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0 + 0 = 0

0 000 0000

1 + 1 = 2

1 001 0010

2 + 2 = 4

2 010 0100

3 + 3 = 6

3 011 0110

4 + 4 = 8

4 100 1000

5 + 5 = 10

5 101 1010

6 + 6 = 12

6 110 1100

7 + 7 = 14

7 111 1110

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11y ago
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13y ago

first of all you can not have the square of a 3bit number in 3 bits. The result would have 6 bits in general

Multiplication scheme:

abc

x abc

_______

abc (if c=1)

abc (if b=1)

+ abc (if a=1)

_______

EDCBAc

sorry about the format of the above

multiplication schem, i tried , i could not find it. Hope you can understand

with

A=(bc+cb) ,

B=(ac+bb+ca+CarryA) ,

C=(ab+ba+CarryB) ,

D=a+CarryC and

E=CarryD

where '+' i mean the addition , not the OR operation

and ac for example means AND operation

and for the carries, they take the value '1' if two or more one's took place in the addition

If you have partial or full adders in your hands it's easy from now on.

If you have available only nand,nor,xor etc reply to the post I'll help you

Another way to see it is that you add the number with a left shifting. Be careful, only when you have '1' in the collateral possition

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Q: Design a combination circuit that accepts a 3 bit number and generates a 3 bit binary number equal to the square of the input number?
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