# Fill out the Truth Table for the following specification.

#### shamieh

##### Active member
Fill out the truthe table for the following specification. Given a system with 3 inputs(y2,y1,y0) and a single output (f), where Y = y2,y1,y0 represents a 3 bit unsigned integer (Y is the decimal equivalent), determine the truth table for f such that f = 1 if and only if 1 < Y <= 6 (Y is greater than 1 and less than or equal to 6).

• y2 y1 y0 | f
• 0 0 0 |
• 0 0 1 |
• 0 1 0 |
• 0 1 1 |
• 1 0 0 |
• 1 0 1 |
• 1 1 0 |
• 1 1 1 |

How in the world am I supposed to evaluate this? I am so confused.

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Do you know how to convert binary numbers to decimals? The truth table lists 8 3-bit binary numbers (the decimal equivalents are from 0 to 7), and you need to write 1 next to those numbers that are greater than 1 and less than or equal to 6.

#### shamieh

##### Active member

$f$
0
0
1
1
1
1
0
0

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
The required inequality 1 < Y ≤ 6 is strict only on one side. The number of different Y's satisfying it is 6 - 1 = 5.

#### shamieh

##### Active member
The required inequality 1 < Y ≤ 6 is strict only on one side. The number of different Y's satisfying it is 6 - 1 = 5.
So I'm missing a 1 then - correct?

so my 7th row or [6th row ] whatever you'd like to call it; should be a 1 as well because I didn't account for <= 6. and 2^1 + 2^2 = 6; 6 <= 6 = 1

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
so my 7th row or [6th row ] whatever you'd like to call it; should be a 1 as well because I didn't account for <= 6. and 2^1 + 2^2 = 6; 6 <= 6 = 1
Yes.