Proofs in Mathematics

(i) This statement is always false, as there are 12 months in a year.

(ii) This statement is ambigious, because it is not necessary that Diwali falls on a Friday every year.

(iii) This statement is ambiguous, because temperature does not remains fixed. At some time in the year, the temperature in Magadi, could be less than or greater than 26°C.

(iv) This statement is always true, because the earth has only one moon.

(v) This statement is always false, because dogs cannot fly.

(vi) This statement is ambiguous, because in a leap year, February has 29 days and in non-leap years February has 28 days.

(i) This statement is false because the sum of the interior angles of a quadrilateral is 360°.

(ii) This statement is true for all real numbers (x), x² ≥ 0.

(iii) This statement is true because in a rhombus, both pairs of opposite sides are parallel and hence every rhombus is a parallelogram.

(iv) This statement is true because this is true for any two even numbers.

(v) This statement is false because the sum of any two odd numbers is always an even number.

(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 1.5

On 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 > $\dfrac{3}{3}$

⇒ 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.

(i) Since, all humans are mammals and all mammals are vertebrates.

∴ We can conclude that,

Humans are vertebrates.

(ii) Since, Anthony is a barber and Dinesh had his hair cut.

∴ We can conclude that,

It is not necessary that Antony has cut Dinesh’s hair, as Dinesh could have got his hair cut by another barber.

(iii) Since, Martians have red tongues and Gulag is a Martian.

∴ We can conclude that,

Gulag has a red tongue.

(iv) Given,

If it rains for more than four hours on a particular day, the gutters will have to be cleaned the next day and it has rained for 6 hours today.

∴ We can conclude that,

The gutters will have to be cleaned tomorrow.

(v) We know that,

All animals have tails need not be dogs. For example, animals such as buffaloes, monkeys, cats, etc. have tails but are not dogs.

We need to turn over B and 8, in order to check whether the rule holds.

Rule is broken, if any of the one condition occurs :

1. If B has an even number on the other side.

2. If 8 has a consonant on the other side.

Let three consecutive numbers be 8, 10 and 12.

Product = 8 × 10 × 12 = 960.

Three possible conjectures are:

(i) The product of any three consecutive even numbers is even.

(ii) The product of any three consecutive even numbers is divisible by 4.

(iii) The product of any three consecutive even numbers is divisible by 6.

Given,

Line 1 : 1 = 11⁰

Line 2 : 1 1 = 11¹

Line 3 : 1 2 1 = 11²

Looking at the above pascal triangle,

Possible conjecture :

Line 4

1 3 3 1 = 11³

Line 5

1 4 6 4 1 = 11⁴

Since, The conjecture holds for line 4 and line 5.

Line 6 :

161051 = 11⁵

The conjecture does not holds for above line.

Hence, conjecture holds for line 4 and 5 but does not holds for line 6.

Given,

⇒ T₁ + T₂ = 4 = 2²

⇒ T₂ + T₃ = 9 = 3²

⇒ T₃ + T₄ = 16 = 4²

Conjecture :

On adding two triangular numbers the resultant value is the square of the larger number.

By using conjecture :

⇒ T₄ + T₅ = 5² = 25.

Hence, T_{n - 1} + T_n = n².

Looking at above pattern,

Possible conjecture :

The resultant will be series of numbers from one till number of 1's in the digit to be squared and then reversing the order.

By following conjecture, we can write:

⇒ 111111² = 12345654321 and

⇒ 1111111² = 1234567654321

On actual multiplication, we get :

⇒ 111111² = 12345654321 and

⇒ 1111111² = 1234567654321.

Hence, conjecture is true.

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.

(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 : $\sqrt{a^2 + b^2}$ = a + b

Counter example : Let a = 3 and b = 4

Substituting value of a and b in L.H.S. of the equation, we get :

$\Rightarrow \sqrt{3^2 + 4^2} \\[1em] \Rightarrow \sqrt{9 + 16} \\[1em] \Rightarrow \sqrt{25} = 5$

Substituting 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 $\sqrt{a^2 + b^2}$ = a + b is not true.

(v) Given,

Equation : 2n² + 11

Counter 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 + 41

Counter 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.

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 = ∠4

Hence, 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 :

1. We begin with a triangle and one exterior angle.

2. 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)

3. 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°.

4. Here we use Euclid's axiom which states that: “Things which are equal to the same thing are equal to each other.”

5. Here we use Euclid‟s axiom which states that: “If equals are subtracted from equals, the remainders are equal.”

Let x and y be two odd numbers.

Then x = 2k + 1 for some natural number and y = 2l + 1 for some natural number l.

Adding x and y, we get :

⇒ x + y = 2k + 1 + 2l + 1

= 2k + 2l + 2

= 2(k + l + 1).

We know that,

Any natural number on multiplying by 2 is an even number.

Hence, proved that the sum of two odd numbers is even.

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) + 1

Let (kl + l + 1) be M.

⇒ xy = 2M + 1

We 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.

Let three consecutive even numbers be : 2n, 2n + 2, 2n + 4

Their sum = 2n + 2n + 2 + 2n + 4

⇒ (2n + 2n + 2n) + (2 + 4)

⇒ 6n + 6 = 6(n + 1) = 6k, where k = n + 1.

Clearly 6k is divisible by 6.

Hence, proved that the sum of three consecutive even numbers is divisible by 6.

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.

(i) Steps :

1. Let our original number be n.

2. Double the number (2n).

3. Add nine to the number (2n + 9).

4. Add original number (2n + 9 + n = 3n + 9).

5. Divide the number by 3 = $\dfrac{3n + 9}{3} = \dfrac{3(n + 3)}{3}$ = n + 3.

6. Add four (n + 3 + 4 = n + 7).

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 = 1001

Take any three digits number say, xyz.

On multiplying,

⇒ xyz x 1001 = xyzxyz

Since, 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.