Question 1

Question 3Restate the following statements with appropriate conditions, so that they become true statements.(i) All prime numbers are odd.(ii) Two times a real number is always even.(iii) For any x, 3x + 1 > 4.(iv) For any x, x3 ≥ 0.(v) In every triangle, a median is also an angle bisector.

Accepted Answer

(i) We know that,A natural number that has exactly two factors, i.e., 1 and the number itself, is a prime number. Example - 2, 3, 5, 7,.......Hence, all prime numbers greater than 2 are odd.(ii) Let real number be 1.5On multiplying by 2, we get :1.5 × 2 = 3, which is an odd number.∴ The statement, two times a real number is always even, is incorrect.But if we multiply, any natural number by 2, we always get an even number.Hence, two times a natural number is always even.(iii) Given,⇒ 3x + 1 > 4⇒ 3x > 4 - 1⇒ 3x > 3⇒ x > 33\dfrac{3}{3}33​⇒ x > 1.Hence, for x > 1, 3x + 1 > 4.(iv) For any x ≥ 0, x3 ≥ 0.(v) We know that,A median is also the angle bisector in equilateral triangle.Hence, in equilateral triangle, a median is also an angle bisector.

Question 2

Question 5List five axioms (postulates) used in this book.

Accepted Answer

Euclid's postulates are:Axiom 5.1- Things which are equal to the same thing are equal to one another.Axiom 5.2 - If equals are added to equals, the wholes are equal.Axiom 5.3 - If equals are subtracted from equals, the remainders are equal.Axiom 5.4 - Things which coincide with one another are equal to one another.Axiom 5.5 - The whole is greater than the part.

Question 3

Question 1Find counter-examples to disprove the following statements:(i) If the corresponding angles in two triangles are equal, then the triangles are congruent.(ii) A quadrilateral with all sides equal is a square.(iii) A quadrilateral with all angles equal is a square.(iv) For integers a and b, a2+b2\sqrt{a^2 + b^2}a2+b2​ = a + b(v) 2n2 + 11 is a prime for all whole numbers n.(vi) n2 – n + 41 is a prime for all positive integers n.

Accepted Answer

(i) Counter example :Consider two equilateral triangle Δ ABC and Δ DEF of length 3 cm and 6 cm respectively.From figure,All corresponding angles are same but sides are of different length.∴ Δ ABC and Δ DEF are not congruent.(ii) Counter example :A quadrilateral with all sides equal is a rhombus which may not be a square.(iii) Counter example :A rectangle has all angles equal, but may not be a square.(iv) Given,Equation : a2+b2\sqrt{a^2 + b^2}a2+b2​ = a + bCounter example : Let a = 3 and b = 4Substituting value of a and b in L.H.S. of the equation, we get :⇒32+42⇒9+16⇒25=5\Rightarrow \sqrt{3^2 + 4^2} \[1em] \Rightarrow \sqrt{9 + 16} \[1em] \Rightarrow \sqrt{25} = 5⇒32+42​⇒9+16​⇒25​=5Substituting value of a and b in R.H.S. of the equation, we get :⇒ 3 + 4 = 7.Since,L.H.S. ≠ R.H.S.Hence, for a = 3 and b = 4, the statement a2+b2\sqrt{a^2 + b^2}a2+b2​ = a + b is not true.(v) Given,Equation : 2n2 + 11Counter example :Let n = 11 (by hit and trial method)Substituting value of n in given equation.2n2 + 11 = 2(11)2 + 11= 2 x 121 + 11= 242 + 11 = 253.253 is divisible by 11.Therefore, 253 is not a prime number so given statement is false for n = 11.Hence, for n = 11, 2n2 + 11 = 253, which is not a prime number.(vi) Given,Equation : n2 - n + 41Counter example :Let n = 41.Substituting value of n in given equation, we get :n2 - n + 41 = (41)2 - 41 + 41= 1681.1681 is not prime.Hence, for n = 41, n2 – n + 41 is not prime.

Question 4

Question 2Take your favourite proof and analyse it step-by-step along the lines discussed in Section A1.5 (what is given, what has been proved, what theorems and axioms have been used, and so on).

Accepted Answer

If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.In Δ ABC,⇒ ∠1 + ∠2 + ∠3 = 180° (By angle sum property) .....(1)⇒ ∠3 + ∠4 = 180° (Linear pair) ....(2)By comparing equation (1) and (2), we get :⇒ ∠1 + ∠2 + ∠3 = ∠3 + ∠4⇒ ∠1 + ∠2 = ∠4Hence, proved that if a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.Analysis of proof :We begin with a triangle and one exterior angle.Here we use angle sum property of triangle that is sum of three interior angles of a triangle is 180° to show ∠1 + ∠2 + ∠3 = 180°.(By theorem)Here we use linear pair axiom, which states that the angles on a straight line add up to 180° to show that ∠3 + ∠4 = 180°.Here we use Euclid's axiom which states that: “Things which are equal to the same thing are equal to each other.”Here we use Euclid‟s axiom which states that: “If equals are subtracted from equals, the remainders are equal.”

Question 5

Question 4Prove that the product of two odd numbers is odd.

Accepted Answer

Let x and y be two odd numbers.Then x = 2k + 1 for some natural number k and y = 2l + 1 for some natural number l.Multipying both x and y, we get :⇒ xy = (2k + 1)(2l + 1)⇒ xy = 2k(2l + 1) + 1(2l + 1)⇒ xy = 4kl + 2k + 2l + 1⇒ xy = 2(kl + k + l) + 1Let (kl + l + 1) be M.⇒ xy = 2M + 1We know that,Any natural number on multiplying by 2 is an even number.On adding 1 to an even number we an odd number.So, product of two odd numbers is odd.Hence, proved that the product of two odd numbers is odd.

Question 6

Question 6Prove that infinitely many points lie on the line whose equation is y = 2x. (Hint : Consider the point (n, 2n) for any integer n.)

Accepted Answer

Consider the point (n, 2n) for any integer n.Given equation y = 2x ....(1)Putting x = n and y = 2n in equation (1) we get :⇒ 2n = 2(n)⇒ 2n = 2n, which is always true.So, (n, 2n) lies on equation (1).Now for infinite values of n, we can obtain infinite points which lie on line whose equation is y = 2x.Hence, proved that infinitely many points lie on the line whose equation is y = 2x.

Question 7

Question 7You must have had a friend who must have told you to think of a number and do various things to it, and then without knowing your original number, telling you what number you ended up with. Here are two examples. Examine why they work.(i) Choose a number. Double it. Add nine. Add your original number. Divide by three. Add four. Subtract your original number. Your result is seven.(ii) Write down any three-digit number (for example, 425). Make a six-digit number by repeating these digits in the same order (425425). Your new number is divisible by 7, 11 and 13.

Accepted Answer

(i) Steps :Let our original number be n.Double the number (2n).Add nine to the number (2n + 9).Add original number (2n + 9 + n = 3n + 9).Divide the number by 3 = 3n+93=3(n+3)3\dfrac{3n + 9}{3} = \dfrac{3(n + 3)}{3}33n+9​=33(n+3)​ = n + 3.Add four (n + 3 + 4 = n + 7).Subtract original number (n + 7 - n = 7).From above steps,It is clear that on selecting any number, the answer remain same, i.e., 7.(ii) Multiplying 7, 11 and 13 we get :7 x 11 x 13 = 1001Take any three digits number say, xyz.On multiplying,⇒ xyz x 1001 = xyzxyzSince, 7, 11 and 13 are factor of 1001 and 1001 is a factor of xyzxyz.∴ 7, 11 and 13 are factor of xyzxyz.Hence, any six digit number of form xyzxyz formed from a three digit number of form xyz is divisible by 7, 11 and 13.

Proofs in Mathematics

Exercise A11

Exercise A13

Exercise A14