CG Board Class 10 Maths Chapter 14 Solutions: Proof of Mathematical Statements (गणितीय कथनों की जाँच) (2026-27)

गणितीय कथन वह वाक्य है जो या तो निश्चित रूप से सत्य होता है या निश्चित रूप से असत्य होता है, परंतु दोनों एक साथ नहीं हो सकता। केवल मापन या प्रयोगों (Empirical observation) द्वारा सभी स्थितियों के लिए गणितीय कथन को सिद्ध नहीं किया जा सकता; इसके लिए व्यापक तार्किक उपपत्ति (Proof) की आवश्यकता होती है।

असत्य**.

• Justification: The sum of interior angles of any quadrilateral is always (4 - 2) × 180°= 360°, not 350°.

1. x² ≥ 0 for a real number x.

सत्य.

• Justification: The square of any real number (positive, negative, or zero) is always non-negative.

1. The sum of two even numbers is always an even number.

सत्य.

• Justification: Let a = 2k₁ and b = 2k₂. Then a + b = 2(k₁ + k₂) = 2m, which is divisible by 2.

1. All prime numbers are odd.

असत्य.

• Justification: Counter-example: 2 is a prime number, but it is even.

1. 3n + 1 > 4, where n is a natural number.

असत्य.

• Justification: Counter-example: For natural number n = 1, 3(1) + 1 = 4, which is equal to 4, not strictly greater than 4.

1. x² > 0, where x is a real number.

असत्य.

• Justification: Counter-example: For real number x = 0, 0² = 0, which is not strictly greater than 0.

1. (a + b) + c = a + (b + c) ∀ a, b, c ∈ N

सत्य.

• Justification: Natural numbers satisfy the associative property under addition.

1. (p - q) + r = p - (q + r) ∀ p, q, r ∈ Q

असत्य.

• Justification: Left side is p - q + r while Right side is p - q - r. Since r ≠ -r for non-zero r, LHS ≠ RHS.

1. (x + y) - z = x + (y - z) ∀ x, y, z ∈ R

सत्य.

• Justification: LHS = x + y - z and RHS = x + y - z, which are identical for all real numbers.

Some axioms, theorems, and definitions are given in the following table:

1. Whole is greater than part (Axiom)
2. If all three sides of a triangle are different in measure, then that triangle is called a scalene triangle (Definition)
3. If n is an odd integer then it can be expressed as n = 2k + 1, where k is any integer (Definition)
4. Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle (Theorem)
5. If two things are equal to the same third thing then they are also equal to one another (Axiom)

Write possible conclusions on the basis of the above statements for the following given statements:

(i) In triangle RST and XYZ, RS = XY, ST = YZ and TR = ZX.

• Conclusion: Δ RST ≅ Δ XYZ (By Statement 4 - SSS Congruence Theorem).

(ii) AB/2 = AC

• Conclusion: AB > AC (By Statement 1 - Whole AB is greater than its part AC).

(iii) l = k+5/2 and 2m = k + 5, where k, l and m ∈ R

• Conclusion: k + 5 = 2l and k + 5 = 2m ⇒ 2l = 2m ⇒ l = m (By Statement 5 - Things equal to the same third thing are equal to one another).

(iv) In Δ DEF, DE ≠ EF ≠ FD

• Conclusion: Δ DEF is a scalene triangle (By Statement 2 - Definition of scalene triangle).

(v) 141 is an odd integer.

• Conclusion: 141 = 2 × 70 + 1 (By Statement 3 - Definition of odd integer where k = 70).

n₁ = 2k₁ and n₂ = 2k₂**.

(ii) Write the product n₁ n₂ in the form of k₁ and k₂.

n₁ n₂ = (2k₁)(2k₂) = 4k_1k₂ = 2(2k_1k₂).

(iii) Write n₁ + n₂ in the form of k₁ and k₂.

n₁ + n₂ = 2k₁ + 2k₂ = 2(k₁ + k₂).

(iv) Is n₁ × n₂ an even number or odd number? Why?

सम संख्या.

• Reason: n₁ × n₂ = 2(2k_1k₂) = 2m where m = 2k_1k₂ is an integer, which is divisible by 2.

(v) Is n₁ + n₂ an even number or odd number? Why?

सम संख्या.

• Reason: n₁ + n₂ = 2(k₁ + k₂) = 2p where p = k₁ + k₂ is an integer, which is divisible by 2.

द्विघात समीकरण है**.

• Reason: Degree of variable x is 2 and coefficient a ≠ 0.

(ii) bx + c = 0

द्विघात समीकरण नहीं है.

• Reason: Here का गुणांक x² is 0 (a = 0), so degree is 1 (linear equation).

(iii) ax² + c = 0

द्विघात समीकरण है.

• Reason: Degree of variable x is 2 and coefficient a ≠ 0.

(iv) ax² = 0

द्विघात समीकरण है.

• Reason: Degree of variable x is 2 and coefficient a ≠ 0.

(v) bx = 0

द्विघात समीकरण नहीं है.

• Reason: Here a = 0, so the degree of the equation is 1.

A rational number is any number that can be written in the form p/q, where p and q are integers and q is not equal to zero.

नहीं.

• Reason: Here denominator q = 0, which violates the definition q ≠ 0 (division by zero is undefined).

(iii) Is 81/1 a rational number? Give reason on the basis of given definition.

हाँ.

• Reason: Both p = 81 and q = 1 are integers (p, q ∈ I) and denominator q = 1 ≠ 0.

(iv) If b+9/a-5 is a rational number, where a, b ∈ N, then which value of a is not valid here? Why?

a = 5 is not valid.

• Reason: If a = 5, denominator a - 5 = 5 - 5 = 0, making the fraction undefined.

(v) If (p² + 7)/(q² - 25) is a rational expression, then here 5 and -5 cannot be values of q. (Use definition).

If q = 5 or q = -5, then q² - 25 = ( ± 5)² - 25 = 25 - 25 = 0. Since denominator cannot be zero, q cannot equal 5 or -5 (q ∈ I - {5, -5}).

Use mathematical symbols to write the following statements:

A. Integers are closed under multiplication.

• Mathematical Statement: a · b = c ∀ a, b, c ∈ I (or a × b ∈ I ∀ a, b ∈ I).

B. The subtraction of rational numbers is not commutative.

• Mathematical Statement: p - q ≠ q - p ∀ p, q ∈ Q.

Cube of a rational number is greater than or equal to that number.

असत्य है.

(iii) For which numbers is the statement true?

It is true for natural numbers n ∈ N (or n ≥ 1 and n = 0).

(iv) Can we take n = √2/7 in this statement?

नहीं.

• Reason: n = √2/7 is an irrational number, so n ∉ Q.

(v) Is statement true for negative numbers?

No, because the cube of a negative number is smaller than the number itself (e.g., (-2)³ = -8 < -2).

B. Statement: x² = 1 ⇒ x = 1

(i) Write this statement in the form of 'if-then'.

If the square of a number is 1, then the value of that number is 1.

असत्य है. (For real numbers, x² = 1 ⇒ x = 1 or x = -1).

(iii) Is the statement true when x ∉ N?

नहीं. (If x ∈ R, x can be -1).

असत्य**.

• Reason: Triangles and quadrilaterals are polygons, but they are not pentagons.

(ii) All hexagons are polygons.

सत्य.

• Reason: By definition, a hexagon is a 6-sided closed polygon.

(iii) All even numbers are not divisible by 2.

असत्य.

• Reason: By definition, every even number is divisible by 2.

(iv) Some real numbers are irrational.

सत्य.

• Reason: Real numbers consist of both rational and irrational numbers (√2, π are real and irrational).

(v) All real numbers are not rational.

सत्य.

• Reason: Irrational numbers are real numbers, but they are not rational.

In a parallelogram ABCD:

1. Opposite angles are equal: ∠ D = ∠ B = 80°.
2. Adjacent angles are supplementary (Sum = 180°):

∠ A + ∠ B = 180° ⇒ ∠ A + 80°= 180° ⇒ ∠ A = 100°∠ C = ∠ A = 100°* Final Answer: ∠ A = 100°, ∠ C = 100°, ∠ D = 80°.

Prove that 4m + 9 is an odd integer where m is an integer.

• Proof / उपपत्ति:

Given expression n = 4m + 9 where m ∈ I.

We can rewrite 9 as 8 + 1:

n = 4m + 8 + 1

n = 2(2m + 4) + 1

Let k = 2m + 4. Since m is an integer, k is also an integer (k ∈ I).

n = 2k + 1

By definition of odd integers, any integer of the form 2k + 1 is an odd integer.

Therefore, 4m + 9 is an odd integer for all m ∈ I. (Proved)

Write negation for following statements by using appropriate symbols:

A. M : √7 is an irrational number.

• Negation ( ∼ M): ∼ M : √7 is not an irrational number.

B. A : 6 + 3 = 9

• Negation ( ∼ A): ∼ A : 6 + 3 ≠ 9

C. D : Some rational numbers are integers.

• Negation ( ∼ D): ∼ D : Some rational numbers are not integers.

D. P : Triangle PQR is equilateral.

• Negation ( ∼ P): ∼ P : Triangle PQR is not equilateral.

Find logical connections in the following statements and write them by using ( ⇒ ):

A. A : All interior angles of a triangle Δ ABC are equal.

B : Triangle Δ ABC is an equilateral triangle.

• Logical Connection: A ⇒ B and B ⇒ A (i.e., A ⇔ B).

B. T : P(a) = 0

S : (x - a) का एक गुणनखंड है the polynomial P(x).

• Logical Connection: S ⇒ T and T ⇒ S (i.e., S ⇔ T).

C. P : x and y are two odd numbers.

Q : x + y is an even number.

• Logical Connection: P ⇒ Q.

### Exercise - 3 (प्रश्नावली - 3)

#### Question 1:

Prove that the sum of interior angles of a polygon with n sides whose all angles are equal is n [ 180 - 360/n ]°where n ≥ 3.

• Proof / उपपत्ति:

1. We know that the sum of all interior angles of an n-sided convex polygon is given by the formula:

Sum of interior angles = (n - 2) × 180°2. Expanding the expression:

Sum = n × 180°- 2 × 180°= n × 180°- 360°3. Factoring out n from the terms:

Sum = n ( (n × 180°)/(n) - (360°)/(n) ) = n [ 180 - 360/n ]°Hence proved for n ≥ 3. (Proved)

Prove that the sum of n terms of an A.P. is 3n² + 4n, if the n^th term of that AP is 6n + 1.

• Proof / उपपत्ति:

Given n^th term a_n = 6n + 1.

• For n = 1 (First term a₁):

a₁ = 6(1) + 1 = 7

• Using formula for sum of n terms of an AP:

S_n = n/2 [a₁ + a_n]

• Substituting a₁ = 7 and a_n = 6n + 1:

S_n = n/2 [7 + (6n + 1)] = n/2 [6n + 8]

S_n = n/2 · 2(3n + 4) = n(3n + 4) = 3n² + 4n

Hence proved. (Proved)

Prove that the sum of three successive even integers is always a multiple of 6.

• Proof / उपपत्ति:

1. Let three successive even integers be 2k, 2k + 2, and 2k + 4, where k is any integer.
2. Their sum S is:

S = (2k) + (2k + 2) + (2k + 4)

S = 6k + 6 = 6(k + 1)

1. Let m = k + 1. Since k ∈ I, m ∈ I.

S = 6m

Since S is divisible by 6, the sum of three successive even integers is always a multiple of 6. (Proved)

Prove that 8 का एक गुणनखंड है (2n + 3)² - (2n - 3)², where n is a natural number.

• Proof / उपपत्ति:

Given expression E = (2n + 3)² - (2n - 3)².

Using algebraic identity a² - b² = (a + b)(a - b):

Here a = 2n + 3 and b = 2n - 3.

a + b = (2n + 3) + (2n - 3) = 4n

a - b = (2n + 3) - (2n - 3) = 6

E = (4n)(6) = 24n

Factoring out 8:

E = 8(3n)

Since 3n is an integer for all n ∈ N, 8 का एक गुणनखंड है (2n + 3)² - (2n - 3)². (Proved)

Prove that if the sum of squares of two successive whole numbers is divided by 4, then the शेषफल always comes as 1.

• Proof / उपपत्ति:

1. Let two successive whole numbers be n and n + 1 where n ∈ W.
2. Sum of their squares S:

S = n² + (n + 1)² = n² + (n² + 2n + 1) = 2n² + 2n + 1

S = 2n(n + 1) + 1

1. Since n and n + 1 are two consecutive whole numbers, one of them must be even. Therefore, their product n(n + 1) is always an even number, i.e., n(n + 1) = 2k for some integer k.
2. Substituting n(n + 1) = 2k:

S = 2(2k) + 1 = 4k + 1

1. Dividing S = 4k + 1 by 4 gives भागफल k and शेषफल 1.

Hence, when the sum of squares of two successive whole numbers is divided by 4, the शेषफल is always 1. (Proved)