-0.000 000 077 194 272 307 679 057 121 276 855 461 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 077 194 272 307 679 057 121 276 855 461(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
-0.000 000 077 194 272 307 679 057 121 276 855 461(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. Start with the positive version of the number:

|-0.000 000 077 194 272 307 679 057 121 276 855 461| = 0.000 000 077 194 272 307 679 057 121 276 855 461


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: 0.000 000 077 194 272 307 679 057 121 276 855 461.

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) 0.000 000 077 194 272 307 679 057 121 276 855 461 × 2 = 0 + 0.000 000 154 388 544 615 358 114 242 553 710 922;
  • 2) 0.000 000 154 388 544 615 358 114 242 553 710 922 × 2 = 0 + 0.000 000 308 777 089 230 716 228 485 107 421 844;
  • 3) 0.000 000 308 777 089 230 716 228 485 107 421 844 × 2 = 0 + 0.000 000 617 554 178 461 432 456 970 214 843 688;
  • 4) 0.000 000 617 554 178 461 432 456 970 214 843 688 × 2 = 0 + 0.000 001 235 108 356 922 864 913 940 429 687 376;
  • 5) 0.000 001 235 108 356 922 864 913 940 429 687 376 × 2 = 0 + 0.000 002 470 216 713 845 729 827 880 859 374 752;
  • 6) 0.000 002 470 216 713 845 729 827 880 859 374 752 × 2 = 0 + 0.000 004 940 433 427 691 459 655 761 718 749 504;
  • 7) 0.000 004 940 433 427 691 459 655 761 718 749 504 × 2 = 0 + 0.000 009 880 866 855 382 919 311 523 437 499 008;
  • 8) 0.000 009 880 866 855 382 919 311 523 437 499 008 × 2 = 0 + 0.000 019 761 733 710 765 838 623 046 874 998 016;
  • 9) 0.000 019 761 733 710 765 838 623 046 874 998 016 × 2 = 0 + 0.000 039 523 467 421 531 677 246 093 749 996 032;
  • 10) 0.000 039 523 467 421 531 677 246 093 749 996 032 × 2 = 0 + 0.000 079 046 934 843 063 354 492 187 499 992 064;
  • 11) 0.000 079 046 934 843 063 354 492 187 499 992 064 × 2 = 0 + 0.000 158 093 869 686 126 708 984 374 999 984 128;
  • 12) 0.000 158 093 869 686 126 708 984 374 999 984 128 × 2 = 0 + 0.000 316 187 739 372 253 417 968 749 999 968 256;
  • 13) 0.000 316 187 739 372 253 417 968 749 999 968 256 × 2 = 0 + 0.000 632 375 478 744 506 835 937 499 999 936 512;
  • 14) 0.000 632 375 478 744 506 835 937 499 999 936 512 × 2 = 0 + 0.001 264 750 957 489 013 671 874 999 999 873 024;
  • 15) 0.001 264 750 957 489 013 671 874 999 999 873 024 × 2 = 0 + 0.002 529 501 914 978 027 343 749 999 999 746 048;
  • 16) 0.002 529 501 914 978 027 343 749 999 999 746 048 × 2 = 0 + 0.005 059 003 829 956 054 687 499 999 999 492 096;
  • 17) 0.005 059 003 829 956 054 687 499 999 999 492 096 × 2 = 0 + 0.010 118 007 659 912 109 374 999 999 998 984 192;
  • 18) 0.010 118 007 659 912 109 374 999 999 998 984 192 × 2 = 0 + 0.020 236 015 319 824 218 749 999 999 997 968 384;
  • 19) 0.020 236 015 319 824 218 749 999 999 997 968 384 × 2 = 0 + 0.040 472 030 639 648 437 499 999 999 995 936 768;
  • 20) 0.040 472 030 639 648 437 499 999 999 995 936 768 × 2 = 0 + 0.080 944 061 279 296 874 999 999 999 991 873 536;
  • 21) 0.080 944 061 279 296 874 999 999 999 991 873 536 × 2 = 0 + 0.161 888 122 558 593 749 999 999 999 983 747 072;
  • 22) 0.161 888 122 558 593 749 999 999 999 983 747 072 × 2 = 0 + 0.323 776 245 117 187 499 999 999 999 967 494 144;
  • 23) 0.323 776 245 117 187 499 999 999 999 967 494 144 × 2 = 0 + 0.647 552 490 234 374 999 999 999 999 934 988 288;
  • 24) 0.647 552 490 234 374 999 999 999 999 934 988 288 × 2 = 1 + 0.295 104 980 468 749 999 999 999 999 869 976 576;
  • 25) 0.295 104 980 468 749 999 999 999 999 869 976 576 × 2 = 0 + 0.590 209 960 937 499 999 999 999 999 739 953 152;
  • 26) 0.590 209 960 937 499 999 999 999 999 739 953 152 × 2 = 1 + 0.180 419 921 874 999 999 999 999 999 479 906 304;
  • 27) 0.180 419 921 874 999 999 999 999 999 479 906 304 × 2 = 0 + 0.360 839 843 749 999 999 999 999 998 959 812 608;
  • 28) 0.360 839 843 749 999 999 999 999 998 959 812 608 × 2 = 0 + 0.721 679 687 499 999 999 999 999 997 919 625 216;
  • 29) 0.721 679 687 499 999 999 999 999 997 919 625 216 × 2 = 1 + 0.443 359 374 999 999 999 999 999 995 839 250 432;
  • 30) 0.443 359 374 999 999 999 999 999 995 839 250 432 × 2 = 0 + 0.886 718 749 999 999 999 999 999 991 678 500 864;
  • 31) 0.886 718 749 999 999 999 999 999 991 678 500 864 × 2 = 1 + 0.773 437 499 999 999 999 999 999 983 357 001 728;
  • 32) 0.773 437 499 999 999 999 999 999 983 357 001 728 × 2 = 1 + 0.546 874 999 999 999 999 999 999 966 714 003 456;
  • 33) 0.546 874 999 999 999 999 999 999 966 714 003 456 × 2 = 1 + 0.093 749 999 999 999 999 999 999 933 428 006 912;
  • 34) 0.093 749 999 999 999 999 999 999 933 428 006 912 × 2 = 0 + 0.187 499 999 999 999 999 999 999 866 856 013 824;
  • 35) 0.187 499 999 999 999 999 999 999 866 856 013 824 × 2 = 0 + 0.374 999 999 999 999 999 999 999 733 712 027 648;
  • 36) 0.374 999 999 999 999 999 999 999 733 712 027 648 × 2 = 0 + 0.749 999 999 999 999 999 999 999 467 424 055 296;
  • 37) 0.749 999 999 999 999 999 999 999 467 424 055 296 × 2 = 1 + 0.499 999 999 999 999 999 999 998 934 848 110 592;
  • 38) 0.499 999 999 999 999 999 999 998 934 848 110 592 × 2 = 0 + 0.999 999 999 999 999 999 999 997 869 696 221 184;
  • 39) 0.999 999 999 999 999 999 999 997 869 696 221 184 × 2 = 1 + 0.999 999 999 999 999 999 999 995 739 392 442 368;
  • 40) 0.999 999 999 999 999 999 999 995 739 392 442 368 × 2 = 1 + 0.999 999 999 999 999 999 999 991 478 784 884 736;
  • 41) 0.999 999 999 999 999 999 999 991 478 784 884 736 × 2 = 1 + 0.999 999 999 999 999 999 999 982 957 569 769 472;
  • 42) 0.999 999 999 999 999 999 999 982 957 569 769 472 × 2 = 1 + 0.999 999 999 999 999 999 999 965 915 139 538 944;
  • 43) 0.999 999 999 999 999 999 999 965 915 139 538 944 × 2 = 1 + 0.999 999 999 999 999 999 999 931 830 279 077 888;
  • 44) 0.999 999 999 999 999 999 999 931 830 279 077 888 × 2 = 1 + 0.999 999 999 999 999 999 999 863 660 558 155 776;
  • 45) 0.999 999 999 999 999 999 999 863 660 558 155 776 × 2 = 1 + 0.999 999 999 999 999 999 999 727 321 116 311 552;
  • 46) 0.999 999 999 999 999 999 999 727 321 116 311 552 × 2 = 1 + 0.999 999 999 999 999 999 999 454 642 232 623 104;
  • 47) 0.999 999 999 999 999 999 999 454 642 232 623 104 × 2 = 1 + 0.999 999 999 999 999 999 998 909 284 465 246 208;

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:


0.000 000 077 194 272 307 679 057 121 276 855 461(10) =

0.0000 0000 0000 0000 0000 0001 0100 1011 1000 1011 1111 111(2)

6. Positive number before normalization:

0.000 000 077 194 272 307 679 057 121 276 855 461(10) =

0.0000 0000 0000 0000 0000 0001 0100 1011 1000 1011 1111 111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 24 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 077 194 272 307 679 057 121 276 855 461(10) =

0.0000 0000 0000 0000 0000 0001 0100 1011 1000 1011 1111 111(2) =

0.0000 0000 0000 0000 0000 0001 0100 1011 1000 1011 1111 111(2) × 20 =

1.0100 1011 1000 1011 1111 111(2) × 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 representation:

Sign 1 (a negative number)

Exponent (unadjusted): -24

Mantissa (not normalized):
1.0100 1011 1000 1011 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-24 + 2(8-1) - 1 =

(-24 + 127)(10) =

103(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;
  • 103 ÷ 2 = 51 + 1;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

103(10) =

0110 0111(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. 010 0101 1100 0101 1111 1111 =

010 0101 1100 0101 1111 1111


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) =
0110 0111

Mantissa (23 bits) =
010 0101 1100 0101 1111 1111


Decimal number -0.000 000 077 194 272 307 679 057 121 276 855 461 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 010 0101 1100 0101 1111 1111

Share

» Convert decimal -0.000 000 077 194 272 307 679 057 121 276 855 5 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: My Manager Only Says “We” When Referring to “My Work”. Gemini Notebook YouTube Video of 11.13 minutes

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