-0.000 000 000 742 147 676 646 337 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 337(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
-0.000 000 000 742 147 676 646 337(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:

|-0.000 000 000 742 147 676 646 337| = 0.000 000 000 742 147 676 646 337


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: 0.000 000 000 742 147 676 646 337.

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) 0.000 000 000 742 147 676 646 337 × 2 = 0 + 0.000 000 001 484 295 353 292 674;
  • 2) 0.000 000 001 484 295 353 292 674 × 2 = 0 + 0.000 000 002 968 590 706 585 348;
  • 3) 0.000 000 002 968 590 706 585 348 × 2 = 0 + 0.000 000 005 937 181 413 170 696;
  • 4) 0.000 000 005 937 181 413 170 696 × 2 = 0 + 0.000 000 011 874 362 826 341 392;
  • 5) 0.000 000 011 874 362 826 341 392 × 2 = 0 + 0.000 000 023 748 725 652 682 784;
  • 6) 0.000 000 023 748 725 652 682 784 × 2 = 0 + 0.000 000 047 497 451 305 365 568;
  • 7) 0.000 000 047 497 451 305 365 568 × 2 = 0 + 0.000 000 094 994 902 610 731 136;
  • 8) 0.000 000 094 994 902 610 731 136 × 2 = 0 + 0.000 000 189 989 805 221 462 272;
  • 9) 0.000 000 189 989 805 221 462 272 × 2 = 0 + 0.000 000 379 979 610 442 924 544;
  • 10) 0.000 000 379 979 610 442 924 544 × 2 = 0 + 0.000 000 759 959 220 885 849 088;
  • 11) 0.000 000 759 959 220 885 849 088 × 2 = 0 + 0.000 001 519 918 441 771 698 176;
  • 12) 0.000 001 519 918 441 771 698 176 × 2 = 0 + 0.000 003 039 836 883 543 396 352;
  • 13) 0.000 003 039 836 883 543 396 352 × 2 = 0 + 0.000 006 079 673 767 086 792 704;
  • 14) 0.000 006 079 673 767 086 792 704 × 2 = 0 + 0.000 012 159 347 534 173 585 408;
  • 15) 0.000 012 159 347 534 173 585 408 × 2 = 0 + 0.000 024 318 695 068 347 170 816;
  • 16) 0.000 024 318 695 068 347 170 816 × 2 = 0 + 0.000 048 637 390 136 694 341 632;
  • 17) 0.000 048 637 390 136 694 341 632 × 2 = 0 + 0.000 097 274 780 273 388 683 264;
  • 18) 0.000 097 274 780 273 388 683 264 × 2 = 0 + 0.000 194 549 560 546 777 366 528;
  • 19) 0.000 194 549 560 546 777 366 528 × 2 = 0 + 0.000 389 099 121 093 554 733 056;
  • 20) 0.000 389 099 121 093 554 733 056 × 2 = 0 + 0.000 778 198 242 187 109 466 112;
  • 21) 0.000 778 198 242 187 109 466 112 × 2 = 0 + 0.001 556 396 484 374 218 932 224;
  • 22) 0.001 556 396 484 374 218 932 224 × 2 = 0 + 0.003 112 792 968 748 437 864 448;
  • 23) 0.003 112 792 968 748 437 864 448 × 2 = 0 + 0.006 225 585 937 496 875 728 896;
  • 24) 0.006 225 585 937 496 875 728 896 × 2 = 0 + 0.012 451 171 874 993 751 457 792;
  • 25) 0.012 451 171 874 993 751 457 792 × 2 = 0 + 0.024 902 343 749 987 502 915 584;
  • 26) 0.024 902 343 749 987 502 915 584 × 2 = 0 + 0.049 804 687 499 975 005 831 168;
  • 27) 0.049 804 687 499 975 005 831 168 × 2 = 0 + 0.099 609 374 999 950 011 662 336;
  • 28) 0.099 609 374 999 950 011 662 336 × 2 = 0 + 0.199 218 749 999 900 023 324 672;
  • 29) 0.199 218 749 999 900 023 324 672 × 2 = 0 + 0.398 437 499 999 800 046 649 344;
  • 30) 0.398 437 499 999 800 046 649 344 × 2 = 0 + 0.796 874 999 999 600 093 298 688;
  • 31) 0.796 874 999 999 600 093 298 688 × 2 = 1 + 0.593 749 999 999 200 186 597 376;
  • 32) 0.593 749 999 999 200 186 597 376 × 2 = 1 + 0.187 499 999 998 400 373 194 752;
  • 33) 0.187 499 999 998 400 373 194 752 × 2 = 0 + 0.374 999 999 996 800 746 389 504;
  • 34) 0.374 999 999 996 800 746 389 504 × 2 = 0 + 0.749 999 999 993 601 492 779 008;
  • 35) 0.749 999 999 993 601 492 779 008 × 2 = 1 + 0.499 999 999 987 202 985 558 016;
  • 36) 0.499 999 999 987 202 985 558 016 × 2 = 0 + 0.999 999 999 974 405 971 116 032;
  • 37) 0.999 999 999 974 405 971 116 032 × 2 = 1 + 0.999 999 999 948 811 942 232 064;
  • 38) 0.999 999 999 948 811 942 232 064 × 2 = 1 + 0.999 999 999 897 623 884 464 128;
  • 39) 0.999 999 999 897 623 884 464 128 × 2 = 1 + 0.999 999 999 795 247 768 928 256;
  • 40) 0.999 999 999 795 247 768 928 256 × 2 = 1 + 0.999 999 999 590 495 537 856 512;
  • 41) 0.999 999 999 590 495 537 856 512 × 2 = 1 + 0.999 999 999 180 991 075 713 024;
  • 42) 0.999 999 999 180 991 075 713 024 × 2 = 1 + 0.999 999 998 361 982 151 426 048;
  • 43) 0.999 999 998 361 982 151 426 048 × 2 = 1 + 0.999 999 996 723 964 302 852 096;
  • 44) 0.999 999 996 723 964 302 852 096 × 2 = 1 + 0.999 999 993 447 928 605 704 192;
  • 45) 0.999 999 993 447 928 605 704 192 × 2 = 1 + 0.999 999 986 895 857 211 408 384;
  • 46) 0.999 999 986 895 857 211 408 384 × 2 = 1 + 0.999 999 973 791 714 422 816 768;
  • 47) 0.999 999 973 791 714 422 816 768 × 2 = 1 + 0.999 999 947 583 428 845 633 536;
  • 48) 0.999 999 947 583 428 845 633 536 × 2 = 1 + 0.999 999 895 166 857 691 267 072;
  • 49) 0.999 999 895 166 857 691 267 072 × 2 = 1 + 0.999 999 790 333 715 382 534 144;
  • 50) 0.999 999 790 333 715 382 534 144 × 2 = 1 + 0.999 999 580 667 430 765 068 288;
  • 51) 0.999 999 580 667 430 765 068 288 × 2 = 1 + 0.999 999 161 334 861 530 136 576;
  • 52) 0.999 999 161 334 861 530 136 576 × 2 = 1 + 0.999 998 322 669 723 060 273 152;
  • 53) 0.999 998 322 669 723 060 273 152 × 2 = 1 + 0.999 996 645 339 446 120 546 304;
  • 54) 0.999 996 645 339 446 120 546 304 × 2 = 1 + 0.999 993 290 678 892 241 092 608;

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:


0.000 000 000 742 147 676 646 337(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 337(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 337(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 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) =

96(10) =

0110 0000(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. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 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 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 337 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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