0.000 000 000 000 013 33 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 013 33(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
0.000 000 000 000 013 33(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: 0.000 000 000 000 013 33.

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) 0.000 000 000 000 013 33 × 2 = 0 + 0.000 000 000 000 026 66;
  • 2) 0.000 000 000 000 026 66 × 2 = 0 + 0.000 000 000 000 053 32;
  • 3) 0.000 000 000 000 053 32 × 2 = 0 + 0.000 000 000 000 106 64;
  • 4) 0.000 000 000 000 106 64 × 2 = 0 + 0.000 000 000 000 213 28;
  • 5) 0.000 000 000 000 213 28 × 2 = 0 + 0.000 000 000 000 426 56;
  • 6) 0.000 000 000 000 426 56 × 2 = 0 + 0.000 000 000 000 853 12;
  • 7) 0.000 000 000 000 853 12 × 2 = 0 + 0.000 000 000 001 706 24;
  • 8) 0.000 000 000 001 706 24 × 2 = 0 + 0.000 000 000 003 412 48;
  • 9) 0.000 000 000 003 412 48 × 2 = 0 + 0.000 000 000 006 824 96;
  • 10) 0.000 000 000 006 824 96 × 2 = 0 + 0.000 000 000 013 649 92;
  • 11) 0.000 000 000 013 649 92 × 2 = 0 + 0.000 000 000 027 299 84;
  • 12) 0.000 000 000 027 299 84 × 2 = 0 + 0.000 000 000 054 599 68;
  • 13) 0.000 000 000 054 599 68 × 2 = 0 + 0.000 000 000 109 199 36;
  • 14) 0.000 000 000 109 199 36 × 2 = 0 + 0.000 000 000 218 398 72;
  • 15) 0.000 000 000 218 398 72 × 2 = 0 + 0.000 000 000 436 797 44;
  • 16) 0.000 000 000 436 797 44 × 2 = 0 + 0.000 000 000 873 594 88;
  • 17) 0.000 000 000 873 594 88 × 2 = 0 + 0.000 000 001 747 189 76;
  • 18) 0.000 000 001 747 189 76 × 2 = 0 + 0.000 000 003 494 379 52;
  • 19) 0.000 000 003 494 379 52 × 2 = 0 + 0.000 000 006 988 759 04;
  • 20) 0.000 000 006 988 759 04 × 2 = 0 + 0.000 000 013 977 518 08;
  • 21) 0.000 000 013 977 518 08 × 2 = 0 + 0.000 000 027 955 036 16;
  • 22) 0.000 000 027 955 036 16 × 2 = 0 + 0.000 000 055 910 072 32;
  • 23) 0.000 000 055 910 072 32 × 2 = 0 + 0.000 000 111 820 144 64;
  • 24) 0.000 000 111 820 144 64 × 2 = 0 + 0.000 000 223 640 289 28;
  • 25) 0.000 000 223 640 289 28 × 2 = 0 + 0.000 000 447 280 578 56;
  • 26) 0.000 000 447 280 578 56 × 2 = 0 + 0.000 000 894 561 157 12;
  • 27) 0.000 000 894 561 157 12 × 2 = 0 + 0.000 001 789 122 314 24;
  • 28) 0.000 001 789 122 314 24 × 2 = 0 + 0.000 003 578 244 628 48;
  • 29) 0.000 003 578 244 628 48 × 2 = 0 + 0.000 007 156 489 256 96;
  • 30) 0.000 007 156 489 256 96 × 2 = 0 + 0.000 014 312 978 513 92;
  • 31) 0.000 014 312 978 513 92 × 2 = 0 + 0.000 028 625 957 027 84;
  • 32) 0.000 028 625 957 027 84 × 2 = 0 + 0.000 057 251 914 055 68;
  • 33) 0.000 057 251 914 055 68 × 2 = 0 + 0.000 114 503 828 111 36;
  • 34) 0.000 114 503 828 111 36 × 2 = 0 + 0.000 229 007 656 222 72;
  • 35) 0.000 229 007 656 222 72 × 2 = 0 + 0.000 458 015 312 445 44;
  • 36) 0.000 458 015 312 445 44 × 2 = 0 + 0.000 916 030 624 890 88;
  • 37) 0.000 916 030 624 890 88 × 2 = 0 + 0.001 832 061 249 781 76;
  • 38) 0.001 832 061 249 781 76 × 2 = 0 + 0.003 664 122 499 563 52;
  • 39) 0.003 664 122 499 563 52 × 2 = 0 + 0.007 328 244 999 127 04;
  • 40) 0.007 328 244 999 127 04 × 2 = 0 + 0.014 656 489 998 254 08;
  • 41) 0.014 656 489 998 254 08 × 2 = 0 + 0.029 312 979 996 508 16;
  • 42) 0.029 312 979 996 508 16 × 2 = 0 + 0.058 625 959 993 016 32;
  • 43) 0.058 625 959 993 016 32 × 2 = 0 + 0.117 251 919 986 032 64;
  • 44) 0.117 251 919 986 032 64 × 2 = 0 + 0.234 503 839 972 065 28;
  • 45) 0.234 503 839 972 065 28 × 2 = 0 + 0.469 007 679 944 130 56;
  • 46) 0.469 007 679 944 130 56 × 2 = 0 + 0.938 015 359 888 261 12;
  • 47) 0.938 015 359 888 261 12 × 2 = 1 + 0.876 030 719 776 522 24;
  • 48) 0.876 030 719 776 522 24 × 2 = 1 + 0.752 061 439 553 044 48;
  • 49) 0.752 061 439 553 044 48 × 2 = 1 + 0.504 122 879 106 088 96;
  • 50) 0.504 122 879 106 088 96 × 2 = 1 + 0.008 245 758 212 177 92;
  • 51) 0.008 245 758 212 177 92 × 2 = 0 + 0.016 491 516 424 355 84;
  • 52) 0.016 491 516 424 355 84 × 2 = 0 + 0.032 983 032 848 711 68;
  • 53) 0.032 983 032 848 711 68 × 2 = 0 + 0.065 966 065 697 423 36;
  • 54) 0.065 966 065 697 423 36 × 2 = 0 + 0.131 932 131 394 846 72;
  • 55) 0.131 932 131 394 846 72 × 2 = 0 + 0.263 864 262 789 693 44;
  • 56) 0.263 864 262 789 693 44 × 2 = 0 + 0.527 728 525 579 386 88;
  • 57) 0.527 728 525 579 386 88 × 2 = 1 + 0.055 457 051 158 773 76;
  • 58) 0.055 457 051 158 773 76 × 2 = 0 + 0.110 914 102 317 547 52;
  • 59) 0.110 914 102 317 547 52 × 2 = 0 + 0.221 828 204 635 095 04;
  • 60) 0.221 828 204 635 095 04 × 2 = 0 + 0.443 656 409 270 190 08;
  • 61) 0.443 656 409 270 190 08 × 2 = 0 + 0.887 312 818 540 380 16;
  • 62) 0.887 312 818 540 380 16 × 2 = 1 + 0.774 625 637 080 760 32;
  • 63) 0.774 625 637 080 760 32 × 2 = 1 + 0.549 251 274 161 520 64;
  • 64) 0.549 251 274 161 520 64 × 2 = 1 + 0.098 502 548 323 041 28;
  • 65) 0.098 502 548 323 041 28 × 2 = 0 + 0.197 005 096 646 082 56;
  • 66) 0.197 005 096 646 082 56 × 2 = 0 + 0.394 010 193 292 165 12;
  • 67) 0.394 010 193 292 165 12 × 2 = 0 + 0.788 020 386 584 330 24;
  • 68) 0.788 020 386 584 330 24 × 2 = 1 + 0.576 040 773 168 660 48;
  • 69) 0.576 040 773 168 660 48 × 2 = 1 + 0.152 081 546 337 320 96;
  • 70) 0.152 081 546 337 320 96 × 2 = 0 + 0.304 163 092 674 641 92;

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:


0.000 000 000 000 013 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1100 0000 1000 0111 0001 10(2)

5. Positive number before normalization:

0.000 000 000 000 013 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1100 0000 1000 0111 0001 10(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 47 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 013 33(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1100 0000 1000 0111 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 1100 0000 1000 0111 0001 10(2) × 20 =

1.1110 0000 0100 0011 1000 110(2) × 2-47


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): -47

Mantissa (not normalized):
1.1110 0000 0100 0011 1000 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-47 + 2(8-1) - 1 =

(-47 + 127)(10) =

80(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;
  • 80 ÷ 2 = 40 + 0;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 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) =

80(10) =

0101 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. 111 0000 0010 0001 1100 0110 =

111 0000 0010 0001 1100 0110


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 0000

Mantissa (23 bits) =
111 0000 0010 0001 1100 0110


Decimal number 0.000 000 000 000 013 33 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 0000 - 111 0000 0010 0001 1100 0110

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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