0.003 921 561 874 449 240 1 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.003 921 561 874 449 240 1(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
0.003 921 561 874 449 240 1(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. 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;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.003 921 561 874 449 240 1.

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) 0.003 921 561 874 449 240 1 × 2 = 0 + 0.007 843 123 748 898 480 2;
  • 2) 0.007 843 123 748 898 480 2 × 2 = 0 + 0.015 686 247 497 796 960 4;
  • 3) 0.015 686 247 497 796 960 4 × 2 = 0 + 0.031 372 494 995 593 920 8;
  • 4) 0.031 372 494 995 593 920 8 × 2 = 0 + 0.062 744 989 991 187 841 6;
  • 5) 0.062 744 989 991 187 841 6 × 2 = 0 + 0.125 489 979 982 375 683 2;
  • 6) 0.125 489 979 982 375 683 2 × 2 = 0 + 0.250 979 959 964 751 366 4;
  • 7) 0.250 979 959 964 751 366 4 × 2 = 0 + 0.501 959 919 929 502 732 8;
  • 8) 0.501 959 919 929 502 732 8 × 2 = 1 + 0.003 919 839 859 005 465 6;
  • 9) 0.003 919 839 859 005 465 6 × 2 = 0 + 0.007 839 679 718 010 931 2;
  • 10) 0.007 839 679 718 010 931 2 × 2 = 0 + 0.015 679 359 436 021 862 4;
  • 11) 0.015 679 359 436 021 862 4 × 2 = 0 + 0.031 358 718 872 043 724 8;
  • 12) 0.031 358 718 872 043 724 8 × 2 = 0 + 0.062 717 437 744 087 449 6;
  • 13) 0.062 717 437 744 087 449 6 × 2 = 0 + 0.125 434 875 488 174 899 2;
  • 14) 0.125 434 875 488 174 899 2 × 2 = 0 + 0.250 869 750 976 349 798 4;
  • 15) 0.250 869 750 976 349 798 4 × 2 = 0 + 0.501 739 501 952 699 596 8;
  • 16) 0.501 739 501 952 699 596 8 × 2 = 1 + 0.003 479 003 905 399 193 6;
  • 17) 0.003 479 003 905 399 193 6 × 2 = 0 + 0.006 958 007 810 798 387 2;
  • 18) 0.006 958 007 810 798 387 2 × 2 = 0 + 0.013 916 015 621 596 774 4;
  • 19) 0.013 916 015 621 596 774 4 × 2 = 0 + 0.027 832 031 243 193 548 8;
  • 20) 0.027 832 031 243 193 548 8 × 2 = 0 + 0.055 664 062 486 387 097 6;
  • 21) 0.055 664 062 486 387 097 6 × 2 = 0 + 0.111 328 124 972 774 195 2;
  • 22) 0.111 328 124 972 774 195 2 × 2 = 0 + 0.222 656 249 945 548 390 4;
  • 23) 0.222 656 249 945 548 390 4 × 2 = 0 + 0.445 312 499 891 096 780 8;
  • 24) 0.445 312 499 891 096 780 8 × 2 = 0 + 0.890 624 999 782 193 561 6;
  • 25) 0.890 624 999 782 193 561 6 × 2 = 1 + 0.781 249 999 564 387 123 2;
  • 26) 0.781 249 999 564 387 123 2 × 2 = 1 + 0.562 499 999 128 774 246 4;
  • 27) 0.562 499 999 128 774 246 4 × 2 = 1 + 0.124 999 998 257 548 492 8;
  • 28) 0.124 999 998 257 548 492 8 × 2 = 0 + 0.249 999 996 515 096 985 6;
  • 29) 0.249 999 996 515 096 985 6 × 2 = 0 + 0.499 999 993 030 193 971 2;
  • 30) 0.499 999 993 030 193 971 2 × 2 = 0 + 0.999 999 986 060 387 942 4;
  • 31) 0.999 999 986 060 387 942 4 × 2 = 1 + 0.999 999 972 120 775 884 8;

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).


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:


0.003 921 561 874 449 240 1(10) =

0.0000 0001 0000 0001 0000 0000 1110 001(2)

5. Positive number before normalization:

0.003 921 561 874 449 240 1(10) =

0.0000 0001 0000 0001 0000 0000 1110 001(2)

6. 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:


0.003 921 561 874 449 240 1(10) =

0.0000 0001 0000 0001 0000 0000 1110 001(2) =

0.0000 0001 0000 0001 0000 0000 1110 001(2) × 20 =

1.0000 0001 0000 0000 1110 001(2) × 2-8


7. 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 0 (a positive number)

Exponent (unadjusted): -8

Mantissa (not normalized):
1.0000 0001 0000 0000 1110 001


8. 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)


9. 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;

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) =

119(10) =

0111 0111(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 000 0000 1000 0000 0111 0001 =

000 0000 1000 0000 0111 0001


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

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

Exponent (8 bits) =
0111 0111

Mantissa (23 bits) =
000 0000 1000 0000 0111 0001


Decimal number 0.003 921 561 874 449 240 1 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 0111 - 000 0000 1000 0000 0111 0001

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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