0.000 104 719 755 119 659 774 615 776 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 104 719 755 119 659 774 615 776(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
0.000 104 719 755 119 659 774 615 776(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: 0.000 104 719 755 119 659 774 615 776.

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) 0.000 104 719 755 119 659 774 615 776 × 2 = 0 + 0.000 209 439 510 239 319 549 231 552;
  • 2) 0.000 209 439 510 239 319 549 231 552 × 2 = 0 + 0.000 418 879 020 478 639 098 463 104;
  • 3) 0.000 418 879 020 478 639 098 463 104 × 2 = 0 + 0.000 837 758 040 957 278 196 926 208;
  • 4) 0.000 837 758 040 957 278 196 926 208 × 2 = 0 + 0.001 675 516 081 914 556 393 852 416;
  • 5) 0.001 675 516 081 914 556 393 852 416 × 2 = 0 + 0.003 351 032 163 829 112 787 704 832;
  • 6) 0.003 351 032 163 829 112 787 704 832 × 2 = 0 + 0.006 702 064 327 658 225 575 409 664;
  • 7) 0.006 702 064 327 658 225 575 409 664 × 2 = 0 + 0.013 404 128 655 316 451 150 819 328;
  • 8) 0.013 404 128 655 316 451 150 819 328 × 2 = 0 + 0.026 808 257 310 632 902 301 638 656;
  • 9) 0.026 808 257 310 632 902 301 638 656 × 2 = 0 + 0.053 616 514 621 265 804 603 277 312;
  • 10) 0.053 616 514 621 265 804 603 277 312 × 2 = 0 + 0.107 233 029 242 531 609 206 554 624;
  • 11) 0.107 233 029 242 531 609 206 554 624 × 2 = 0 + 0.214 466 058 485 063 218 413 109 248;
  • 12) 0.214 466 058 485 063 218 413 109 248 × 2 = 0 + 0.428 932 116 970 126 436 826 218 496;
  • 13) 0.428 932 116 970 126 436 826 218 496 × 2 = 0 + 0.857 864 233 940 252 873 652 436 992;
  • 14) 0.857 864 233 940 252 873 652 436 992 × 2 = 1 + 0.715 728 467 880 505 747 304 873 984;
  • 15) 0.715 728 467 880 505 747 304 873 984 × 2 = 1 + 0.431 456 935 761 011 494 609 747 968;
  • 16) 0.431 456 935 761 011 494 609 747 968 × 2 = 0 + 0.862 913 871 522 022 989 219 495 936;
  • 17) 0.862 913 871 522 022 989 219 495 936 × 2 = 1 + 0.725 827 743 044 045 978 438 991 872;
  • 18) 0.725 827 743 044 045 978 438 991 872 × 2 = 1 + 0.451 655 486 088 091 956 877 983 744;
  • 19) 0.451 655 486 088 091 956 877 983 744 × 2 = 0 + 0.903 310 972 176 183 913 755 967 488;
  • 20) 0.903 310 972 176 183 913 755 967 488 × 2 = 1 + 0.806 621 944 352 367 827 511 934 976;
  • 21) 0.806 621 944 352 367 827 511 934 976 × 2 = 1 + 0.613 243 888 704 735 655 023 869 952;
  • 22) 0.613 243 888 704 735 655 023 869 952 × 2 = 1 + 0.226 487 777 409 471 310 047 739 904;
  • 23) 0.226 487 777 409 471 310 047 739 904 × 2 = 0 + 0.452 975 554 818 942 620 095 479 808;
  • 24) 0.452 975 554 818 942 620 095 479 808 × 2 = 0 + 0.905 951 109 637 885 240 190 959 616;
  • 25) 0.905 951 109 637 885 240 190 959 616 × 2 = 1 + 0.811 902 219 275 770 480 381 919 232;
  • 26) 0.811 902 219 275 770 480 381 919 232 × 2 = 1 + 0.623 804 438 551 540 960 763 838 464;
  • 27) 0.623 804 438 551 540 960 763 838 464 × 2 = 1 + 0.247 608 877 103 081 921 527 676 928;
  • 28) 0.247 608 877 103 081 921 527 676 928 × 2 = 0 + 0.495 217 754 206 163 843 055 353 856;
  • 29) 0.495 217 754 206 163 843 055 353 856 × 2 = 0 + 0.990 435 508 412 327 686 110 707 712;
  • 30) 0.990 435 508 412 327 686 110 707 712 × 2 = 1 + 0.980 871 016 824 655 372 221 415 424;
  • 31) 0.980 871 016 824 655 372 221 415 424 × 2 = 1 + 0.961 742 033 649 310 744 442 830 848;
  • 32) 0.961 742 033 649 310 744 442 830 848 × 2 = 1 + 0.923 484 067 298 621 488 885 661 696;
  • 33) 0.923 484 067 298 621 488 885 661 696 × 2 = 1 + 0.846 968 134 597 242 977 771 323 392;
  • 34) 0.846 968 134 597 242 977 771 323 392 × 2 = 1 + 0.693 936 269 194 485 955 542 646 784;
  • 35) 0.693 936 269 194 485 955 542 646 784 × 2 = 1 + 0.387 872 538 388 971 911 085 293 568;
  • 36) 0.387 872 538 388 971 911 085 293 568 × 2 = 0 + 0.775 745 076 777 943 822 170 587 136;
  • 37) 0.775 745 076 777 943 822 170 587 136 × 2 = 1 + 0.551 490 153 555 887 644 341 174 272;

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:


0.000 104 719 755 119 659 774 615 776(10) =

0.0000 0000 0000 0110 1101 1100 1110 0111 1110 1(2)

5. Positive number before normalization:

0.000 104 719 755 119 659 774 615 776(10) =

0.0000 0000 0000 0110 1101 1100 1110 0111 1110 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 14 positions to the right, so that only one non zero digit remains to the left of it:


0.000 104 719 755 119 659 774 615 776(10) =

0.0000 0000 0000 0110 1101 1100 1110 0111 1110 1(2) =

0.0000 0000 0000 0110 1101 1100 1110 0111 1110 1(2) × 20 =

1.1011 0111 0011 1001 1111 101(2) × 2-14


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): -14

Mantissa (not normalized):
1.1011 0111 0011 1001 1111 101


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-14 + 2(8-1) - 1 =

(-14 + 127)(10) =

113(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;
  • 113 ÷ 2 = 56 + 1;
  • 56 ÷ 2 = 28 + 0;
  • 28 ÷ 2 = 14 + 0;
  • 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) =

113(10) =

0111 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. 101 1011 1001 1100 1111 1101 =

101 1011 1001 1100 1111 1101


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 0001

Mantissa (23 bits) =
101 1011 1001 1100 1111 1101


Decimal number 0.000 104 719 755 119 659 774 615 776 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 0001 - 101 1011 1001 1100 1111 1101

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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