d the relations S 2 for i = 1, 2, 3,4, , 6i’here S 1 = { ( a , b ) ∈ Z 2 | a > b } , the greater than relation, S 2 = { ( a , b ) ∈ Z 2 | a ≥ b } , the greater than or equal to relation, S 3 = { ( a , b ) ∈ Z 2 | a < b } , the less than relation, S 4 = { ( a , b ) ∈ Z 2 | a ≤ b } , the less than or equal to relation, S 5 = { ( a , b ) ∈ Z 2 | a = b } , the equal to relation, S 6 = { ( a , b ) ∈ Z 2 | a ≠ b } , the unequal to relation.
d the relations S 2 for i = 1, 2, 3,4, , 6i’here S 1 = { ( a , b ) ∈ Z 2 | a > b } , the greater than relation, S 2 = { ( a , b ) ∈ Z 2 | a ≥ b } , the greater than or equal to relation, S 3 = { ( a , b ) ∈ Z 2 | a < b } , the less than relation, S 4 = { ( a , b ) ∈ Z 2 | a ≤ b } , the less than or equal to relation, S 5 = { ( a , b ) ∈ Z 2 | a = b } , the equal to relation, S 6 = { ( a , b ) ∈ Z 2 | a ≠ b } , the unequal to relation.
Solution Summary: The author explains the relation R_i2 for I=1,2,
For each real-valued nonprincipal character x mod k, let
A(n) = x(d) and F(x) = Σ
:
dn
* Prove that
F(x) = L(1,x) log x + O(1).
n
By considering appropriate series expansions,
e². e²²/2. e²³/3.
....
=
= 1 + x + x² + ·
...
when |x| < 1.
By expanding each individual exponential term on the left-hand side
the coefficient of x- 19 has the form
and multiplying out,
1/19!1/19+r/s,
where 19 does not divide s. Deduce that
18! 1 (mod 19).
Proof: LN⎯⎯⎯⎯⎯LN¯ divides quadrilateral KLMN into two triangles. The sum of the angle measures in each triangle is ˚, so the sum of the angle measures for both triangles is ˚. So, m∠K+m∠L+m∠M+m∠N=m∠K+m∠L+m∠M+m∠N=˚. Because ∠K≅∠M∠K≅∠M and ∠N≅∠L, m∠K=m∠M∠N≅∠L, m∠K=m∠M and m∠N=m∠Lm∠N=m∠L by the definition of congruence. By the Substitution Property of Equality, m∠K+m∠L+m∠K+m∠L=m∠K+m∠L+m∠K+m∠L=°,°, so (m∠K)+ m∠K+ (m∠L)= m∠L= ˚. Dividing each side by gives m∠K+m∠L=m∠K+m∠L= °.°. The consecutive angles are supplementary, so KN⎯⎯⎯⎯⎯⎯∥LM⎯⎯⎯⎯⎯⎯KN¯∥LM¯ by the Converse of the Consecutive Interior Angles Theorem. Likewise, (m∠K)+m∠K+ (m∠N)=m∠N= ˚, or m∠K+m∠N=m∠K+m∠N= ˚. So these consecutive angles are supplementary and KL⎯⎯⎯⎯⎯∥NM⎯⎯⎯⎯⎯⎯KL¯∥NM¯ by the Converse of the Consecutive Interior Angles Theorem. Opposite sides are parallel, so quadrilateral KLMN is a parallelogram.
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RELATIONS-DOMAIN, RANGE AND CO-DOMAIN (RELATIONS AND FUNCTIONS CBSE/ ISC MATHS); Author: Neha Agrawal Mathematically Inclined;https://www.youtube.com/watch?v=u4IQh46VoU4;License: Standard YouTube License, CC-BY