Compute the following decimal arithmetic operation using 8-bit 2's complement: i) t5- 7; ii) -10+6 ; iii) -23-7.

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

i) t5 - 7: Assuming 't5' is a typo and you meant '15', let's proceed:

First, we need to represent 15 and 7 in 8-bit 2's complement form:

15 in binary is 00001111 7 in binary is 00000111

Subtraction in binary is done by adding the 2's complement of the second number. So, we need to find the 2's complement of 7:

1's complement of 7 (00000111) is 11111000 Adding 1 to the 1's complement gives the 2's complement: 11111001

Now, we add 15 (00001111) and the 2's complement of 7 (11111001):

00001111

• 11111001

---

100001100

Since we're working with 8 bits, we discard the carry, and the result is 00001100, which is 8 in decimal. So, 15 - 7 = 8.

ii) -10 + 6:

First, represent -10 and 6 in 8-bit 2's complement form:

-10 in 2's complement is 11110110 6 in binary is 00000110

Adding these together:

11110110

• 00000110

---

11111100

This is a negative number (since the leftmost bit is 1). To find its value, we take the 2's complement:

1's complement of 11111100 is 00000011 Adding 1 gives the 2's complement: 00000100, which is 4 in decimal. So, -10 + 6 = -4.

iii) -23 - 7:

This is the same as -23 + (-7). So, represent -23 and -7 in 8-bit 2's complement form:

-23 in 2's complement is 11101001 -7 in 2's complement is 11111001

Adding these together:

11101001

• 11111001

---

111100010

Discarding the carry, the result is 11100010. This is a negative number, so take the 2's complement to find its value:

1's complement of 11100010 is 00011101 Adding 1 gives the 2's complement: 00011110, which is 30 in decimal. So, -23 - 7 = -30.