-0.000 000 000 742 147 676 618 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 618 9(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
-0.000 000 000 742 147 676 618 9(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:

|-0.000 000 000 742 147 676 618 9| = 0.000 000 000 742 147 676 618 9


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: 0.000 000 000 742 147 676 618 9.

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) 0.000 000 000 742 147 676 618 9 × 2 = 0 + 0.000 000 001 484 295 353 237 8;
  • 2) 0.000 000 001 484 295 353 237 8 × 2 = 0 + 0.000 000 002 968 590 706 475 6;
  • 3) 0.000 000 002 968 590 706 475 6 × 2 = 0 + 0.000 000 005 937 181 412 951 2;
  • 4) 0.000 000 005 937 181 412 951 2 × 2 = 0 + 0.000 000 011 874 362 825 902 4;
  • 5) 0.000 000 011 874 362 825 902 4 × 2 = 0 + 0.000 000 023 748 725 651 804 8;
  • 6) 0.000 000 023 748 725 651 804 8 × 2 = 0 + 0.000 000 047 497 451 303 609 6;
  • 7) 0.000 000 047 497 451 303 609 6 × 2 = 0 + 0.000 000 094 994 902 607 219 2;
  • 8) 0.000 000 094 994 902 607 219 2 × 2 = 0 + 0.000 000 189 989 805 214 438 4;
  • 9) 0.000 000 189 989 805 214 438 4 × 2 = 0 + 0.000 000 379 979 610 428 876 8;
  • 10) 0.000 000 379 979 610 428 876 8 × 2 = 0 + 0.000 000 759 959 220 857 753 6;
  • 11) 0.000 000 759 959 220 857 753 6 × 2 = 0 + 0.000 001 519 918 441 715 507 2;
  • 12) 0.000 001 519 918 441 715 507 2 × 2 = 0 + 0.000 003 039 836 883 431 014 4;
  • 13) 0.000 003 039 836 883 431 014 4 × 2 = 0 + 0.000 006 079 673 766 862 028 8;
  • 14) 0.000 006 079 673 766 862 028 8 × 2 = 0 + 0.000 012 159 347 533 724 057 6;
  • 15) 0.000 012 159 347 533 724 057 6 × 2 = 0 + 0.000 024 318 695 067 448 115 2;
  • 16) 0.000 024 318 695 067 448 115 2 × 2 = 0 + 0.000 048 637 390 134 896 230 4;
  • 17) 0.000 048 637 390 134 896 230 4 × 2 = 0 + 0.000 097 274 780 269 792 460 8;
  • 18) 0.000 097 274 780 269 792 460 8 × 2 = 0 + 0.000 194 549 560 539 584 921 6;
  • 19) 0.000 194 549 560 539 584 921 6 × 2 = 0 + 0.000 389 099 121 079 169 843 2;
  • 20) 0.000 389 099 121 079 169 843 2 × 2 = 0 + 0.000 778 198 242 158 339 686 4;
  • 21) 0.000 778 198 242 158 339 686 4 × 2 = 0 + 0.001 556 396 484 316 679 372 8;
  • 22) 0.001 556 396 484 316 679 372 8 × 2 = 0 + 0.003 112 792 968 633 358 745 6;
  • 23) 0.003 112 792 968 633 358 745 6 × 2 = 0 + 0.006 225 585 937 266 717 491 2;
  • 24) 0.006 225 585 937 266 717 491 2 × 2 = 0 + 0.012 451 171 874 533 434 982 4;
  • 25) 0.012 451 171 874 533 434 982 4 × 2 = 0 + 0.024 902 343 749 066 869 964 8;
  • 26) 0.024 902 343 749 066 869 964 8 × 2 = 0 + 0.049 804 687 498 133 739 929 6;
  • 27) 0.049 804 687 498 133 739 929 6 × 2 = 0 + 0.099 609 374 996 267 479 859 2;
  • 28) 0.099 609 374 996 267 479 859 2 × 2 = 0 + 0.199 218 749 992 534 959 718 4;
  • 29) 0.199 218 749 992 534 959 718 4 × 2 = 0 + 0.398 437 499 985 069 919 436 8;
  • 30) 0.398 437 499 985 069 919 436 8 × 2 = 0 + 0.796 874 999 970 139 838 873 6;
  • 31) 0.796 874 999 970 139 838 873 6 × 2 = 1 + 0.593 749 999 940 279 677 747 2;
  • 32) 0.593 749 999 940 279 677 747 2 × 2 = 1 + 0.187 499 999 880 559 355 494 4;
  • 33) 0.187 499 999 880 559 355 494 4 × 2 = 0 + 0.374 999 999 761 118 710 988 8;
  • 34) 0.374 999 999 761 118 710 988 8 × 2 = 0 + 0.749 999 999 522 237 421 977 6;
  • 35) 0.749 999 999 522 237 421 977 6 × 2 = 1 + 0.499 999 999 044 474 843 955 2;
  • 36) 0.499 999 999 044 474 843 955 2 × 2 = 0 + 0.999 999 998 088 949 687 910 4;
  • 37) 0.999 999 998 088 949 687 910 4 × 2 = 1 + 0.999 999 996 177 899 375 820 8;
  • 38) 0.999 999 996 177 899 375 820 8 × 2 = 1 + 0.999 999 992 355 798 751 641 6;
  • 39) 0.999 999 992 355 798 751 641 6 × 2 = 1 + 0.999 999 984 711 597 503 283 2;
  • 40) 0.999 999 984 711 597 503 283 2 × 2 = 1 + 0.999 999 969 423 195 006 566 4;
  • 41) 0.999 999 969 423 195 006 566 4 × 2 = 1 + 0.999 999 938 846 390 013 132 8;
  • 42) 0.999 999 938 846 390 013 132 8 × 2 = 1 + 0.999 999 877 692 780 026 265 6;
  • 43) 0.999 999 877 692 780 026 265 6 × 2 = 1 + 0.999 999 755 385 560 052 531 2;
  • 44) 0.999 999 755 385 560 052 531 2 × 2 = 1 + 0.999 999 510 771 120 105 062 4;
  • 45) 0.999 999 510 771 120 105 062 4 × 2 = 1 + 0.999 999 021 542 240 210 124 8;
  • 46) 0.999 999 021 542 240 210 124 8 × 2 = 1 + 0.999 998 043 084 480 420 249 6;
  • 47) 0.999 998 043 084 480 420 249 6 × 2 = 1 + 0.999 996 086 168 960 840 499 2;
  • 48) 0.999 996 086 168 960 840 499 2 × 2 = 1 + 0.999 992 172 337 921 680 998 4;
  • 49) 0.999 992 172 337 921 680 998 4 × 2 = 1 + 0.999 984 344 675 843 361 996 8;
  • 50) 0.999 984 344 675 843 361 996 8 × 2 = 1 + 0.999 968 689 351 686 723 993 6;
  • 51) 0.999 968 689 351 686 723 993 6 × 2 = 1 + 0.999 937 378 703 373 447 987 2;
  • 52) 0.999 937 378 703 373 447 987 2 × 2 = 1 + 0.999 874 757 406 746 895 974 4;
  • 53) 0.999 874 757 406 746 895 974 4 × 2 = 1 + 0.999 749 514 813 493 791 948 8;
  • 54) 0.999 749 514 813 493 791 948 8 × 2 = 1 + 0.999 499 029 626 987 583 897 6;

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:


0.000 000 000 742 147 676 618 9(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 618 9(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:


0.000 000 000 742 147 676 618 9(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 -0.000 000 000 742 147 676 618 9 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 615 9 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 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 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