Convert the Number 737 099.537 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations

Number 737 099.537(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

The first steps we'll go through to make the conversion:

Convert to binary (to base 2) the integer part of the number.

Convert to binary the fractional part of the number.


1. First, convert to binary (in base 2) the integer part: 737 099.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 737 099 ÷ 2 = 368 549 + 1;
  • 368 549 ÷ 2 = 184 274 + 1;
  • 184 274 ÷ 2 = 92 137 + 0;
  • 92 137 ÷ 2 = 46 068 + 1;
  • 46 068 ÷ 2 = 23 034 + 0;
  • 23 034 ÷ 2 = 11 517 + 0;
  • 11 517 ÷ 2 = 5 758 + 1;
  • 5 758 ÷ 2 = 2 879 + 0;
  • 2 879 ÷ 2 = 1 439 + 1;
  • 1 439 ÷ 2 = 719 + 1;
  • 719 ÷ 2 = 359 + 1;
  • 359 ÷ 2 = 179 + 1;
  • 179 ÷ 2 = 89 + 1;
  • 89 ÷ 2 = 44 + 1;
  • 44 ÷ 2 = 22 + 0;
  • 22 ÷ 2 = 11 + 0;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


737 099(10) =


1011 0011 1111 0100 1011(2)


3. Convert to binary (base 2) the fractional part: 0.537.

Multiply it repeatedly by 2.


Keep track of each integer part of the results.


Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.537 × 2 = 1 + 0.074;
  • 2) 0.074 × 2 = 0 + 0.148;
  • 3) 0.148 × 2 = 0 + 0.296;
  • 4) 0.296 × 2 = 0 + 0.592;
  • 5) 0.592 × 2 = 1 + 0.184;
  • 6) 0.184 × 2 = 0 + 0.368;
  • 7) 0.368 × 2 = 0 + 0.736;
  • 8) 0.736 × 2 = 1 + 0.472;
  • 9) 0.472 × 2 = 0 + 0.944;
  • 10) 0.944 × 2 = 1 + 0.888;
  • 11) 0.888 × 2 = 1 + 0.776;
  • 12) 0.776 × 2 = 1 + 0.552;
  • 13) 0.552 × 2 = 1 + 0.104;
  • 14) 0.104 × 2 = 0 + 0.208;
  • 15) 0.208 × 2 = 0 + 0.416;
  • 16) 0.416 × 2 = 0 + 0.832;
  • 17) 0.832 × 2 = 1 + 0.664;
  • 18) 0.664 × 2 = 1 + 0.328;
  • 19) 0.328 × 2 = 0 + 0.656;
  • 20) 0.656 × 2 = 1 + 0.312;
  • 21) 0.312 × 2 = 0 + 0.624;
  • 22) 0.624 × 2 = 1 + 0.248;
  • 23) 0.248 × 2 = 0 + 0.496;
  • 24) 0.496 × 2 = 0 + 0.992;
  • 25) 0.992 × 2 = 1 + 0.984;
  • 26) 0.984 × 2 = 1 + 0.968;
  • 27) 0.968 × 2 = 1 + 0.936;
  • 28) 0.936 × 2 = 1 + 0.872;
  • 29) 0.872 × 2 = 1 + 0.744;
  • 30) 0.744 × 2 = 1 + 0.488;
  • 31) 0.488 × 2 = 0 + 0.976;
  • 32) 0.976 × 2 = 1 + 0.952;
  • 33) 0.952 × 2 = 1 + 0.904;
  • 34) 0.904 × 2 = 1 + 0.808;
  • 35) 0.808 × 2 = 1 + 0.616;
  • 36) 0.616 × 2 = 1 + 0.232;
  • 37) 0.232 × 2 = 0 + 0.464;
  • 38) 0.464 × 2 = 0 + 0.928;
  • 39) 0.928 × 2 = 1 + 0.856;
  • 40) 0.856 × 2 = 1 + 0.712;
  • 41) 0.712 × 2 = 1 + 0.424;
  • 42) 0.424 × 2 = 0 + 0.848;
  • 43) 0.848 × 2 = 1 + 0.696;
  • 44) 0.696 × 2 = 1 + 0.392;
  • 45) 0.392 × 2 = 0 + 0.784;
  • 46) 0.784 × 2 = 1 + 0.568;
  • 47) 0.568 × 2 = 1 + 0.136;
  • 48) 0.136 × 2 = 0 + 0.272;
  • 49) 0.272 × 2 = 0 + 0.544;
  • 50) 0.544 × 2 = 1 + 0.088;
  • 51) 0.088 × 2 = 0 + 0.176;
  • 52) 0.176 × 2 = 0 + 0.352;
  • 53) 0.352 × 2 = 0 + 0.704;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.537(10) =


0.1000 1001 0111 1000 1101 0100 1111 1101 1111 0011 1011 0110 0100 0(2)


5. Positive number before normalization:

737 099.537(10) =


1011 0011 1111 0100 1011.1000 1001 0111 1000 1101 0100 1111 1101 1111 0011 1011 0110 0100 0(2)


The last steps we'll go through to make the conversion:

Normalize the binary representation of the number.

Adjust the exponent.

Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Normalize the mantissa.


6. Normalize the binary representation of the number.

Shift the decimal mark 19 positions to the left, so that only one non zero digit remains to the left of it:


737 099.537(10) =


1011 0011 1111 0100 1011.1000 1001 0111 1000 1101 0100 1111 1101 1111 0011 1011 0110 0100 0(2) =


1011 0011 1111 0100 1011.1000 1001 0111 1000 1101 0100 1111 1101 1111 0011 1011 0110 0100 0(2) × 20 =


1.0110 0111 1110 1001 0111 0001 0010 1111 0001 1010 1001 1111 1011 1110 0111 0110 1100 1000(2) × 219


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)


Exponent (unadjusted): 19


Mantissa (not normalized):
1.0110 0111 1110 1001 0111 0001 0010 1111 0001 1010 1001 1111 1011 1110 0111 0110 1100 1000


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


Exponent (unadjusted) + 2(11-1) - 1 =


19 + 2(11-1) - 1 =


(19 + 1 023)(10) =


1 042(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 042 ÷ 2 = 521 + 0;
  • 521 ÷ 2 = 260 + 1;
  • 260 ÷ 2 = 130 + 0;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1042(10) =


100 0001 0010(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.


b) Adjust its length to 52 bits, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 0110 0111 1110 1001 0111 0001 0010 1111 0001 1010 1001 1111 1011 1110 0111 0110 1100 1000 =


0110 0111 1110 1001 0111 0001 0010 1111 0001 1010 1001 1111 1011


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (11 bits) =
100 0001 0010


Mantissa (52 bits) =
0110 0111 1110 1001 0111 0001 0010 1111 0001 1010 1001 1111 1011


The base ten decimal number 737 099.537 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 0010 - 0110 0111 1110 1001 0111 0001 0010 1111 0001 1010 1001 1111 1011

(64 bits IEEE 754)

Number 737 099.536 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

Number 737 099.538 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

Convert to 64 bit double precision IEEE 754 binary floating point representation standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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