-0.000 000 000 742 147 673 31 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 31(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
-0.000 000 000 742 147 673 31(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:

|-0.000 000 000 742 147 673 31| = 0.000 000 000 742 147 673 31


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: 0.000 000 000 742 147 673 31.

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) 0.000 000 000 742 147 673 31 × 2 = 0 + 0.000 000 001 484 295 346 62;
  • 2) 0.000 000 001 484 295 346 62 × 2 = 0 + 0.000 000 002 968 590 693 24;
  • 3) 0.000 000 002 968 590 693 24 × 2 = 0 + 0.000 000 005 937 181 386 48;
  • 4) 0.000 000 005 937 181 386 48 × 2 = 0 + 0.000 000 011 874 362 772 96;
  • 5) 0.000 000 011 874 362 772 96 × 2 = 0 + 0.000 000 023 748 725 545 92;
  • 6) 0.000 000 023 748 725 545 92 × 2 = 0 + 0.000 000 047 497 451 091 84;
  • 7) 0.000 000 047 497 451 091 84 × 2 = 0 + 0.000 000 094 994 902 183 68;
  • 8) 0.000 000 094 994 902 183 68 × 2 = 0 + 0.000 000 189 989 804 367 36;
  • 9) 0.000 000 189 989 804 367 36 × 2 = 0 + 0.000 000 379 979 608 734 72;
  • 10) 0.000 000 379 979 608 734 72 × 2 = 0 + 0.000 000 759 959 217 469 44;
  • 11) 0.000 000 759 959 217 469 44 × 2 = 0 + 0.000 001 519 918 434 938 88;
  • 12) 0.000 001 519 918 434 938 88 × 2 = 0 + 0.000 003 039 836 869 877 76;
  • 13) 0.000 003 039 836 869 877 76 × 2 = 0 + 0.000 006 079 673 739 755 52;
  • 14) 0.000 006 079 673 739 755 52 × 2 = 0 + 0.000 012 159 347 479 511 04;
  • 15) 0.000 012 159 347 479 511 04 × 2 = 0 + 0.000 024 318 694 959 022 08;
  • 16) 0.000 024 318 694 959 022 08 × 2 = 0 + 0.000 048 637 389 918 044 16;
  • 17) 0.000 048 637 389 918 044 16 × 2 = 0 + 0.000 097 274 779 836 088 32;
  • 18) 0.000 097 274 779 836 088 32 × 2 = 0 + 0.000 194 549 559 672 176 64;
  • 19) 0.000 194 549 559 672 176 64 × 2 = 0 + 0.000 389 099 119 344 353 28;
  • 20) 0.000 389 099 119 344 353 28 × 2 = 0 + 0.000 778 198 238 688 706 56;
  • 21) 0.000 778 198 238 688 706 56 × 2 = 0 + 0.001 556 396 477 377 413 12;
  • 22) 0.001 556 396 477 377 413 12 × 2 = 0 + 0.003 112 792 954 754 826 24;
  • 23) 0.003 112 792 954 754 826 24 × 2 = 0 + 0.006 225 585 909 509 652 48;
  • 24) 0.006 225 585 909 509 652 48 × 2 = 0 + 0.012 451 171 819 019 304 96;
  • 25) 0.012 451 171 819 019 304 96 × 2 = 0 + 0.024 902 343 638 038 609 92;
  • 26) 0.024 902 343 638 038 609 92 × 2 = 0 + 0.049 804 687 276 077 219 84;
  • 27) 0.049 804 687 276 077 219 84 × 2 = 0 + 0.099 609 374 552 154 439 68;
  • 28) 0.099 609 374 552 154 439 68 × 2 = 0 + 0.199 218 749 104 308 879 36;
  • 29) 0.199 218 749 104 308 879 36 × 2 = 0 + 0.398 437 498 208 617 758 72;
  • 30) 0.398 437 498 208 617 758 72 × 2 = 0 + 0.796 874 996 417 235 517 44;
  • 31) 0.796 874 996 417 235 517 44 × 2 = 1 + 0.593 749 992 834 471 034 88;
  • 32) 0.593 749 992 834 471 034 88 × 2 = 1 + 0.187 499 985 668 942 069 76;
  • 33) 0.187 499 985 668 942 069 76 × 2 = 0 + 0.374 999 971 337 884 139 52;
  • 34) 0.374 999 971 337 884 139 52 × 2 = 0 + 0.749 999 942 675 768 279 04;
  • 35) 0.749 999 942 675 768 279 04 × 2 = 1 + 0.499 999 885 351 536 558 08;
  • 36) 0.499 999 885 351 536 558 08 × 2 = 0 + 0.999 999 770 703 073 116 16;
  • 37) 0.999 999 770 703 073 116 16 × 2 = 1 + 0.999 999 541 406 146 232 32;
  • 38) 0.999 999 541 406 146 232 32 × 2 = 1 + 0.999 999 082 812 292 464 64;
  • 39) 0.999 999 082 812 292 464 64 × 2 = 1 + 0.999 998 165 624 584 929 28;
  • 40) 0.999 998 165 624 584 929 28 × 2 = 1 + 0.999 996 331 249 169 858 56;
  • 41) 0.999 996 331 249 169 858 56 × 2 = 1 + 0.999 992 662 498 339 717 12;
  • 42) 0.999 992 662 498 339 717 12 × 2 = 1 + 0.999 985 324 996 679 434 24;
  • 43) 0.999 985 324 996 679 434 24 × 2 = 1 + 0.999 970 649 993 358 868 48;
  • 44) 0.999 970 649 993 358 868 48 × 2 = 1 + 0.999 941 299 986 717 736 96;
  • 45) 0.999 941 299 986 717 736 96 × 2 = 1 + 0.999 882 599 973 435 473 92;
  • 46) 0.999 882 599 973 435 473 92 × 2 = 1 + 0.999 765 199 946 870 947 84;
  • 47) 0.999 765 199 946 870 947 84 × 2 = 1 + 0.999 530 399 893 741 895 68;
  • 48) 0.999 530 399 893 741 895 68 × 2 = 1 + 0.999 060 799 787 483 791 36;
  • 49) 0.999 060 799 787 483 791 36 × 2 = 1 + 0.998 121 599 574 967 582 72;
  • 50) 0.998 121 599 574 967 582 72 × 2 = 1 + 0.996 243 199 149 935 165 44;
  • 51) 0.996 243 199 149 935 165 44 × 2 = 1 + 0.992 486 398 299 870 330 88;
  • 52) 0.992 486 398 299 870 330 88 × 2 = 1 + 0.984 972 796 599 740 661 76;
  • 53) 0.984 972 796 599 740 661 76 × 2 = 1 + 0.969 945 593 199 481 323 52;
  • 54) 0.969 945 593 199 481 323 52 × 2 = 1 + 0.939 891 186 398 962 647 04;

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:


0.000 000 000 742 147 673 31(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 673 31(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:


0.000 000 000 742 147 673 31(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 -0.000 000 000 742 147 673 31 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 673 66 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