-0.000 000 000 742 147 676 173 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 173(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
-0.000 000 000 742 147 676 173(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:

|-0.000 000 000 742 147 676 173| = 0.000 000 000 742 147 676 173


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: 0.000 000 000 742 147 676 173.

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) 0.000 000 000 742 147 676 173 × 2 = 0 + 0.000 000 001 484 295 352 346;
  • 2) 0.000 000 001 484 295 352 346 × 2 = 0 + 0.000 000 002 968 590 704 692;
  • 3) 0.000 000 002 968 590 704 692 × 2 = 0 + 0.000 000 005 937 181 409 384;
  • 4) 0.000 000 005 937 181 409 384 × 2 = 0 + 0.000 000 011 874 362 818 768;
  • 5) 0.000 000 011 874 362 818 768 × 2 = 0 + 0.000 000 023 748 725 637 536;
  • 6) 0.000 000 023 748 725 637 536 × 2 = 0 + 0.000 000 047 497 451 275 072;
  • 7) 0.000 000 047 497 451 275 072 × 2 = 0 + 0.000 000 094 994 902 550 144;
  • 8) 0.000 000 094 994 902 550 144 × 2 = 0 + 0.000 000 189 989 805 100 288;
  • 9) 0.000 000 189 989 805 100 288 × 2 = 0 + 0.000 000 379 979 610 200 576;
  • 10) 0.000 000 379 979 610 200 576 × 2 = 0 + 0.000 000 759 959 220 401 152;
  • 11) 0.000 000 759 959 220 401 152 × 2 = 0 + 0.000 001 519 918 440 802 304;
  • 12) 0.000 001 519 918 440 802 304 × 2 = 0 + 0.000 003 039 836 881 604 608;
  • 13) 0.000 003 039 836 881 604 608 × 2 = 0 + 0.000 006 079 673 763 209 216;
  • 14) 0.000 006 079 673 763 209 216 × 2 = 0 + 0.000 012 159 347 526 418 432;
  • 15) 0.000 012 159 347 526 418 432 × 2 = 0 + 0.000 024 318 695 052 836 864;
  • 16) 0.000 024 318 695 052 836 864 × 2 = 0 + 0.000 048 637 390 105 673 728;
  • 17) 0.000 048 637 390 105 673 728 × 2 = 0 + 0.000 097 274 780 211 347 456;
  • 18) 0.000 097 274 780 211 347 456 × 2 = 0 + 0.000 194 549 560 422 694 912;
  • 19) 0.000 194 549 560 422 694 912 × 2 = 0 + 0.000 389 099 120 845 389 824;
  • 20) 0.000 389 099 120 845 389 824 × 2 = 0 + 0.000 778 198 241 690 779 648;
  • 21) 0.000 778 198 241 690 779 648 × 2 = 0 + 0.001 556 396 483 381 559 296;
  • 22) 0.001 556 396 483 381 559 296 × 2 = 0 + 0.003 112 792 966 763 118 592;
  • 23) 0.003 112 792 966 763 118 592 × 2 = 0 + 0.006 225 585 933 526 237 184;
  • 24) 0.006 225 585 933 526 237 184 × 2 = 0 + 0.012 451 171 867 052 474 368;
  • 25) 0.012 451 171 867 052 474 368 × 2 = 0 + 0.024 902 343 734 104 948 736;
  • 26) 0.024 902 343 734 104 948 736 × 2 = 0 + 0.049 804 687 468 209 897 472;
  • 27) 0.049 804 687 468 209 897 472 × 2 = 0 + 0.099 609 374 936 419 794 944;
  • 28) 0.099 609 374 936 419 794 944 × 2 = 0 + 0.199 218 749 872 839 589 888;
  • 29) 0.199 218 749 872 839 589 888 × 2 = 0 + 0.398 437 499 745 679 179 776;
  • 30) 0.398 437 499 745 679 179 776 × 2 = 0 + 0.796 874 999 491 358 359 552;
  • 31) 0.796 874 999 491 358 359 552 × 2 = 1 + 0.593 749 998 982 716 719 104;
  • 32) 0.593 749 998 982 716 719 104 × 2 = 1 + 0.187 499 997 965 433 438 208;
  • 33) 0.187 499 997 965 433 438 208 × 2 = 0 + 0.374 999 995 930 866 876 416;
  • 34) 0.374 999 995 930 866 876 416 × 2 = 0 + 0.749 999 991 861 733 752 832;
  • 35) 0.749 999 991 861 733 752 832 × 2 = 1 + 0.499 999 983 723 467 505 664;
  • 36) 0.499 999 983 723 467 505 664 × 2 = 0 + 0.999 999 967 446 935 011 328;
  • 37) 0.999 999 967 446 935 011 328 × 2 = 1 + 0.999 999 934 893 870 022 656;
  • 38) 0.999 999 934 893 870 022 656 × 2 = 1 + 0.999 999 869 787 740 045 312;
  • 39) 0.999 999 869 787 740 045 312 × 2 = 1 + 0.999 999 739 575 480 090 624;
  • 40) 0.999 999 739 575 480 090 624 × 2 = 1 + 0.999 999 479 150 960 181 248;
  • 41) 0.999 999 479 150 960 181 248 × 2 = 1 + 0.999 998 958 301 920 362 496;
  • 42) 0.999 998 958 301 920 362 496 × 2 = 1 + 0.999 997 916 603 840 724 992;
  • 43) 0.999 997 916 603 840 724 992 × 2 = 1 + 0.999 995 833 207 681 449 984;
  • 44) 0.999 995 833 207 681 449 984 × 2 = 1 + 0.999 991 666 415 362 899 968;
  • 45) 0.999 991 666 415 362 899 968 × 2 = 1 + 0.999 983 332 830 725 799 936;
  • 46) 0.999 983 332 830 725 799 936 × 2 = 1 + 0.999 966 665 661 451 599 872;
  • 47) 0.999 966 665 661 451 599 872 × 2 = 1 + 0.999 933 331 322 903 199 744;
  • 48) 0.999 933 331 322 903 199 744 × 2 = 1 + 0.999 866 662 645 806 399 488;
  • 49) 0.999 866 662 645 806 399 488 × 2 = 1 + 0.999 733 325 291 612 798 976;
  • 50) 0.999 733 325 291 612 798 976 × 2 = 1 + 0.999 466 650 583 225 597 952;
  • 51) 0.999 466 650 583 225 597 952 × 2 = 1 + 0.998 933 301 166 451 195 904;
  • 52) 0.998 933 301 166 451 195 904 × 2 = 1 + 0.997 866 602 332 902 391 808;
  • 53) 0.997 866 602 332 902 391 808 × 2 = 1 + 0.995 733 204 665 804 783 616;
  • 54) 0.995 733 204 665 804 783 616 × 2 = 1 + 0.991 466 409 331 609 567 232;

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:


0.000 000 000 742 147 676 173(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 173(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:


0.000 000 000 742 147 676 173(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 -0.000 000 000 742 147 676 173 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