32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.000 000 001 011 110 14 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.000 000 001 011 110 14(10) converted and written in 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: 0.000 000 001 011 110 14.

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) 0.000 000 001 011 110 14 × 2 = 0 + 0.000 000 002 022 220 28;
  • 2) 0.000 000 002 022 220 28 × 2 = 0 + 0.000 000 004 044 440 56;
  • 3) 0.000 000 004 044 440 56 × 2 = 0 + 0.000 000 008 088 881 12;
  • 4) 0.000 000 008 088 881 12 × 2 = 0 + 0.000 000 016 177 762 24;
  • 5) 0.000 000 016 177 762 24 × 2 = 0 + 0.000 000 032 355 524 48;
  • 6) 0.000 000 032 355 524 48 × 2 = 0 + 0.000 000 064 711 048 96;
  • 7) 0.000 000 064 711 048 96 × 2 = 0 + 0.000 000 129 422 097 92;
  • 8) 0.000 000 129 422 097 92 × 2 = 0 + 0.000 000 258 844 195 84;
  • 9) 0.000 000 258 844 195 84 × 2 = 0 + 0.000 000 517 688 391 68;
  • 10) 0.000 000 517 688 391 68 × 2 = 0 + 0.000 001 035 376 783 36;
  • 11) 0.000 001 035 376 783 36 × 2 = 0 + 0.000 002 070 753 566 72;
  • 12) 0.000 002 070 753 566 72 × 2 = 0 + 0.000 004 141 507 133 44;
  • 13) 0.000 004 141 507 133 44 × 2 = 0 + 0.000 008 283 014 266 88;
  • 14) 0.000 008 283 014 266 88 × 2 = 0 + 0.000 016 566 028 533 76;
  • 15) 0.000 016 566 028 533 76 × 2 = 0 + 0.000 033 132 057 067 52;
  • 16) 0.000 033 132 057 067 52 × 2 = 0 + 0.000 066 264 114 135 04;
  • 17) 0.000 066 264 114 135 04 × 2 = 0 + 0.000 132 528 228 270 08;
  • 18) 0.000 132 528 228 270 08 × 2 = 0 + 0.000 265 056 456 540 16;
  • 19) 0.000 265 056 456 540 16 × 2 = 0 + 0.000 530 112 913 080 32;
  • 20) 0.000 530 112 913 080 32 × 2 = 0 + 0.001 060 225 826 160 64;
  • 21) 0.001 060 225 826 160 64 × 2 = 0 + 0.002 120 451 652 321 28;
  • 22) 0.002 120 451 652 321 28 × 2 = 0 + 0.004 240 903 304 642 56;
  • 23) 0.004 240 903 304 642 56 × 2 = 0 + 0.008 481 806 609 285 12;
  • 24) 0.008 481 806 609 285 12 × 2 = 0 + 0.016 963 613 218 570 24;
  • 25) 0.016 963 613 218 570 24 × 2 = 0 + 0.033 927 226 437 140 48;
  • 26) 0.033 927 226 437 140 48 × 2 = 0 + 0.067 854 452 874 280 96;
  • 27) 0.067 854 452 874 280 96 × 2 = 0 + 0.135 708 905 748 561 92;
  • 28) 0.135 708 905 748 561 92 × 2 = 0 + 0.271 417 811 497 123 84;
  • 29) 0.271 417 811 497 123 84 × 2 = 0 + 0.542 835 622 994 247 68;
  • 30) 0.542 835 622 994 247 68 × 2 = 1 + 0.085 671 245 988 495 36;
  • 31) 0.085 671 245 988 495 36 × 2 = 0 + 0.171 342 491 976 990 72;
  • 32) 0.171 342 491 976 990 72 × 2 = 0 + 0.342 684 983 953 981 44;
  • 33) 0.342 684 983 953 981 44 × 2 = 0 + 0.685 369 967 907 962 88;
  • 34) 0.685 369 967 907 962 88 × 2 = 1 + 0.370 739 935 815 925 76;
  • 35) 0.370 739 935 815 925 76 × 2 = 0 + 0.741 479 871 631 851 52;
  • 36) 0.741 479 871 631 851 52 × 2 = 1 + 0.482 959 743 263 703 04;
  • 37) 0.482 959 743 263 703 04 × 2 = 0 + 0.965 919 486 527 406 08;
  • 38) 0.965 919 486 527 406 08 × 2 = 1 + 0.931 838 973 054 812 16;
  • 39) 0.931 838 973 054 812 16 × 2 = 1 + 0.863 677 946 109 624 32;
  • 40) 0.863 677 946 109 624 32 × 2 = 1 + 0.727 355 892 219 248 64;
  • 41) 0.727 355 892 219 248 64 × 2 = 1 + 0.454 711 784 438 497 28;
  • 42) 0.454 711 784 438 497 28 × 2 = 0 + 0.909 423 568 876 994 56;
  • 43) 0.909 423 568 876 994 56 × 2 = 1 + 0.818 847 137 753 989 12;
  • 44) 0.818 847 137 753 989 12 × 2 = 1 + 0.637 694 275 507 978 24;
  • 45) 0.637 694 275 507 978 24 × 2 = 1 + 0.275 388 551 015 956 48;
  • 46) 0.275 388 551 015 956 48 × 2 = 0 + 0.550 777 102 031 912 96;
  • 47) 0.550 777 102 031 912 96 × 2 = 1 + 0.101 554 204 063 825 92;
  • 48) 0.101 554 204 063 825 92 × 2 = 0 + 0.203 108 408 127 651 84;
  • 49) 0.203 108 408 127 651 84 × 2 = 0 + 0.406 216 816 255 303 68;
  • 50) 0.406 216 816 255 303 68 × 2 = 0 + 0.812 433 632 510 607 36;
  • 51) 0.812 433 632 510 607 36 × 2 = 1 + 0.624 867 265 021 214 72;
  • 52) 0.624 867 265 021 214 72 × 2 = 1 + 0.249 734 530 042 429 44;
  • 53) 0.249 734 530 042 429 44 × 2 = 0 + 0.499 469 060 084 858 88;

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...)


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:


0.000 000 001 011 110 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0101 0111 1011 1010 0011 0(2)


5. Positive number before normalization:

0.000 000 001 011 110 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0101 0111 1011 1010 0011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 30 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 001 011 110 14(10) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0101 0111 1011 1010 0011 0(2) =

0.0000 0000 0000 0000 0000 0000 0000 0100 0101 0111 1011 1010 0011 0(2) × 20 =

1.0001 0101 1110 1110 1000 110(2) × 2-30


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): -30

Mantissa (not normalized):
1.0001 0101 1110 1110 1000 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-30 + 2(8-1) - 1 =

(-30 + 127)(10) =

97(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;
  • 97 ÷ 2 = 48 + 1;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

97(10) =

0110 0001(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 1010 1111 0111 0100 0110 =

000 1010 1111 0111 0100 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) =
0110 0001

Mantissa (23 bits) =
000 1010 1111 0111 0100 0110


The base ten decimal number 0.000 000 001 011 110 14 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0110 0001 - 000 1010 1111 0111 0100 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