32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: 0.002 394 522 356 732 212 773 527 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number 0.002 394 522 356 732 212 773 527(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: 0.002 394 522 356 732 212 773 527.

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) 0.002 394 522 356 732 212 773 527 × 2 = 0 + 0.004 789 044 713 464 425 547 054;
  • 2) 0.004 789 044 713 464 425 547 054 × 2 = 0 + 0.009 578 089 426 928 851 094 108;
  • 3) 0.009 578 089 426 928 851 094 108 × 2 = 0 + 0.019 156 178 853 857 702 188 216;
  • 4) 0.019 156 178 853 857 702 188 216 × 2 = 0 + 0.038 312 357 707 715 404 376 432;
  • 5) 0.038 312 357 707 715 404 376 432 × 2 = 0 + 0.076 624 715 415 430 808 752 864;
  • 6) 0.076 624 715 415 430 808 752 864 × 2 = 0 + 0.153 249 430 830 861 617 505 728;
  • 7) 0.153 249 430 830 861 617 505 728 × 2 = 0 + 0.306 498 861 661 723 235 011 456;
  • 8) 0.306 498 861 661 723 235 011 456 × 2 = 0 + 0.612 997 723 323 446 470 022 912;
  • 9) 0.612 997 723 323 446 470 022 912 × 2 = 1 + 0.225 995 446 646 892 940 045 824;
  • 10) 0.225 995 446 646 892 940 045 824 × 2 = 0 + 0.451 990 893 293 785 880 091 648;
  • 11) 0.451 990 893 293 785 880 091 648 × 2 = 0 + 0.903 981 786 587 571 760 183 296;
  • 12) 0.903 981 786 587 571 760 183 296 × 2 = 1 + 0.807 963 573 175 143 520 366 592;
  • 13) 0.807 963 573 175 143 520 366 592 × 2 = 1 + 0.615 927 146 350 287 040 733 184;
  • 14) 0.615 927 146 350 287 040 733 184 × 2 = 1 + 0.231 854 292 700 574 081 466 368;
  • 15) 0.231 854 292 700 574 081 466 368 × 2 = 0 + 0.463 708 585 401 148 162 932 736;
  • 16) 0.463 708 585 401 148 162 932 736 × 2 = 0 + 0.927 417 170 802 296 325 865 472;
  • 17) 0.927 417 170 802 296 325 865 472 × 2 = 1 + 0.854 834 341 604 592 651 730 944;
  • 18) 0.854 834 341 604 592 651 730 944 × 2 = 1 + 0.709 668 683 209 185 303 461 888;
  • 19) 0.709 668 683 209 185 303 461 888 × 2 = 1 + 0.419 337 366 418 370 606 923 776;
  • 20) 0.419 337 366 418 370 606 923 776 × 2 = 0 + 0.838 674 732 836 741 213 847 552;
  • 21) 0.838 674 732 836 741 213 847 552 × 2 = 1 + 0.677 349 465 673 482 427 695 104;
  • 22) 0.677 349 465 673 482 427 695 104 × 2 = 1 + 0.354 698 931 346 964 855 390 208;
  • 23) 0.354 698 931 346 964 855 390 208 × 2 = 0 + 0.709 397 862 693 929 710 780 416;
  • 24) 0.709 397 862 693 929 710 780 416 × 2 = 1 + 0.418 795 725 387 859 421 560 832;
  • 25) 0.418 795 725 387 859 421 560 832 × 2 = 0 + 0.837 591 450 775 718 843 121 664;
  • 26) 0.837 591 450 775 718 843 121 664 × 2 = 1 + 0.675 182 901 551 437 686 243 328;
  • 27) 0.675 182 901 551 437 686 243 328 × 2 = 1 + 0.350 365 803 102 875 372 486 656;
  • 28) 0.350 365 803 102 875 372 486 656 × 2 = 0 + 0.700 731 606 205 750 744 973 312;
  • 29) 0.700 731 606 205 750 744 973 312 × 2 = 1 + 0.401 463 212 411 501 489 946 624;
  • 30) 0.401 463 212 411 501 489 946 624 × 2 = 0 + 0.802 926 424 823 002 979 893 248;
  • 31) 0.802 926 424 823 002 979 893 248 × 2 = 1 + 0.605 852 849 646 005 959 786 496;
  • 32) 0.605 852 849 646 005 959 786 496 × 2 = 1 + 0.211 705 699 292 011 919 572 992;

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:


0.002 394 522 356 732 212 773 527(10) =

0.0000 0000 1001 1100 1110 1101 0110 1011(2)


5. Positive number before normalization:

0.002 394 522 356 732 212 773 527(10) =

0.0000 0000 1001 1100 1110 1101 0110 1011(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 9 positions to the right, so that only one non zero digit remains to the left of it:


0.002 394 522 356 732 212 773 527(10) =

0.0000 0000 1001 1100 1110 1101 0110 1011(2) =

0.0000 0000 1001 1100 1110 1101 0110 1011(2) × 20 =

1.0011 1001 1101 1010 1101 011(2) × 2-9


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): -9

Mantissa (not normalized):
1.0011 1001 1101 1010 1101 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-9 + 2(8-1) - 1 =

(-9 + 127)(10) =

118(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;
  • 118 ÷ 2 = 59 + 0;
  • 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) =

118(10) =

0111 0110(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. 001 1100 1110 1101 0110 1011 =

001 1100 1110 1101 0110 1011


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 0110

Mantissa (23 bits) =
001 1100 1110 1101 0110 1011


The base ten decimal number 0.002 394 522 356 732 212 773 527 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
0 - 0111 0110 - 001 1100 1110 1101 0110 1011

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number 0.002 394 522 356 732 212 773 526 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number 0.002 394 522 356 732 212 773 528 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

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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