0.000 031 003 892 217 995 565 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 031 003 892 217 995 565(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
0.000 031 003 892 217 995 565(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: 0.000 031 003 892 217 995 565.

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) 0.000 031 003 892 217 995 565 × 2 = 0 + 0.000 062 007 784 435 991 13;
  • 2) 0.000 062 007 784 435 991 13 × 2 = 0 + 0.000 124 015 568 871 982 26;
  • 3) 0.000 124 015 568 871 982 26 × 2 = 0 + 0.000 248 031 137 743 964 52;
  • 4) 0.000 248 031 137 743 964 52 × 2 = 0 + 0.000 496 062 275 487 929 04;
  • 5) 0.000 496 062 275 487 929 04 × 2 = 0 + 0.000 992 124 550 975 858 08;
  • 6) 0.000 992 124 550 975 858 08 × 2 = 0 + 0.001 984 249 101 951 716 16;
  • 7) 0.001 984 249 101 951 716 16 × 2 = 0 + 0.003 968 498 203 903 432 32;
  • 8) 0.003 968 498 203 903 432 32 × 2 = 0 + 0.007 936 996 407 806 864 64;
  • 9) 0.007 936 996 407 806 864 64 × 2 = 0 + 0.015 873 992 815 613 729 28;
  • 10) 0.015 873 992 815 613 729 28 × 2 = 0 + 0.031 747 985 631 227 458 56;
  • 11) 0.031 747 985 631 227 458 56 × 2 = 0 + 0.063 495 971 262 454 917 12;
  • 12) 0.063 495 971 262 454 917 12 × 2 = 0 + 0.126 991 942 524 909 834 24;
  • 13) 0.126 991 942 524 909 834 24 × 2 = 0 + 0.253 983 885 049 819 668 48;
  • 14) 0.253 983 885 049 819 668 48 × 2 = 0 + 0.507 967 770 099 639 336 96;
  • 15) 0.507 967 770 099 639 336 96 × 2 = 1 + 0.015 935 540 199 278 673 92;
  • 16) 0.015 935 540 199 278 673 92 × 2 = 0 + 0.031 871 080 398 557 347 84;
  • 17) 0.031 871 080 398 557 347 84 × 2 = 0 + 0.063 742 160 797 114 695 68;
  • 18) 0.063 742 160 797 114 695 68 × 2 = 0 + 0.127 484 321 594 229 391 36;
  • 19) 0.127 484 321 594 229 391 36 × 2 = 0 + 0.254 968 643 188 458 782 72;
  • 20) 0.254 968 643 188 458 782 72 × 2 = 0 + 0.509 937 286 376 917 565 44;
  • 21) 0.509 937 286 376 917 565 44 × 2 = 1 + 0.019 874 572 753 835 130 88;
  • 22) 0.019 874 572 753 835 130 88 × 2 = 0 + 0.039 749 145 507 670 261 76;
  • 23) 0.039 749 145 507 670 261 76 × 2 = 0 + 0.079 498 291 015 340 523 52;
  • 24) 0.079 498 291 015 340 523 52 × 2 = 0 + 0.158 996 582 030 681 047 04;
  • 25) 0.158 996 582 030 681 047 04 × 2 = 0 + 0.317 993 164 061 362 094 08;
  • 26) 0.317 993 164 061 362 094 08 × 2 = 0 + 0.635 986 328 122 724 188 16;
  • 27) 0.635 986 328 122 724 188 16 × 2 = 1 + 0.271 972 656 245 448 376 32;
  • 28) 0.271 972 656 245 448 376 32 × 2 = 0 + 0.543 945 312 490 896 752 64;
  • 29) 0.543 945 312 490 896 752 64 × 2 = 1 + 0.087 890 624 981 793 505 28;
  • 30) 0.087 890 624 981 793 505 28 × 2 = 0 + 0.175 781 249 963 587 010 56;
  • 31) 0.175 781 249 963 587 010 56 × 2 = 0 + 0.351 562 499 927 174 021 12;
  • 32) 0.351 562 499 927 174 021 12 × 2 = 0 + 0.703 124 999 854 348 042 24;
  • 33) 0.703 124 999 854 348 042 24 × 2 = 1 + 0.406 249 999 708 696 084 48;
  • 34) 0.406 249 999 708 696 084 48 × 2 = 0 + 0.812 499 999 417 392 168 96;
  • 35) 0.812 499 999 417 392 168 96 × 2 = 1 + 0.624 999 998 834 784 337 92;
  • 36) 0.624 999 998 834 784 337 92 × 2 = 1 + 0.249 999 997 669 568 675 84;
  • 37) 0.249 999 997 669 568 675 84 × 2 = 0 + 0.499 999 995 339 137 351 68;
  • 38) 0.499 999 995 339 137 351 68 × 2 = 0 + 0.999 999 990 678 274 703 36;

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:


0.000 031 003 892 217 995 565(10) =

0.0000 0000 0000 0010 0000 1000 0010 1000 1011 00(2)

5. Positive number before normalization:

0.000 031 003 892 217 995 565(10) =

0.0000 0000 0000 0010 0000 1000 0010 1000 1011 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the right, so that only one non zero digit remains to the left of it:


0.000 031 003 892 217 995 565(10) =

0.0000 0000 0000 0010 0000 1000 0010 1000 1011 00(2) =

0.0000 0000 0000 0010 0000 1000 0010 1000 1011 00(2) × 20 =

1.0000 0100 0001 0100 0101 100(2) × 2-15


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): -15

Mantissa (not normalized):
1.0000 0100 0001 0100 0101 100


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-15 + 2(8-1) - 1 =

(-15 + 127)(10) =

112(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;
  • 112 ÷ 2 = 56 + 0;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 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) =

112(10) =

0111 0000(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 0010 0000 1010 0010 1100 =

000 0010 0000 1010 0010 1100


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 0000

Mantissa (23 bits) =
000 0010 0000 1010 0010 1100


Decimal number 0.000 031 003 892 217 995 565 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 0000 - 000 0010 0000 1010 0010 1100

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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