Unsolved Problems In Mathematics!
Ever wondered “Math is complete or not?” If you have thought Math is Complete
than in some cases you may be wrong. Wrong in the sense that there are still
some problems in Mathematics, which are still remains to be solved or we can
say still they are unsolved. Yeah you are getting it right. So, what are the
problems they are still remains to be solved?
A few of
them are as follows:
1. The Riemann’s Hypothesis
2. The P vs. NP problem
3. The Navier-Stokes Equation
4. The Hodge Conjecture
5. Yang-Mills Theory and mass Gap
Hypothesis
To solve these problems, you need a good Mathematical
background. And also, in 2000 “The Clay Mathematics Institute” situated in Peterborough, New Hampshire, United States has announced $1 million prize to the person who ever solve any one of these
problems.
We are not going to
talk about any of these problems but there is a problem in Mathematics which
can be understood by any sixth-grade student and still remains to be solved.
And the problem is named as “Collatz Conjecture” or sometimes known as “3n+1”
problem.
Collatz conjecture:
This problem
or we can simply say Conjecture is named upon German Mathematician Lothar
Collatz (July6, 1910- Sep. 26, 1990). It states that every whole number either
even or odd eventually goes down to 1. And for this we have to consider the
following operation on any positive Integer n;
1. If the number is even divide it by
two that is
(n/2)
2. If n is odd triple it and add one
that is
(3n+1)
In Arithmetic notation we define a function f as follow:
The sequence of number involved is
known as Hailstone sequence or Hailstone numbers (because the value usually
goes up and down like hailstone in cloud.) or as wondrous number. People have
tried this function for the number up to 260 and that still get back
to 1.
Now the question
may arise what is the problem on this function, we will talk about it on the
last first let’s see some illustrative examples:
For example,
take no. 12
12 is even
so divide it by 2.
12/2=6
(even)
6/2=3 (odd)
(3*3)+1=10
(even)
10/2=5 (Odd)
(5*3)+1=16
(even)
16/2=8
(even)
8/2=4 (even)
4/2=2 (even)
2/2=1
Here again
if you think 1 is odd, then we can do it again for 1
(1*3)+1=4
(even)
4/2=2 (even)
2/2=1
Back to 1
again, so the cycle
4 2 1
This cycle keeps repeating after we
got to 1.
Now take a number less than 12, take
9 (smaller than 12)
Back to 1 again, here we can see that 12 greater than 9 took
more step to become 1. But also if we take 8 there will be more less steps.
Taking example of those numbers which will take more steps to
descend to 1:
1.
Between
1 and 100:
· 27 take 111 steps to become 1; it climbs
up to 9,232 before descending to 1.
· 54 and 55 both takes 112 steps to
descend to 1; both of them also climb up to 9,232.
· 97 take 118 steps to descend to 1.
2.
Between
1 and 1000:
· 703 take 170 steps to descend to 1 it
climbs up to 250,504.
· 937 take 173 steps to descend to 1 it
also climbs up to 250,504.
· 871 take 178 steps to descend to 1 it
climbs up to 190996.
3.
Between
1,000 and 10,000 is 6,171 which take 261 steps to descend to 1, It climbs up to
975,400.
4.
Between
10,000 and 100,000 is 77,031 it take 350 steps to descend to 1, it climbs up to
21,933,016.
5.
Between
100,000 and 1 million is 837,799 it takes 524 steps.
6.
Between
1 million and 10 million is 8,400,511 it takes 685 steps.
7.
Between
10 billion and 100 billion is 75,128,138,247 it takes 1,228 steps.
Actually, the problem in this Conjecture is that there is no
certain rule that, the larger number will take more steps or smaller number
will take fewer steps to become 1. For example, the number 909 take 15 steps to
descend to 1 where as the number 7 take 16 steps. There are so many other
examples like this. Also, we cannot find any pattern of numbers descending to
1.
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