning formulas include the algebra of statements.

Algebra of Statements

Mathematical reasoning laws are,

Idempotent Law

1. $p ∨ p ≡ p$

2. $p ∧ p ≡ p$

$\begin{array}{|c|c|c|}\hline \mathrm{\;\;\;\;\;}p\mathrm{\;\;\;\;\;}&\mathrm{\;\;\;\;\;}p\vee p\mathrm{\;\;\;\;\;} &\mathrm{\;\;\;\;\;}p\wedge p\mathrm{\;\;\;\;\;} \\\hline \hline \mathrm{T}& \mathrm{T}&\mathrm{T} \\ \hline \mathrm{F}&\mathrm{F}&\mathrm{F} \\ \hline\end{array}$

Associative Law

1. $( p ∨ q ) ∨ r ≡ p ∨ (q ∨ r )$

2. $( p ∧ q ) ∧ r ≡ p ∧ (q ∧ r )$

$\begin{array}{|c|c|c|c|c|c|c|}
\hline \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{r} & \boldsymbol{p} \vee \boldsymbol{q} & \boldsymbol{q} \vee \boldsymbol{r} & (\boldsymbol{p} \vee \boldsymbol{q}) \vee \boldsymbol{r} & \boldsymbol{p} \vee(\boldsymbol{q} \vee \boldsymbol{r}) \\
\hline T & T & T & T & T& T & T \\
\hline T & T & F & T & T & T & T \\
\hline T & F & T & T & T & T & T \\
\hline T & F & F & T & F & T & T \\
\hline F & T & T & T & T & T & T \\
\hline F & T & F & T & T & T & T \\
\hline F & F & T & F & T & T & T \\
\hline F & F & F& F & F & F & F \\
\hline
\end{array}$

Distributive Law

1. $p ∧ (q ∨ r ) ≡ ( p ∧ q ) ∨ ( p ∧ r ) $

2. $p ∨ (q ∧ r ) ≡ ( p ∨ q ) ∧ ( p ∨ r ) $

\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{r} & \boldsymbol{q} \wedge \boldsymbol{r} & \boldsymbol{p} \vee(\boldsymbol{q} \wedge \boldsymbol{r}) & \boldsymbol{p} \vee \boldsymbol{q} & \boldsymbol{p} \vee \boldsymbol{r} & (\boldsymbol{p} \vee \boldsymbol{q}) \wedge(\boldsymbol{p} \vee \boldsymbol{r}) \\
\hline T & T & T & T & T & T & T & T \\
\hline T & T & F & F & T & T & T & T \\
\hline T & F & T & F & T & T & T & T \\
\hline T & F & F & F & T & T & T & T \\
\hline F & T & T & T & T & T & T & T \\
\hline F & T & F & F & F & T & F & F \\
\hline F & F & T & F & F & F & T & F \\
\hline F & F & F & F & F & F & F & F \\
\hline
\end{array}

Commutative Law

1. $p ∨ q ≡ q ∨ p$

2. $p ∧ q ≡ q ∧ p$

$\begin{array}{|c|c|c|c|}
\hline \boldsymbol{p} & \boldsymbol{q} & \boldsymbol{p} \vee \boldsymbol{q} & \boldsymbol{q} \vee \boldsymbol{p} \\
\hline T & T & T & T \\
\hline T & F & T & T \\
\hline F & T & T & T \\
\hline F & F & F & F \\
\hline
\end{array}$

Identity Law

1. $p ∧ T ≡ p$

2. $p ∧ F ≡ F$

3. $p ∨ T ≡ T$

4. $p ∨ F ≡ p$

$\begin{array}{|c|c|c|c|c|}
\hline \boldsymbol{p} & \mathbb{T} & \mathbb{F} & \boldsymbol{p} \vee \mathbb{T} & \boldsymbol{p} \vee \mathbb{F} \\
\hline T & T & F & T & T \\
\hline F & T & F & T & F \\
\hline
\end{array}$

Complement Law

5. $p ∨ ～p ≡ T$

6. $p ∧ ～p ≡ F$

7. $\sim (\sim p) ≡ p$

8. $～T ≡ F$

9. $～F ≡ T$

$\begin{array}{|c|c|c|c|c|c|c|c|}
\hline \boldsymbol{p} & \neg \boldsymbol{p} & \mathbb{T} & \neg \mathbb{T} & \mathbb{F} & \neg \mathbb{F} & \boldsymbol{p} \vee \neg \boldsymbol{p} & \boldsymbol{p} \wedge \neg \boldsymbol{p} \\
\hline T & F & T & F & F & T & T & F \\
\hline F & T & T & F & F & T & T & F \\
\hline
\end{array}$

De-Morgan’s Law

1. $～ ( p ∨ q ) ≡ ～p ∧ ～q$

2. $～ ( p ∧ q ) ≡ ～p ∨ ～q$

Truth table for $～ ( p ∨ q ) and ～p ∧ ～q$

$\begin{array}{|c|c|c|c|c|c|c|}
\hline p & q & \sim p & \sim q & p \vee q & \sim(p \vee q) & \sim p \wedge \sim q \\
\hline \hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline
\end{array}$

Truth table for $～ ( p ∧ q ) and ～p ∨ ～q$

$\begin{array}{|c|c|c|c|c|c|c|}
\hline p & q & \sim p & \sim q & p \wedge q & \sim(p \wedge q) & \sim p \vee \sim q \\
\hline \hline \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{F} \\
\hline \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline \mathrm{F} & \mathrm{F} & \mathrm{T} & \mathrm{T} & \mathrm{F} & \mathrm{T} & \mathrm{T} \\
\hline
\end{array}$

Mathematical Reasoning Questions with Solutions

Example 1: Let $S$ be a non-empty subset of $R$. Consider the following statement:
$P$ : There is a rational number $x \in S$ such that $x>0$.

Which of the following statements is the negation of the statement $P$ ?

1) There is a rational number $x \in S$ such that $x \leq 0$.
2) There is no rational number $x \in S$ such that $x \leq 0$.
3) Every rational number $x \in S$ satisfies $x \leq 0$.
4) $x \in S$ and $x \leq 0 \Rightarrow x$ is not rational.

Solution:

We write it up as $\sim p$
$P$ : There is rational numbers $x \epsilon S$ such that $x>0$. negation is every rational number $x \epsilon S$ satisfies $x \leq 0$.

Since negation of $x>0$ is $x \leq 0$.

The answer is option (3).

Example 2: Consider the following statements :
P : Ramu is intelligent.
Q : Ramu is rich.
R :Ramu is not honest.
The negation of the statement "Ramu is intelligent and honest if and only if Ramu is not rich" can be expressed as :

1) $
((P \wedge(\sim R)) \wedge Q) \wedge((\sim Q) \wedge((\sim P) \vee R))
$

2) $
((P \wedge R) \wedge Q) \vee((\sim Q) \wedge((\sim P) \vee(\sim R)))
$

3) $
((P \wedge R) \wedge Q) \wedge((\sim Q) \wedge((\sim P) \vee(\sim R)))
$

4) $
((\mathrm{P} \wedge(\sim \mathrm{R})) \wedge \mathrm{Q}) \vee((\sim \mathrm{Q}) \wedge((\sim \mathrm{P}) \vee \mathrm{R}))
$

Solution:

Given statement is $(\mathrm{p} \wedge \sim \mathrm{r}) \longleftrightarrow \sim \mathrm{q}$

$
\begin{aligned}
& \equiv((\mathrm{p} \wedge \sim \mathrm{r}) \rightarrow \sim \mathrm{q}) \wedge(\sim \mathrm{q} \rightarrow(\mathrm{p} \wedge \sim \mathrm{r})) \\
& \equiv(\sim(\mathrm{p} \wedge \sim \mathrm{r}) \vee \sim \mathrm{q}) \wedge(\mathrm{q} \vee(\mathrm{p} \wedge \sim \mathrm{r}))
\end{aligned}
$

It's negation is

$
\begin{aligned}
& \sim(\sim(\mathrm{p} \wedge \sim \mathrm{r}) \vee \wedge \mathrm{q}) \vee \sim(\mathrm{q} \vee(\mathrm{p} \wedge \sim \mathrm{r})) \\
& \equiv((\mathrm{p} \wedge \sim \mathrm{r}) \wedge \mathrm{q}) \vee(\sim \mathrm{q} \wedge \sim(\mathrm{p} \wedge \sim \mathrm{r}))
\end{aligned}
$

Hence correct option is 4

Example 3: If the truth value of the statement $(\mathrm{P} \wedge(\sim \mathrm{R})) \rightarrow((\sim \mathrm{R}) \wedge \mathrm{Q})$ is $F$, then the truth value of which of the following is $F$?

1) $
\mathrm{P} \vee \mathrm{Q} \rightarrow \sim \mathrm{R}
$

2) $
\mathrm{R} \vee \mathrm{Q} \rightarrow \sim \mathrm{P}
$

3) $
\sim(\mathrm{P} \vee \mathrm{Q}) \rightarrow \sim \mathrm{R}
$

4) $
\sim(\mathrm{R} \vee \mathrm{Q}) \rightarrow \sim \mathrm{P}
$

Solution:

$\mathrm{P} \wedge \sim \mathrm{R} \rightarrow(\sim \mathrm{R} \wedge \mathrm{Q})$ is false
$\therefore \mathrm{P} \wedge \sim \mathrm{R}$ is true and $((\sim \mathrm{R}) \wedge \mathrm{Q})$ is false $\Rightarrow P$ is true $R$ is false, $Q$ is false
(A) : $\mathrm{P} \vee \mathrm{Q} \rightarrow \sim \mathrm{R} \Rightarrow \mathrm{T} \rightarrow \mathrm{T} \Rightarrow \mathrm{T}$
(B) : $\mathrm{R} \vee \mathrm{Q} \rightarrow \sim \mathrm{P} \Rightarrow \mathrm{F} \rightarrow \mathrm{F} \Rightarrow \mathrm{T}$
(C) $: \sim(\mathrm{P} \vee \mathrm{Q}) \rightarrow \sim \mathrm{R} \Rightarrow \mathrm{F} \rightarrow \mathrm{T} \Rightarrow \mathrm{T}$
(D) $\sim(\mathrm{R} \vee \mathrm{Q}) \rightarrow \sim \mathrm{P} \Rightarrow \mathrm{T} \rightarrow \mathrm{F} \Rightarrow \mathrm{F}$
$\therefore$ option (D)

Example 4: Which of the following is an example of a disjunction?

1) If it rains, I will stay home.
2) It is raining and I am staying home.
3) It is raining or I am staying home.
4) I am staying home only if it is raining.

Solution:

Option (3) is correct as a disjunction is a compound proposition that connects two or more propositions using the logical operator "or". In this case, the two propositions being connected are "It is raining" and "I am staying home". The resulting compound proposition is true if at least one of the propositions is true, which is the case when it is either raining or the speaker is staying home.

Option (1) is incorrect as it is an example of a conditional statement, not a disjunction.

A conditional statement is a proposition of the form "If $A$, then $B$", where $A$ is the antecedent and $B$ is the consequent.

Option (2) is incorrect as it is an example of a conjunction, which is a compound proposition that connects two or more propositions using the logical operator "and".

Option (4) is incorrect as it is an example of a conditional statement, but the order of the antecedent and consequent are reversed.

Example 5: Which of the following is an example of a contradiction?

1) If it is raining, then it is sunny.
2) It is raining and it is not raining.
3) Either it is raining or it is not raining.
4) If it is raining, then it is not raining.

Solution:

A contradiction is a compound proposition that is always false, regardless of the truth values of its components.

Option (2) is correct as in this case, the proposition is always false because it cannot be both raining and not raining at the same time.

Option (1) is incorrect as it is a conditional statement that is not a contradiction, but it is false in situations when it is raining and not sunny.

Option (3) is incorrect as it is a tautology, not a contradiction.

Option (4) is incorrect as it is a conditional statement that is not a contradiction, but it is false in situations when it is both raining and not raining.

List of Topics According to NCERT/JEE MAIN

Importance of Mathematical Reasoning Class 11

Algebra at the JEE level is very interesting. All topics are more or less independent of each other. And one of the interesting and important topics is class 11 maths Sequences and Series and every year you will get 1-2 question in JEE Main exam as well as in other engineering entrance exams. JEE question paper is highly unpredictable, you never know questions from which topic will be asked. A general trend noticed in Mathematics paper is that a question involving multiple concepts are asked. As compared to other chapters in maths, Mathematical reasonoing questions for JEE MAINS requires less effort to prepare for the examination.

Every competitive exams include the topic mathematical reasoning and aptitude. Apart from competitive exams, almost every interview has mathematical reasoning and aptitude questions in the written test.

How to Study Mathematical Reasoning Class 11?

Mathematical Reasoning is one of the easiest topics, you can prepare this topic without applying many efforts. Start with basic theory, understand all the statemetns like negation, conjunction, disjunction, conditional and bi-conditional statements. Learn all the mathematical reasoning formulas and remember standard results. Give importance to the concepts like tautology and contradiction. Practice many problems from each topic for better understanding. Practice from the previous year mathematical reasoning questions for JEE MAINS.

If you are preparing for competitive exams then solve as many problems as you can. Do not jump on the solution right away. Remember if your basics are clear you should be able to solve any question on this topic.

Important Books for Mathematical Reasoning Class 11

Start from NCERT Books, the illustration is simple and lucid. You should be able to understand most of the things. Solve all problems (including miscellaneous problem) of NCERT. If you do this, your basic level of preparation will be completed.

Then you can refer to the book Arihant Algebra Textbook by SK Goyal or Cengage Algebra Textbook by G. Tewani but make sure you follow any one of these not all. Mathematical reasoning is explained very well in these books and there are an ample amount of questions with crystal clear concepts. Choice of reference book depends on person to person, find the book that best suits you the best, depending on how well you are clear with the concepts and the difficulty of the questions you require.

NCERT Solutions Subject wise link:

• NCERT solutions for class 11 Physics.

• NCERT solutions for class 11 Chemistry.

• NCERT solutions for class 11 Mathematics.

• NCERT solutions for class 11 Biology.