Perform the following conversions in the stated number bases; (a) AB1\(_{16}\) to Octal system (b) 1110111.101\(_2\) to Denary system (c) 507\(_2\) to Binar...

(a) Convert \(AB1_{16}\) to the Octal system

Step 1: Write each hexadecimal digit as 4 binary bits.

\[A=1010,\quad B=1011,\quad 1=0001\]

\[AB1_{16}=1010\,1011\,0001_2\]

Step 2: Regroup the binary digits in threes from the right.

\[101\;010\;110\;001\]

Step 3: Convert each group of three to its octal digit.

\[101=5,\;010=2,\;110=6,\;001=1\]

\[\therefore AB1_{16}=5261_8\]

(b) Convert \(1110111.101_2\) to the Denary (base 10) system

Integer part, using place values \(64,32,16,8,4,2,1\):

\[1110111_2=64+32+16+0+4+2+1=119\]

Fractional part, using place values \(\tfrac{1}{2},\tfrac{1}{4},\tfrac{1}{8}\):

\[.101_2=1(0.5)+0(0.25)+1(0.125)=0.625\]

\[\therefore 1110111.101_2=119.625_{10}\]

(c) Convert \(507_8\) to the Binary system

Since the digit 7 is not valid in base 2, the number is in the octal system. Write each octal digit as 3 binary bits.

\[5=101,\quad 0=000,\quad 7=111\]

\[\therefore 507_8=101\,000\,111_2=101000111_2\]

Check by way of denary: \(507_8=5\times8^2+0\times8^1+7\times8^0=320+0+7=327_{10}\), and repeated division of \(327\) by 2 gives \(101000111_2\), confirming the result.