Answer to Question #147015 in Algebra for Sourav Mondal

Question

Which of the following statements are true, and
which are false ? Justify your answers.
(a) Any polynomial of degree n >= 1 over C can be
expressed as a product of polynomials of
degree 1 over C.
(b) The argument of any purely imaginary
number is 0 or π

(d) Given any n numbers, their AM is larger
than or equal to their GM.
(e) Any system of two or more linear equations
has a solution.

Answer 1

(a) false. Because not any n $\geq$ 1. If we multiply two polynomial of degree 1, it would be 2.

(b) false. The general form of purely imaginary number: $(2*\kappa+1) *\pi/2, \kappa\isin Z$. (2k+1) is a

general form form of odd numbers. So (2k+1)/2 can not be appropriate to Z like 0 or 1 (1*$\pi$ )

(d) true. Consider two numbers a and b. let's find their AM and GM.

AM = (a+b)/2 and GM = $\sqrt{\smash[b]{ab}}$ .

As given in the condition of the matter AM>=GM. ------> (a+b)/2$\geq$ $\sqrt{\smash[b]{ab}}$

we multiply by 2 both sides. (a+b)$\geq$ 2*$\sqrt{\smash[b]{ab}}$

Then we square both sides of equality. (a+b)²$\geq$ 4ab

Do algebraic calculation. a²+2ab+b²-4ab>0.

a²-2ab+b²>0.

(a-b)²>0.

we know that square of something is always non zero number. So unequality is true.

(e) false. Not any system. A system will have a solution when the linear equations of the system intersect or fall on top of each other.

Imagine these linear uqeations parallel but not collinear. There is not any intersection points. So the system has no any solution.

we will see these by the formulas: a₁x+b₁y=c₁

$\begin{cases} ax+by=c \\ dx+ey=f \end{cases}$

the system has no any solution, if $\frac {a}{d} = \frac {b}{e} \cancel{=} \frac {c}{f}$

the system has one solution, if $\frac {a}{d} \cancel{=} \frac {b}{e} \cancel{=} \frac {c}{f}$

the system has infinitive solution, if $\frac {a}{d} = \frac {b}{e} = \frac {c}{f}$