SIMPLE PROOFS ofMATHEMATICAL TRUTHS

V. VENKATA NARAYANA, D.E.E. (R&B),

vvnhighways.blogspot.in , 07799139399

vvn.randb@gmail.com, 9440818440

MATHEMATICAL TRUTHS

1) X*0 = 0

2) X0 = 1

3) 1/0 = ∞

4) 1/∞ = 0

5) π = 22/7 = 355/113

6) Pythagoras theorem.

7) Linear INTERPOLATION

Friends

We know and frequently use the above

MATHEMATICAL truths. I am presenting a

simple proof of them.

1) 0- ZERO ::

> No Quantity,

> No Number,

> Nothing at all.

x-x = 0, 100-100=0.

2) ∞- INFINITY ::

> It is an abstract concept describing something

without any limit.

> Being without finish.

> Unboundedness.

> Another notion is that infinity is a quantity x

such that x + 1 = x . The idea is that the quantity is so

large that increasing its value by 1 does not change it.

DEFINITIONS

1) X*0 = 0

• We Know that

X*(Y+Z) = X*Y + X*Z and

A – A = 0

• Combining the above TWO facts

X*(0) = X(Y –Y)

= XY – XY

= 0

Hence X*0 =0

2) X0 = 1

• We Know that

> a5 = a*a*a*a*a

> a5 / a3 = a*a*a*a*a/a*a*a

= a*a = a2

= a5-3 .

• Hence am / an = a (m-n) .

• Using the above fact

X0 = X(a –a) = Xa/Xa

= 1

Hence X0 = 1

3) 1/0 = ∞• We Know that

> 1/1 = 1

> 1/0.1 = 10

> 1/0.01 = 100

> 1/0.001 = 1000

> 1/0.0001 = 10000

> 1/0.00001 = 100000

> 1/0.000001 = 1000000

> 1/0.0000001 = 10000000

> 1/0.00000001 = 100000000

--- --- --- --- --- --- --- --- --- --- ---

• As the DENOMINATOR decreases, the value of FRACTION increases.

• When the DENOMINATOR approaches to 0, the value of FRACTION

increase and increases to INFINITY.

Hence 1/0 = ∞

4) 1/∞ = 0• We Know that

> 1/1 = 1

> 1/10 = 0.10

> 1/100 = 0.01

> 1/1000 = 0.001

> 1/10000 = 0.0001

> 1/100000 = 0.00001

> 1/1000000 = 0.000001

> 1/10000000 = 0.0000001

> 1/100000000 = 0.00000001

--- --- --- --- --- --- --- --- --- --- ---

• As the DENOMINATOR increases, the value of FRACTION decreases.

• When the DENOMINATOR approaches to INFINITY (∞) , the value of

FRACTION decreases and decreases to ZERO.

Hence 1/ ∞ = 0

5) π = 22/7 = 355/113 = 3.14159

• We Know that

> Area of CIRCLE with radius r = π*r*r .

> Area of SQUARE with side 2r = 2r*2r = 4*r*r .

2r

5) π = 22/7 = 355/113 = 3.14159

• We Know that

> Area of CIRCLE with radius r = π*r*r .

> Area of SQUARE with DIAGONAL 2r = 2*(1/2*2r*r)

=2*r*r

2r

r

r

5) π = 22/7 = 355/113 = 3.14159

Inside SQUARE area < CIRCLE area < Outside SQUARE area

2*r*r < π*r*r < 4*r*r

2 < π < 4

It is proved that the value π lies between 2 and 3

5) π = 22/7 = 355/113 = 3.14159

� Area of ISOSCELES TRIANGLE with

SIDE l and included angle θ

sinθ/2 = (s/2)/l = s/2l

>>> s = 2l* sinθ/2

cosθ/2 = h/l

>>> h = l* cosθ/2

AREA = ½*base*height

= ½*s*h

=1/2*(2l*sinθ/2)*(l*cosθ/2)

=1/2(l*l*2sinθ/2*cosθ/2)

=1/2*l*l*sinθ

s/2

s

s/2

θ/2 θ/2

h

l

5) π = 22/7 = 355/113 = 3.14159

� Area of ISOSCELES TRIANGLE with

Height d and included angle θ

tanθ/2 = (s/2)/d = s/2d

>>> s = 2d* tanθ/2

AREA = ½*base*height

= ½*s*d

=1/2*(2d*tanθ/2)*(d)

=d*d*tanθ/2

s/2

s

s/2

θ/2 θ/2

d

5) π = 22/7 = 355/113 = 3.14159

� A CIRCLE of radius r is drawn.

� A REGULAR POLYGON of n sides is

drawn inside to fit in the CIRCLE.

� Area of this POLYGON is equal to n

times the are of ISOSCELES triangles.

� ISOSCELES TRIANGLE

side = r

included angle = 360/n

� Area of inside REGULAR POLUGON

= n*1/2*r*r*sin(360/n)

5) π = 22/7 = 355/113 = 3.14159

� A CIRCLE of radius r is drawn.

� A REGULAR POLYGON of n sides is

drawn outside to fit in the CIRCLE.

� Area of this POLYGON is equal to n

times the are of ISOSCELES triangles.

� ISOSCELES TRIANGLE

height = r

included angle = 360/n

� Area of outside REGULAR POLUGON

= n*r*r*tan(360/2n)

5) π = 22/7 = 355/113 = 3.14159

Inside POLYGON area < CIRCLE area < Outside POLYGON area

n*1/2*r*r*sin(360/n) < π*r*r < n*r*r*tan(360/2n)

n*sin(360/n) < 2π < 2n*tan(360/2n)

5) π = 22/7 = 355/113 = 3.14159

n*sin(360/n) < 2π < 2n*tan(360/2n)

� n=6 side 360/n=60 degrees

• 6*sin60 < 2π < 12 tan30

• 5.196 < 2π <6.928

• 2.598 < π <3.464

� n=12 side 360/n=30 degrees

• 12*sin30 < 2π < 24 tan15

• 6.00 < 2π <6.43

• 3.00 < π <3.215

5) π = 22/7 = 355/113 = 3.14159

n*sin(360/n) < 2π < 2n*tan(360/2n)

� n=24 side 360/24=15 degrees

• 24*sin15 < 2π < 48 tan7.5

• 6.212 < 2π <6.319

• 3.106 < π <3.159

From the above statements it can be

concluded that as the number of sides of REGULAR

POLYGON n increases, the value π approaches to

3.14159.

6) PYTHAGORAS THEOREM

PYTHAGORAS THEOREM states that

in a RIGHT ANGLE TRIANGLE the square of the

hypotenuse (the side opposite the right angle) is equal

to the sum of the squares of the other two sides.

� ABCD is a Square of side

(x+y).

� A smaller Square PQRS of

side z is inscribed in ABCD.

� Triangles APS, BQP, CRQ and

DSR are 4 RIGHT ANGLED

TRIANGLES with sides x, y, z.

A B

CD

P

Q

R

S

x

x

x

x

y

y

y

y

z

z

z

z

6) PYTHAGORAS THEOREM

FROM THE ABOVE FIGURE

ABCD = PQRS + APS + BQP + CRQ + DSR

(x+y)*(x+y) = z*z + 4*(1/2*x*y)

x2+y2+2xy = z2 +2xy

x2+y2 = z2

The sum of the squares

of the other two sides

The square of

the hypotenuse =

7) Linear INTERPOLATION

• The HEIGHT of an embankment

– At a distance X1 is Y1.

– At a distance X2 is Y2.

– What is it’s HEIGHT at distance X ?

X1

Y1

Y2

X2X

Y= ?

7) Linear INTERPOLATION

• From the Similarity of TRIANGLES

(����)

(����)=(�����)

(�����)

� − �� =(�� − ��)

(�� − ��)(� − ��)

X1

Y1

Y2

X2

X

Y

Y1

(Y-Y1) (Y2-Y1)

(X2-X1)

(X-X1)

� = �� +(�� − ��)

(�� − ��)(� − ��)

8) Squaring of NUMBERS

(� )� --------- 2*3=6 ----------- 625

(� )� --------- 3*4=12 ----------- 1225

(� )� ---------- 4*5=20 ----------- 2025

( )� ----------- 5*6=30 ------------- 3025

(� )� ------------ 6*7=42 ------------- 4225

(� )� ------------ 7*8=56 -------------- 5625

(� )� -------------- 8*9=72 --------------- 7225

(� )� -------------- 9*10=90 --------------- 9025

(�� )� -------------- 12*13=156 ------------- 15625

Got the IDEA of RULE ?

The RULE is applicable to the Numbers ending with ‘5’

FRIENDS

HAVE YOU ENJOYED IT ?