Decimal to 32 Bit IEEE 754 Binary: Convert Number -0.014 459 890 236 287 751 815 702 13 to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From Base Ten Decimal System

Number -0.014 459 890 236 287 751 815 702 13(10) converted and written in 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.014 459 890 236 287 751 815 702 13| = 0.014 459 890 236 287 751 815 702 13


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.014 459 890 236 287 751 815 702 13.

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) 0.014 459 890 236 287 751 815 702 13 × 2 = 0 + 0.028 919 780 472 575 503 631 404 26;
  • 2) 0.028 919 780 472 575 503 631 404 26 × 2 = 0 + 0.057 839 560 945 151 007 262 808 52;
  • 3) 0.057 839 560 945 151 007 262 808 52 × 2 = 0 + 0.115 679 121 890 302 014 525 617 04;
  • 4) 0.115 679 121 890 302 014 525 617 04 × 2 = 0 + 0.231 358 243 780 604 029 051 234 08;
  • 5) 0.231 358 243 780 604 029 051 234 08 × 2 = 0 + 0.462 716 487 561 208 058 102 468 16;
  • 6) 0.462 716 487 561 208 058 102 468 16 × 2 = 0 + 0.925 432 975 122 416 116 204 936 32;
  • 7) 0.925 432 975 122 416 116 204 936 32 × 2 = 1 + 0.850 865 950 244 832 232 409 872 64;
  • 8) 0.850 865 950 244 832 232 409 872 64 × 2 = 1 + 0.701 731 900 489 664 464 819 745 28;
  • 9) 0.701 731 900 489 664 464 819 745 28 × 2 = 1 + 0.403 463 800 979 328 929 639 490 56;
  • 10) 0.403 463 800 979 328 929 639 490 56 × 2 = 0 + 0.806 927 601 958 657 859 278 981 12;
  • 11) 0.806 927 601 958 657 859 278 981 12 × 2 = 1 + 0.613 855 203 917 315 718 557 962 24;
  • 12) 0.613 855 203 917 315 718 557 962 24 × 2 = 1 + 0.227 710 407 834 631 437 115 924 48;
  • 13) 0.227 710 407 834 631 437 115 924 48 × 2 = 0 + 0.455 420 815 669 262 874 231 848 96;
  • 14) 0.455 420 815 669 262 874 231 848 96 × 2 = 0 + 0.910 841 631 338 525 748 463 697 92;
  • 15) 0.910 841 631 338 525 748 463 697 92 × 2 = 1 + 0.821 683 262 677 051 496 927 395 84;
  • 16) 0.821 683 262 677 051 496 927 395 84 × 2 = 1 + 0.643 366 525 354 102 993 854 791 68;
  • 17) 0.643 366 525 354 102 993 854 791 68 × 2 = 1 + 0.286 733 050 708 205 987 709 583 36;
  • 18) 0.286 733 050 708 205 987 709 583 36 × 2 = 0 + 0.573 466 101 416 411 975 419 166 72;
  • 19) 0.573 466 101 416 411 975 419 166 72 × 2 = 1 + 0.146 932 202 832 823 950 838 333 44;
  • 20) 0.146 932 202 832 823 950 838 333 44 × 2 = 0 + 0.293 864 405 665 647 901 676 666 88;
  • 21) 0.293 864 405 665 647 901 676 666 88 × 2 = 0 + 0.587 728 811 331 295 803 353 333 76;
  • 22) 0.587 728 811 331 295 803 353 333 76 × 2 = 1 + 0.175 457 622 662 591 606 706 667 52;
  • 23) 0.175 457 622 662 591 606 706 667 52 × 2 = 0 + 0.350 915 245 325 183 213 413 335 04;
  • 24) 0.350 915 245 325 183 213 413 335 04 × 2 = 0 + 0.701 830 490 650 366 426 826 670 08;
  • 25) 0.701 830 490 650 366 426 826 670 08 × 2 = 1 + 0.403 660 981 300 732 853 653 340 16;
  • 26) 0.403 660 981 300 732 853 653 340 16 × 2 = 0 + 0.807 321 962 601 465 707 306 680 32;
  • 27) 0.807 321 962 601 465 707 306 680 32 × 2 = 1 + 0.614 643 925 202 931 414 613 360 64;
  • 28) 0.614 643 925 202 931 414 613 360 64 × 2 = 1 + 0.229 287 850 405 862 829 226 721 28;
  • 29) 0.229 287 850 405 862 829 226 721 28 × 2 = 0 + 0.458 575 700 811 725 658 453 442 56;
  • 30) 0.458 575 700 811 725 658 453 442 56 × 2 = 0 + 0.917 151 401 623 451 316 906 885 12;

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).


5. 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:


0.014 459 890 236 287 751 815 702 13(10) =

0.0000 0011 1011 0011 1010 0100 1011 00(2)

6. Positive number before normalization:

0.014 459 890 236 287 751 815 702 13(10) =

0.0000 0011 1011 0011 1010 0100 1011 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the right, so that only one non zero digit remains to the left of it:


0.014 459 890 236 287 751 815 702 13(10) =

0.0000 0011 1011 0011 1010 0100 1011 00(2) =

0.0000 0011 1011 0011 1010 0100 1011 00(2) × 20 =

1.1101 1001 1101 0010 0101 100(2) × 2-7


8. 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 1 (a negative number)

Exponent (unadjusted): -7

Mantissa (not normalized):
1.1101 1001 1101 0010 0101 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-7 + 2(8-1) - 1 =

(-7 + 127)(10) =

120(10)


10. 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;
  • 120 ÷ 2 = 60 + 0;
  • 60 ÷ 2 = 30 + 0;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

120(10) =

0111 1000(2)


12. 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. 110 1100 1110 1001 0010 1100 =

110 1100 1110 1001 0010 1100


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (8 bits) =
0111 1000

Mantissa (23 bits) =
110 1100 1110 1001 0010 1100


The base ten decimal number -0.014 459 890 236 287 751 815 702 13 converted and written in 32 bit single precision IEEE 754 binary floating point representation:

1 - 0111 1000 - 110 1100 1110 1001 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