Answer:
Step-by-step explanation:
We are given that [tex]\frac{Z}{20Z}[/tex]
We have to find the invertible elements of [tex]\frac{Z}{20Z}[/tex] along with their inverses under multiplication.
We know that
[tex]\frac{Z}{20Z}^*\approx U(20)[/tex]
U(20)={1,3,7,9,11,13,17,19}
Operation used
[tex]\frac{a\times b}{20}=remainder[/tex]
Identity element=1
Inverse condition :[tex]a\times a^{-1}=1[/tex]
Inverse of 1=1
Inverse of 3
[tex]3x=1mod (20)[/tex]
[tex]3\times 7=1mod(20)[/tex]
[tex]\frac{21-1}{20}=\frac{20}{20}=0 [/tex] ([tex]a=b mod n\implies \frac{a-b}{n}[/tex])
Inverse of 3=7
Inverse of 7=3
Inverse of 9
[tex]9x=1mod (20)[/tex]
Inverse of 9 is 9 .
Inverse of 11
11 is also self inverse element.
[tex] \frac{13\times 17}{20}=1[/tex]
Where 1 is remainder
Hence inverse of 13 is 17 and inverse of 17 is 13.
19 is also self inverse element.
Therefore,
[tex]1^{-1}=1[/tex]
[tex]3^{-1}=7[/tex]
[tex]7^{-1}=3[/tex]
[tex]11^{-1}=11[/tex]
[tex]13^{-1}=17[/tex]
[tex]17^{-1}=13[/tex]
[tex]19^{-1}=19[/tex]
