-0.000 000 077 194 272 307 679 057 121 276 855 564 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 077 194 272 307 679 057 121 276 855 564(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
-0.000 000 077 194 272 307 679 057 121 276 855 564(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:

|-0.000 000 077 194 272 307 679 057 121 276 855 564| = 0.000 000 077 194 272 307 679 057 121 276 855 564


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: 0.000 000 077 194 272 307 679 057 121 276 855 564.

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) 0.000 000 077 194 272 307 679 057 121 276 855 564 × 2 = 0 + 0.000 000 154 388 544 615 358 114 242 553 711 128;
  • 2) 0.000 000 154 388 544 615 358 114 242 553 711 128 × 2 = 0 + 0.000 000 308 777 089 230 716 228 485 107 422 256;
  • 3) 0.000 000 308 777 089 230 716 228 485 107 422 256 × 2 = 0 + 0.000 000 617 554 178 461 432 456 970 214 844 512;
  • 4) 0.000 000 617 554 178 461 432 456 970 214 844 512 × 2 = 0 + 0.000 001 235 108 356 922 864 913 940 429 689 024;
  • 5) 0.000 001 235 108 356 922 864 913 940 429 689 024 × 2 = 0 + 0.000 002 470 216 713 845 729 827 880 859 378 048;
  • 6) 0.000 002 470 216 713 845 729 827 880 859 378 048 × 2 = 0 + 0.000 004 940 433 427 691 459 655 761 718 756 096;
  • 7) 0.000 004 940 433 427 691 459 655 761 718 756 096 × 2 = 0 + 0.000 009 880 866 855 382 919 311 523 437 512 192;
  • 8) 0.000 009 880 866 855 382 919 311 523 437 512 192 × 2 = 0 + 0.000 019 761 733 710 765 838 623 046 875 024 384;
  • 9) 0.000 019 761 733 710 765 838 623 046 875 024 384 × 2 = 0 + 0.000 039 523 467 421 531 677 246 093 750 048 768;
  • 10) 0.000 039 523 467 421 531 677 246 093 750 048 768 × 2 = 0 + 0.000 079 046 934 843 063 354 492 187 500 097 536;
  • 11) 0.000 079 046 934 843 063 354 492 187 500 097 536 × 2 = 0 + 0.000 158 093 869 686 126 708 984 375 000 195 072;
  • 12) 0.000 158 093 869 686 126 708 984 375 000 195 072 × 2 = 0 + 0.000 316 187 739 372 253 417 968 750 000 390 144;
  • 13) 0.000 316 187 739 372 253 417 968 750 000 390 144 × 2 = 0 + 0.000 632 375 478 744 506 835 937 500 000 780 288;
  • 14) 0.000 632 375 478 744 506 835 937 500 000 780 288 × 2 = 0 + 0.001 264 750 957 489 013 671 875 000 001 560 576;
  • 15) 0.001 264 750 957 489 013 671 875 000 001 560 576 × 2 = 0 + 0.002 529 501 914 978 027 343 750 000 003 121 152;
  • 16) 0.002 529 501 914 978 027 343 750 000 003 121 152 × 2 = 0 + 0.005 059 003 829 956 054 687 500 000 006 242 304;
  • 17) 0.005 059 003 829 956 054 687 500 000 006 242 304 × 2 = 0 + 0.010 118 007 659 912 109 375 000 000 012 484 608;
  • 18) 0.010 118 007 659 912 109 375 000 000 012 484 608 × 2 = 0 + 0.020 236 015 319 824 218 750 000 000 024 969 216;
  • 19) 0.020 236 015 319 824 218 750 000 000 024 969 216 × 2 = 0 + 0.040 472 030 639 648 437 500 000 000 049 938 432;
  • 20) 0.040 472 030 639 648 437 500 000 000 049 938 432 × 2 = 0 + 0.080 944 061 279 296 875 000 000 000 099 876 864;
  • 21) 0.080 944 061 279 296 875 000 000 000 099 876 864 × 2 = 0 + 0.161 888 122 558 593 750 000 000 000 199 753 728;
  • 22) 0.161 888 122 558 593 750 000 000 000 199 753 728 × 2 = 0 + 0.323 776 245 117 187 500 000 000 000 399 507 456;
  • 23) 0.323 776 245 117 187 500 000 000 000 399 507 456 × 2 = 0 + 0.647 552 490 234 375 000 000 000 000 799 014 912;
  • 24) 0.647 552 490 234 375 000 000 000 000 799 014 912 × 2 = 1 + 0.295 104 980 468 750 000 000 000 001 598 029 824;
  • 25) 0.295 104 980 468 750 000 000 000 001 598 029 824 × 2 = 0 + 0.590 209 960 937 500 000 000 000 003 196 059 648;
  • 26) 0.590 209 960 937 500 000 000 000 003 196 059 648 × 2 = 1 + 0.180 419 921 875 000 000 000 000 006 392 119 296;
  • 27) 0.180 419 921 875 000 000 000 000 006 392 119 296 × 2 = 0 + 0.360 839 843 750 000 000 000 000 012 784 238 592;
  • 28) 0.360 839 843 750 000 000 000 000 012 784 238 592 × 2 = 0 + 0.721 679 687 500 000 000 000 000 025 568 477 184;
  • 29) 0.721 679 687 500 000 000 000 000 025 568 477 184 × 2 = 1 + 0.443 359 375 000 000 000 000 000 051 136 954 368;
  • 30) 0.443 359 375 000 000 000 000 000 051 136 954 368 × 2 = 0 + 0.886 718 750 000 000 000 000 000 102 273 908 736;
  • 31) 0.886 718 750 000 000 000 000 000 102 273 908 736 × 2 = 1 + 0.773 437 500 000 000 000 000 000 204 547 817 472;
  • 32) 0.773 437 500 000 000 000 000 000 204 547 817 472 × 2 = 1 + 0.546 875 000 000 000 000 000 000 409 095 634 944;
  • 33) 0.546 875 000 000 000 000 000 000 409 095 634 944 × 2 = 1 + 0.093 750 000 000 000 000 000 000 818 191 269 888;
  • 34) 0.093 750 000 000 000 000 000 000 818 191 269 888 × 2 = 0 + 0.187 500 000 000 000 000 000 001 636 382 539 776;
  • 35) 0.187 500 000 000 000 000 000 001 636 382 539 776 × 2 = 0 + 0.375 000 000 000 000 000 000 003 272 765 079 552;
  • 36) 0.375 000 000 000 000 000 000 003 272 765 079 552 × 2 = 0 + 0.750 000 000 000 000 000 000 006 545 530 159 104;
  • 37) 0.750 000 000 000 000 000 000 006 545 530 159 104 × 2 = 1 + 0.500 000 000 000 000 000 000 013 091 060 318 208;
  • 38) 0.500 000 000 000 000 000 000 013 091 060 318 208 × 2 = 1 + 0.000 000 000 000 000 000 000 026 182 120 636 416;
  • 39) 0.000 000 000 000 000 000 000 026 182 120 636 416 × 2 = 0 + 0.000 000 000 000 000 000 000 052 364 241 272 832;
  • 40) 0.000 000 000 000 000 000 000 052 364 241 272 832 × 2 = 0 + 0.000 000 000 000 000 000 000 104 728 482 545 664;
  • 41) 0.000 000 000 000 000 000 000 104 728 482 545 664 × 2 = 0 + 0.000 000 000 000 000 000 000 209 456 965 091 328;
  • 42) 0.000 000 000 000 000 000 000 209 456 965 091 328 × 2 = 0 + 0.000 000 000 000 000 000 000 418 913 930 182 656;
  • 43) 0.000 000 000 000 000 000 000 418 913 930 182 656 × 2 = 0 + 0.000 000 000 000 000 000 000 837 827 860 365 312;
  • 44) 0.000 000 000 000 000 000 000 837 827 860 365 312 × 2 = 0 + 0.000 000 000 000 000 000 001 675 655 720 730 624;
  • 45) 0.000 000 000 000 000 000 001 675 655 720 730 624 × 2 = 0 + 0.000 000 000 000 000 000 003 351 311 441 461 248;
  • 46) 0.000 000 000 000 000 000 003 351 311 441 461 248 × 2 = 0 + 0.000 000 000 000 000 000 006 702 622 882 922 496;
  • 47) 0.000 000 000 000 000 000 006 702 622 882 922 496 × 2 = 0 + 0.000 000 000 000 000 000 013 405 245 765 844 992;

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 - 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:


0.000 000 077 194 272 307 679 057 121 276 855 564(10) =

0.0000 0000 0000 0000 0000 0001 0100 1011 1000 1100 0000 000(2)

6. Positive number before normalization:

0.000 000 077 194 272 307 679 057 121 276 855 564(10) =

0.0000 0000 0000 0000 0000 0001 0100 1011 1000 1100 0000 000(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:


0.000 000 077 194 272 307 679 057 121 276 855 564(10) =

0.0000 0000 0000 0000 0000 0001 0100 1011 1000 1100 0000 000(2) =

0.0000 0000 0000 0000 0000 0001 0100 1011 1000 1100 0000 000(2) × 20 =

1.0100 1011 1000 1100 0000 000(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 1100 0000 000


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 0110 0000 0000 =

010 0101 1100 0110 0000 0000


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 0110 0000 0000


Decimal number -0.000 000 077 194 272 307 679 057 121 276 855 564 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0111 - 010 0101 1100 0110 0000 0000

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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