Maths Problem

This document contains solutions to various maths problems including proving alt-basis, monic FSP polynomials, combinations of polynomials, and properties of operations.

Maths Problem

Problem 1:
a) Prove D = (2n-1) = (1, 3, 7, 15, 31, …)
Proof:
For given sequence –
1 = 1,
2 = -1 + 3
3 = 3
4 = -3 + 7
5 = 1 -3 + 7
6 = -1 + 7
7 = 7
8 = -7 + 15
9 = 1 -7 + 15
Etc…
Therefore, from definition of alt-basis, every positive integer hereby, express in unique way. So,
D = (2n-1) is an alt basis.
(b) Whether E = (2, 3, …) be an alt-basis or not.
Proof:
Constructing an alt-basis as E = (en) with e1 = 2 and e2 = 3
From binomial expansion –
en = 1 + n + n2/!2 + n3/!3 + n2/!4 + …
so, at n = 1
e1 = 2
and, at n = 2
e2 = 3
while, at n = 3
e3 = 1 + 3 + 32/ !2, which cannot obtain result in integer form.
So, E = (2, 3, …) cannot be an alt-basis.
(c) Whether F = (1, 4, …) be an alt-basis or not.
Proof:
F can be constructed as F = (nn) for given values (1, 4, …)
1

From binomial expansion –
F = (1, 4, 9, 14, 25…)
so, 1 = 1
2 =
3 = - 1 + 4
4 = 4
5 = 1 + 4
6 = 1 – 4 + 9
As, 2 cannot be expressed by given expression of F, so, F is not an alt-basis
(d) From above analysis, it has been analysed that any sequence can be determined as an alt-
basis or not, if difference among sequence is neither too less nor too large.
Problem 2
(a) Are x and x – 1 the only monic FSP Polynomials of degree 1?
Solution: Yes, x and x-1 are the only monic Factor-Square Property polynomials, of degree 1
because further terms that are x+1, x+2, x+3 and more, cannot be factors of their squares.
(b) List of all monic FSP polynomials of degree 2.
x2, x2 – 1, x2 – x, x2 + x + 1, x2 – 2x + 1
(c) List of all monic FSP polynomials of degree 3.
Solution: Polynomials of degree 3 are –
x3, x3 – 1, x3 + 6x2 + 12x + 8, x3 – 3x2 + 3x – 1, x3 + 3x2 + 3x + 1, x3 – 6x2 + 12x – 8
These polynomials of degree 3 are new which cannot be express as product of two smaller FSP
polynomials expect x3, which is arise from degree 1 case.
Polynomials of degree 4 are –
x4, x4 – 1, x4 – x
(d) List of monic FSP polynomials of degree 3 which have integer coefficients.
Solution – FSP Polynomials with integer coefficients are x3, x3 – 1, x3 + 6x2 + 12x + 8, x3 – 3x2 +
3x – 1, x3 + 3x2 + 3x + 1, x3 – 6x2 + 12x – 8
2

Problem 3
(a) let P(x) = (1 – 2x)
And, Q(x) = (x2 + 2x -1)
Then,
P(x) (2x + 5) + Q(x) .3
= (1 – 2x) (2x + 5) + (x2 + 2x – 1) 3
= 2x + 5 – 4x2 – 10x + 4x2 + 8x – 3
= 1
This would not give the desired result, therefore, there will be no combination of P(x) and Q(x)
can satisfied the given problem, because the integers 5 and 3 have no common multiples whose
difference is one.
(b) let P(x) = (1 – 2x)
And, Q(x) = (x2 + 2x -1)
Then,
P(x) (2x + 5) + Q(x) .4
= (1 – 2x) (2x + 5) + (x2 + 2x – 1) 4
= 2x + 5 – 4x2 – 10x + 4x2 + 8x – 4
= 1
So, [(1 – 2x) (2x + 5) + (x2 + 2x – 1) 4] is the combination of (2x + 5) and 4 as it satisfies the
relationship.
(c) let P(x) = (5x – 3)
And, Q(x) = (-3x2 +1)
Then,
P(x) (15x + 9) + Q(x). 25
= (5x – 3) (15x + 9) + (-3x2 +1) 25
= 75x2 – 45x – 45x – 27 – 75x2 + 25
= -2
So, there is no combination of (15x + 9) and 25 that gives 1.
Similarly,
let P(x) = (5x – 3)
And, Q(x) = (-3x2 +1)
3

Then,
P(x) (15x + 9) + Q(x). 20
= (5x – 3) (15x + 9) + (-3x2 +1) 20
= 75x2 – 45x – 45x – 27 – 75x2 + 20
= -7
This would also not give the require combination.
Therefore, it has been analysed that integers that are 9 and 25 have no common multiples whose
difference is 1.
(d) From above system of integer polynomials, it has been evaluated that any integer can satisfy
a combination of polynomial as 1 when difference between constant terms is one. In other words,
polynomials (ax + b) and (cx + d) gives one only when (b – d) = 1.
Problem 4
(a) If a ◊ is unital, then it is true only for multiplication
i.e. x * y = y * x
which doesn’t change the answer.
Similarly, a function is called sandwiching both for multiplication and addition –
x * (y * z) = (x * y) * z
or,
x + (y + z) = (x + y) + z
Therefore, if an operation is sandwiching and unital both, then it must be commutative also,
because interchanging the order of operands, doesn’t change the answers in each condition.
(b) If an operation is Sandwiching and unital too, then,
(x ◊ 1) ◊ (1 ◊ 1) = (x ◊ 1) ◊ (1 ◊ 1) = x
Similarly,
(x ◊ 1) = (1 ◊ x) = x
This proves that if an operation is sandwiching and unital then it will associative too.
(c) Given, self-distributive operation as –
x ◊ (y ◊ z) = (x ◊ y) ◊ (x ◊ z)
If it is unital then, one of the variable must be one i.e.
1 ◊ (y ◊ z) = (1 ◊ y) ◊ (1 ◊ z)
This will give the same output, so, it satisfies the associative property too.
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(d) From above numerical analysis, it has been evaluated that if an operation is self-distributive
then it definitely holds the property of associative operation also.
Similarly, if a function is commutative then it consists property of both unital and sandwiching.
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