-0.006 804 177 114 09 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.006 804 177 114 09(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.006 804 177 114 09(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.006 804 177 114 09| = 0.006 804 177 114 09


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.006 804 177 114 09.

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) 0.006 804 177 114 09 × 2 = 0 + 0.013 608 354 228 18;
  • 2) 0.013 608 354 228 18 × 2 = 0 + 0.027 216 708 456 36;
  • 3) 0.027 216 708 456 36 × 2 = 0 + 0.054 433 416 912 72;
  • 4) 0.054 433 416 912 72 × 2 = 0 + 0.108 866 833 825 44;
  • 5) 0.108 866 833 825 44 × 2 = 0 + 0.217 733 667 650 88;
  • 6) 0.217 733 667 650 88 × 2 = 0 + 0.435 467 335 301 76;
  • 7) 0.435 467 335 301 76 × 2 = 0 + 0.870 934 670 603 52;
  • 8) 0.870 934 670 603 52 × 2 = 1 + 0.741 869 341 207 04;
  • 9) 0.741 869 341 207 04 × 2 = 1 + 0.483 738 682 414 08;
  • 10) 0.483 738 682 414 08 × 2 = 0 + 0.967 477 364 828 16;
  • 11) 0.967 477 364 828 16 × 2 = 1 + 0.934 954 729 656 32;
  • 12) 0.934 954 729 656 32 × 2 = 1 + 0.869 909 459 312 64;
  • 13) 0.869 909 459 312 64 × 2 = 1 + 0.739 818 918 625 28;
  • 14) 0.739 818 918 625 28 × 2 = 1 + 0.479 637 837 250 56;
  • 15) 0.479 637 837 250 56 × 2 = 0 + 0.959 275 674 501 12;
  • 16) 0.959 275 674 501 12 × 2 = 1 + 0.918 551 349 002 24;
  • 17) 0.918 551 349 002 24 × 2 = 1 + 0.837 102 698 004 48;
  • 18) 0.837 102 698 004 48 × 2 = 1 + 0.674 205 396 008 96;
  • 19) 0.674 205 396 008 96 × 2 = 1 + 0.348 410 792 017 92;
  • 20) 0.348 410 792 017 92 × 2 = 0 + 0.696 821 584 035 84;
  • 21) 0.696 821 584 035 84 × 2 = 1 + 0.393 643 168 071 68;
  • 22) 0.393 643 168 071 68 × 2 = 0 + 0.787 286 336 143 36;
  • 23) 0.787 286 336 143 36 × 2 = 1 + 0.574 572 672 286 72;
  • 24) 0.574 572 672 286 72 × 2 = 1 + 0.149 145 344 573 44;
  • 25) 0.149 145 344 573 44 × 2 = 0 + 0.298 290 689 146 88;
  • 26) 0.298 290 689 146 88 × 2 = 0 + 0.596 581 378 293 76;
  • 27) 0.596 581 378 293 76 × 2 = 1 + 0.193 162 756 587 52;
  • 28) 0.193 162 756 587 52 × 2 = 0 + 0.386 325 513 175 04;
  • 29) 0.386 325 513 175 04 × 2 = 0 + 0.772 651 026 350 08;
  • 30) 0.772 651 026 350 08 × 2 = 1 + 0.545 302 052 700 16;
  • 31) 0.545 302 052 700 16 × 2 = 1 + 0.090 604 105 400 32;

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 - the converted number we get in the end will be just a very good approximation of the initial one).


5. 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:


0.006 804 177 114 09(10) =

0.0000 0001 1011 1101 1110 1011 0010 011(2)

6. Positive number before normalization:

0.006 804 177 114 09(10) =

0.0000 0001 1011 1101 1110 1011 0010 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the right, so that only one non zero digit remains to the left of it:


0.006 804 177 114 09(10) =

0.0000 0001 1011 1101 1110 1011 0010 011(2) =

0.0000 0001 1011 1101 1110 1011 0010 011(2) × 20 =

1.1011 1101 1110 1011 0010 011(2) × 2-8


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -8

Mantissa (not normalized):
1.1011 1101 1110 1011 0010 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-8 + 2(8-1) - 1 =

(-8 + 127)(10) =

119(10)


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

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 119 ÷ 2 = 59 + 1;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

119(10) =

0111 0111(2)


12. 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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 101 1110 1111 0101 1001 0011 =

101 1110 1111 0101 1001 0011


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0111 0111

Mantissa (23 bits) =
101 1110 1111 0101 1001 0011


Decimal number -0.006 804 177 114 09 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 0111 - 101 1110 1111 0101 1001 0011

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111