Compute the following decimal arithmetic operation using 8-bit l's complement: \,z i) -16+78; ii) -e0-8; iii) 13-26

Sure, let's solve these problems one by one:

i) -16 + 78

First, we need to represent -16 and 78 in 8-bit 1's complement form.

-16 in binary is 10000. To make it 8-bit, we add four 0s in front, so we get 00010000. The 1's complement of 00010000 is 11101111.

78 in binary is 1001110. To make it 8-bit, we add one 0 in front, so we get 01001110.

Now, let's add these two numbers:

11101111

• 01001110

---

= 10111101

But, since we are using 1's complement, if there is a carry out, we need to add it back to the result. In this case, there is no carry out, so the result is 10111101, which is -83 in decimal. So, -16 + 78 = -83.

ii) -80 - 8

First, we need to represent -80 and -8 in 8-bit 1's complement form.

-80 in binary is 1010000. To make it 8-bit, we add one 0 in front, so we get 01010000. The 1's complement of 01010000 is 10101111.

-8 in binary is 1000. To make it 8-bit, we add four 0s in front, so we get 00001000. The 1's complement of 00001000 is 11110111.

Now, let's add these two numbers:

10101111

• 11110111

---

= 10100110

But, since we are using 1's complement, if there is a carry out, we need to add it back to the result. In this case, there is a carry out, so we add it back:

10100110

•   1
  

---

= 10100111

So, -80 - 8 = -39.

iii) 13 - 26

This is equivalent to 13 + (-26).

13 in binary is 1101. To make it 8-bit, we add four 0s in front, so we get 00001101.

-26 in binary is 11010. To make it 8-bit, we add three 0s in front, so we get 00011010. The 1's complement of 00011010 is 11100101.

Now, let's add these two numbers:

00001101

• 11100101

---

= 11110010

But, since we are using 1's complement, if there is a carry out, we need to add it back to the result. In this case, there is no carry out, so the result is 11110010, which is -13 in decimal. So, 13 - 26 = -13.