In: Advanced Math
Let x, y ∈ R. Prove the following:
(a) 0 < 1
(b) For all n ∈ N, if 0 < x < y, then x^n < y^n.
(c) |x · y| = |x| · |y|
Sol.-
Let x ,y R
(a) To prove 0 < 1 we first prove that if a R and a ≠ 0 then a2 > 0
Since a R and a ≠ 0 therefore by the Trichotomy Property either a P or - a P
If a P then a2 = a • a P. [By using fact(2) given below]
Also if -a P then a2 = (-a)•(-a) P [by using fact (2) given below]
So we can conclude that if a≠0 then a2 >0
Now,
Since 1= 12 it follows that 12>0
1 > 0 Hence Proved.
(b) Given 0<x<y and x,y R
To prove:- for all nN, xn < yn
we will prove it by using Mathematical Induction
For n=1 we have x<y [Given]
let the result hold for n=k i.e. xk < yk -------(1)
Now we will prove that the result is also hold for n= k+1
So for n= k+1
We have xk+1 = xk • x < yk • y [ by using induction hypothesis (1) and x<y is given]
xk+1 < yk+1
thus result holds for n= k+1
By using Mathematical Induction
xn < yn for all n N. Hence Proved.
(c) let x , y R
To prove:- | x•y| = |x|•|y|
if either x or y is 0 then both sides are equal to 0. So we are done here.
Now if neither x nor y is 0 then there are four cases.
case(i) if x>0 ,y>0 then x•y> 0
So, |x•y| =x•y = |x|•|y| [ since x>0,y>0 so |x|=x and |y|=y]
|x•y|=|x|•|y|.
Case(ii) if x>0 ,y<0 then x•y<0
So, |x•y|= - x•y = x•(-y) = |x|•|y| [as x>0 ,y<0 so |x|=x and |y|=-y]
|x•y|=|x|•|y|
Case(iii) if x<0 ,y>0 then x•y <0
So, |x•y|= -x •y = |x|•|y|. [As x<0 ,y>0 so |x|= -x and |y|=y]
|x•y|=|x|•|y|
Case(iv) if x<0 ,y<0 the x•y >0
So, |x•y|= x•y = (-x)•(-y) = |x|•|y| [as x<0,y<0 so |x|=-x and |y|=-y]
|x•y|=|x|•|y|
Hence, | x•y|=|x|•|y| Hence Proved.
Some Facts used here:-
(1) Trichotomy Property:- if a belongs to R then exactly one of the following holds-
a P , a=0 , -a P
Where P is the set of positive real numbers.
(2) if a,b belongs to P then ab P
(3) Definition of the absolute value:-
The absolute value of a real number x denoted by |x| , is defined by