Answer:
See proof below
Step-by-step explanation:
An equivalence relation R satisfies
- Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
- Symmetry: For all x,y, if xRy then yRx.
- Transitivity: For all x,y,z, If xRy and yRz then xRz.
Let's check these properties: Let x,y,z be bit strings of length three or more
The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.
If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.
If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.
Answer:
Step-by-step explanation:
Answer:
Angle 1 = 39, Angle 2 = 78 , Angle 3 = 63
Step-by-step explanation:
78/2 = 39
(Angle 2/2 = Angle 1)
78 - 15 = 63
(Angle 2 - 15 = Angle 3)
39 + 78 + 63 = 180
(Angle 1 + Angle 2 + Angle 3 = 180)
As the opposite sides of rhombus are parallel. Also, the opposite sides of rhombus are equal. also it satisfies all other properties of a parallelogram.
Hence , as opposite sides of rhombus are parallel and equal also .
So
A rhombus is ALWAYS a parallelogram.
