SIMPLE
MATHEMATICAL PROOFS
A
NEW APPROACH TOWARDS MATHEMATICS
TABLE
OF CONTENTS
11) INTRODUCTION
22) WHY
(-VE) * (-VE) IS (+VE)
33) WHY
ANYTHING RAISED TO THE POWER 0 IS 1
54) THINGS
TO KNOW ABOUT MATHEMATICS
65) CONCLUSION
The topic I have here is certainly very
intriguing and weird to be presented among other blogs that you might have come across,
filled with jargon terms and technical brilliance, possibly good enough to revolutionize the entire world. At the same time, I assure you dear friends,
that next few minutes will help you to revolutionize yourself. I am going to
present now a series of simple mathematical proofs, some of which you are quiet
likely to have stumbled upon on few occasions and rest are quiet unique.
Has anyone of
you ever realized that we conveniently disregard the necessity of getting a
convincing reason for few mathematical concepts like
1) Why
negative *negative is positive?
2) Why
alternate interior angles are equal?
3) Why
vertically opposite angles are equal?
4) Why
anything raised to the power 0 is 1?
And so forth…..
Even if someone does
ask questions like this most teachers especially in schools and in some cases,
even in colleges ridicule such questions saying it is very evident and needs no
proof. It is as much the mark of an ignorant man, they say, to require
persuasion of evident truths as to believe what is obscure without question.....
In my
personal opinion, as a society, we can't do without mathematics. Mathematics is
in our culture. It's a vehicle of progress of all other sciences. On an
individual level, a small percentage of
professions require mathematical knowledge of various degree. Consumer
mathematics, however useful, is a misnamed term for numeracy. An
average person may be happy without a scruple of mathematical fluency.
1) Why
negative * negative is positive?
A thought occurred to
me in my schooling….
If 3*2 is 6 ……. Then (-3)*(-2) = (-6) right???? Yes, of course I know that it is a
terrible mistake. But still I wasn’t quite convinced.
Now let me try to
convince you that (-3)*(-2) is indeed equal to (+6)…..
When an arithmetic
progression with a common difference d is multiplied with a constant x, we get
another arithmetic progression of common difference xd.
Now I am going to take
an arithmetic progression, say AP1.
3,2,1,0,-1,-2 and so
on………
AP 1 *
constant = AP 2
3 *
2 = 6
2 *
2 = 4
1 *
2 = 2
0 *
2 = 0
-1 *
2 = ?
Now to find what comes in place of that “?” ….. Consider
AP 2.
In AP 2…. First term,
say a1 is 6, a2 is 4….. a4 is 0 then in place of a5 we have “?” .
We know c.d = a5 – a4=
a2- a1…. This implies
a5 = a4+c.d….i.e.
a5= a4+a2-a1
a5= 0+4-6 = (-2)
Therefore, (-2) comes
in place of “?”…..
AP 1 * constant = AP
2
3 *
2 = 6
2 *
2 = 4
1 *
2 = 2
0 *
2 = 0
-1 *
2 = (-2)
Thus we have proved
(-ve) * (+ve) is (-ve)…….
Now to prove that (-ve)
* (-ve) is (+ve) we are going to follow a similar procedure.
Take an arithmetic
progression, say AP1.
3, 2, 1, 0,-1,-2 and so
on………
AP 1 *
constant = AP 2
3 *
- 2 = -6
(Since –ve * +ve is –ve)
2 *
- 2 = -4
1 *
- 2 = -2
0 *
- 2 = 0
-1 *
- 2 = X
To find X, consider AP
2…..
First term a1= -6, a2=
-4… a4=0, a5= X
c.d= a5-a4=a2-a1…. This
implies
a5= a4+c.d … here
common difference is +2.
Therefore a5= 2….. This
helps us to arrive at the result
(-1)*(-2) is (+2) and
hence –ve * -ve is +ve………
Now let us discuss
about a much commonly known proof for proving the above result……..
We know,
•
(-1) + (1) = 0
Now multiply “a” on
both sides of the equation, such that a>0…..
•
a [ (-1) + (1) ] = a*0
•
a * (-1) +
a = 0 ( by distributive property of multiplication
and by identity
that a*0 = 0 )
Now subtract ‘a ‘on
both sides of the equation.
•
a * (-1) + a - a = 0 – a
•
a * (-1) + 0 = -a
•
Therefore a*(-1) = (-a)
mplication
is that ….. +ve * -ve is -ve
We know,
•
(-1) + (1) = 0
Now multiply ‘ –a
‘ on both sides of the equation, such
that a>0 and –a <0…..
•
-a [ (-1) + (1) ] = a*0
-a * (-1) +
-a = 0 ( since we have proved –ve * +ve is –ve we
can write (-a)*1 is –a)
Now add ‘a ‘on both
sides of the equation.
•
-a * (-1) + (-a) + a = 0 + a
•
-a * (-1) + 0 =
+a
•
Therefore -a*(-1)
= (+a)
Implication is that …..
-ve * -ve is +ve
Thus indeed …..(-3) *
(-2) is +6
4) Anything raised to the power 0 is 1????
We very often use this property in mathematics…..
Anything to the power 0 is 1….. But most of us have
no idea why is that so…
Here is a commonly known proof for the benefit of
those who do not know…..
a^0 can be
written as….. a^(m-m)
a^(m-m) = a^m * a^(-m)
= (a^m) / (a^m)
= 1
thus, anything raised to the power 0 is one…..
•
Things to know in mathematics
•
There are few kindergarden topics in
mathematics that we often disregard.
Here is an example…..
if x+3 = 6
then , x = 6
– 3
x = 3…..
Why does + on coming to the right hand side of the
equation becomes minus (-)……???
Why not
x= 6*3 or x = 6/3????
•
Well… as some of you would have figured
out…
The sign “ = “ is like a physical balance.
Operations done on L.H.S should be done on R.H.S to maintain equality.
x + 3 = 6
x + 3 – 3 =
6 – 3
x + 0 = 6 – 3
By identity x+ 0= x = 6-3 =3….
We actually subtract 3 on both sides rather than
bringing the three from L.H.S to R.H.S……..
Same is the case when we right x/2 = 4 as x = 2 * 4…
we actually multiply both sides by 4…….
In fact, math is not always friendly…. There are few
fallacies that have no convincing answer….
For example:
We are
going to evaluate {(1/x)dx by partial integration:
{udv=uv−
{vdu
•
For this example, set u=1/x, v=x, then
du=−1/x^2 dx, dv=dx.
So we obtain,
•
{(1/x)dx=1−{(x*(-1/x^2))dx
•
{(1/x)dx=1-{ (-1/x)dx
•
{(1/x)dx = 1 + {(1/x)dx
•
A
= 1 + A
•
0 = 1
•
Jack Glover wishes to add enough 50%
antifreeze solution to 16 gallons of a 5% antifreeze solution to obtain a 20%
antifreeze solution. How much of the 50% solution should he add?
•
•
I talked to an employee at a service
station. He did this kind of problem in mathematics classes, but on the job
nobody does it this way. They test antifreeze in the car for the temperature at
which it provides protection. Then they add enough antifreeze to get to the
desired temperature.
•
While many people use some mathematics
in their jobs, Dr. Appelbaum feels that mathematics should nonetheless be
taught primarily for its own sake. Applications are fine if they are simple and
appealing; otherwise they should be left to an applied course. She sees the
current emphasis on applications as a response to anti-intellectualism among
students. When students ask what good the mathematics is, she suspects that the
students are really saying that they cannot understand the subject and so hope
that it is no good. She has often met people who are glad that they studied
mathematics, or wish that they had studied more, but never anyone who said that
they were sorry to have learned mathematics.
•
Mathematics contains much that will
neither hurt one if one does not know it nor help one if one does know it.
Stand firm in your refusal to remain conscious
during algebra. In real life, I assure you, there is no such thing as algebra.
Though, mathematics is not entirely useful and
essential to life, learn maths for its pure fun, satisfaction and not just because
it is a 4 credit subject…!!!!
* Have a good day!!!!
….. the end…..
