Question

.

Answer 1

Gmail

Answer 2

I cant see files here xD and i wont be able to get online from pc , so its better if you could send it to gmail please .

Answer 3

Q1 :- Prove that there do not exist prime numbers a,b,c such that a^3 + b^3 = c^3

Proof :-
Since 2 is the only even prime and \(2^3+2^3 \ne 2^3\), the numbers (a,b,c) cannot be all even. Also they cannot be all odd because then the left hand side evaluates to even and right hand side evaluates to odd.

If one of (a, b) is even then clearly \(2^3 < c^3 - b^3\) for any \(c \ne b \ge 3\)

\(\blacksquare \)

Answer 4

Q2 : show that if a,b,and c are odd integers then ax^2+bx+c=0 does not have a rational solution


Proof :-
Consider the discriminant of given quadratic.equation
\[\large b^2 - 4ac \]
For \(x\) to be a rational zero the discriminant must be a perfect square.
Since \(a,b,c\) are odd the discriminant is also odd and so must be of form \(8k+1\)

\(b^2-4ac \equiv 1 - 4(2m+1)(2n+1) \equiv 1 - 4 \equiv 5 \pmod{8} \)

which is not of form 8k+1. that means the discriminant is never a perfect square and consequently the quadratic equation has no rational solutions
\(\blacksquare\)

Answer 5

Q3: prove that if x and y are positive real numbers and then \(x \ne y\) then x+y > 4xy/(x+y)

Proof :-
Since \(x\ne y \) we have
\((x-y)^2 \gt 0 \implies (x+y)^2 -4xy\gt 0 \implies x+y \gt \dfrac{4xy}{x+y} \blacksquare \)

Answer 6

Q4 : for each real number \(x\), let \(A_x = (x, x+1)\).
Find \(\bigcup_{x\in \mathbb{R}} A_x\) and \(\bigcap_{x\in \mathbb{R}} A_x\)

Solution :-
\(\bigcup_{x\in \mathbb{R}} A_x = (-\infty, + \infty)\) because an openball can be defined around each real number and the infinite union of all these openballs contains all real numbers.

\(\bigcap_{x\in \mathbb{R}} A_x = \varnothing \) because the intersection of any set with emptyset is emptyset and it is trivial to see that \((1, 2) \cap (3,4) = \varnothing \)

Answer 7

Q5 : prove that the set of all perfect squares is countably infinite

Proof :-
use \(f :\mathbb{N} \to \mathbb{N} \) where \(f(n) = n^2\) is a bijection between set of natural numbers and the set of perfect squares \(\blacksquare\)

Answer 8

Q6 : The product of two countable sets is countable .

Proof :-
if the two countable sets S and T are finite :
use \(J_n = \{1,2,3,\cdots, n\}\) where \(n = |S| |T|\) to count

else :
Arrange the elements of \(S\times T\) in a matrix \((S_i, T_j)\) where \(i,j \in \mathbb{N}\). By appealing to Cantor's diagonalization argument the set \(S\times T\) is countable.

\(\blacksquare\)

Answer 9

nice so far