Question

part e - additive inverse show work for partial credit / 2 pts each
#9 (mod 16) : inverse of 1?
#10 (mod 40) : inverse of 32?

Explanation:

Step1: Recall additive - inverse definition

The additive inverse of an integer \(a\) modulo \(m\) is an integer \(b\) such that \(a + b\equiv0\pmod{m}\), or \(a + b=km\) for some integer \(k\).

Step2: Solve for the additive inverse of 1 modulo 16

Let the additive inverse of 1 modulo 16 be \(x\). Then \(1 + x\equiv0\pmod{16}\), which means \(1+x = 16k\) for some integer \(k\). When \(k = 1\), \(x=16 - 1=15\) since \(1+15 = 16\equiv0\pmod{16}\).

Step3: Solve for the additive inverse of 32 modulo 40

Let the additive inverse of 32 modulo 40 be \(y\). Then \(32 + y\equiv0\pmod{40}\), which means \(32 + y=40k\) for some integer \(k\). When \(k = 1\), \(y = 40-32 = 8\) since \(32 + 8=40\equiv0\pmod{40}\).

Answer:

#9: 15
#10: 8