0.000 000 000 372 522 71 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 372 522 71(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
0.000 000 000 372 522 71(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: 0.000 000 000 372 522 71.

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) 0.000 000 000 372 522 71 × 2 = 0 + 0.000 000 000 745 045 42;
  • 2) 0.000 000 000 745 045 42 × 2 = 0 + 0.000 000 001 490 090 84;
  • 3) 0.000 000 001 490 090 84 × 2 = 0 + 0.000 000 002 980 181 68;
  • 4) 0.000 000 002 980 181 68 × 2 = 0 + 0.000 000 005 960 363 36;
  • 5) 0.000 000 005 960 363 36 × 2 = 0 + 0.000 000 011 920 726 72;
  • 6) 0.000 000 011 920 726 72 × 2 = 0 + 0.000 000 023 841 453 44;
  • 7) 0.000 000 023 841 453 44 × 2 = 0 + 0.000 000 047 682 906 88;
  • 8) 0.000 000 047 682 906 88 × 2 = 0 + 0.000 000 095 365 813 76;
  • 9) 0.000 000 095 365 813 76 × 2 = 0 + 0.000 000 190 731 627 52;
  • 10) 0.000 000 190 731 627 52 × 2 = 0 + 0.000 000 381 463 255 04;
  • 11) 0.000 000 381 463 255 04 × 2 = 0 + 0.000 000 762 926 510 08;
  • 12) 0.000 000 762 926 510 08 × 2 = 0 + 0.000 001 525 853 020 16;
  • 13) 0.000 001 525 853 020 16 × 2 = 0 + 0.000 003 051 706 040 32;
  • 14) 0.000 003 051 706 040 32 × 2 = 0 + 0.000 006 103 412 080 64;
  • 15) 0.000 006 103 412 080 64 × 2 = 0 + 0.000 012 206 824 161 28;
  • 16) 0.000 012 206 824 161 28 × 2 = 0 + 0.000 024 413 648 322 56;
  • 17) 0.000 024 413 648 322 56 × 2 = 0 + 0.000 048 827 296 645 12;
  • 18) 0.000 048 827 296 645 12 × 2 = 0 + 0.000 097 654 593 290 24;
  • 19) 0.000 097 654 593 290 24 × 2 = 0 + 0.000 195 309 186 580 48;
  • 20) 0.000 195 309 186 580 48 × 2 = 0 + 0.000 390 618 373 160 96;
  • 21) 0.000 390 618 373 160 96 × 2 = 0 + 0.000 781 236 746 321 92;
  • 22) 0.000 781 236 746 321 92 × 2 = 0 + 0.001 562 473 492 643 84;
  • 23) 0.001 562 473 492 643 84 × 2 = 0 + 0.003 124 946 985 287 68;
  • 24) 0.003 124 946 985 287 68 × 2 = 0 + 0.006 249 893 970 575 36;
  • 25) 0.006 249 893 970 575 36 × 2 = 0 + 0.012 499 787 941 150 72;
  • 26) 0.012 499 787 941 150 72 × 2 = 0 + 0.024 999 575 882 301 44;
  • 27) 0.024 999 575 882 301 44 × 2 = 0 + 0.049 999 151 764 602 88;
  • 28) 0.049 999 151 764 602 88 × 2 = 0 + 0.099 998 303 529 205 76;
  • 29) 0.099 998 303 529 205 76 × 2 = 0 + 0.199 996 607 058 411 52;
  • 30) 0.199 996 607 058 411 52 × 2 = 0 + 0.399 993 214 116 823 04;
  • 31) 0.399 993 214 116 823 04 × 2 = 0 + 0.799 986 428 233 646 08;
  • 32) 0.799 986 428 233 646 08 × 2 = 1 + 0.599 972 856 467 292 16;
  • 33) 0.599 972 856 467 292 16 × 2 = 1 + 0.199 945 712 934 584 32;
  • 34) 0.199 945 712 934 584 32 × 2 = 0 + 0.399 891 425 869 168 64;
  • 35) 0.399 891 425 869 168 64 × 2 = 0 + 0.799 782 851 738 337 28;
  • 36) 0.799 782 851 738 337 28 × 2 = 1 + 0.599 565 703 476 674 56;
  • 37) 0.599 565 703 476 674 56 × 2 = 1 + 0.199 131 406 953 349 12;
  • 38) 0.199 131 406 953 349 12 × 2 = 0 + 0.398 262 813 906 698 24;
  • 39) 0.398 262 813 906 698 24 × 2 = 0 + 0.796 525 627 813 396 48;
  • 40) 0.796 525 627 813 396 48 × 2 = 1 + 0.593 051 255 626 792 96;
  • 41) 0.593 051 255 626 792 96 × 2 = 1 + 0.186 102 511 253 585 92;
  • 42) 0.186 102 511 253 585 92 × 2 = 0 + 0.372 205 022 507 171 84;
  • 43) 0.372 205 022 507 171 84 × 2 = 0 + 0.744 410 045 014 343 68;
  • 44) 0.744 410 045 014 343 68 × 2 = 1 + 0.488 820 090 028 687 36;
  • 45) 0.488 820 090 028 687 36 × 2 = 0 + 0.977 640 180 057 374 72;
  • 46) 0.977 640 180 057 374 72 × 2 = 1 + 0.955 280 360 114 749 44;
  • 47) 0.955 280 360 114 749 44 × 2 = 1 + 0.910 560 720 229 498 88;
  • 48) 0.910 560 720 229 498 88 × 2 = 1 + 0.821 121 440 458 997 76;
  • 49) 0.821 121 440 458 997 76 × 2 = 1 + 0.642 242 880 917 995 52;
  • 50) 0.642 242 880 917 995 52 × 2 = 1 + 0.284 485 761 835 991 04;
  • 51) 0.284 485 761 835 991 04 × 2 = 0 + 0.568 971 523 671 982 08;
  • 52) 0.568 971 523 671 982 08 × 2 = 1 + 0.137 943 047 343 964 16;
  • 53) 0.137 943 047 343 964 16 × 2 = 0 + 0.275 886 094 687 928 32;
  • 54) 0.275 886 094 687 928 32 × 2 = 0 + 0.551 772 189 375 856 64;
  • 55) 0.551 772 189 375 856 64 × 2 = 1 + 0.103 544 378 751 713 28;

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:


0.000 000 000 372 522 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1001 1001 1001 0111 1101 001(2)

5. Positive number before normalization:

0.000 000 000 372 522 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1001 1001 1001 0111 1101 001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 32 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 372 522 71(10) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1001 1001 1001 0111 1101 001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0001 1001 1001 1001 0111 1101 001(2) × 20 =

1.1001 1001 1001 0111 1101 001(2) × 2-32


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): -32

Mantissa (not normalized):
1.1001 1001 1001 0111 1101 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-32 + 2(8-1) - 1 =

(-32 + 127)(10) =

95(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;
  • 95 ÷ 2 = 47 + 1;
  • 47 ÷ 2 = 23 + 1;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 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) =

95(10) =

0101 1111(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. 100 1100 1100 1011 1110 1001 =

100 1100 1100 1011 1110 1001


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) =
0101 1111

Mantissa (23 bits) =
100 1100 1100 1011 1110 1001


Decimal number 0.000 000 000 372 522 71 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 1111 - 100 1100 1100 1011 1110 1001

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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