-0.000 000 000 742 147 679 2 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 679 2(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
-0.000 000 000 742 147 679 2(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:

|-0.000 000 000 742 147 679 2| = 0.000 000 000 742 147 679 2


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: 0.000 000 000 742 147 679 2.

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) 0.000 000 000 742 147 679 2 × 2 = 0 + 0.000 000 001 484 295 358 4;
  • 2) 0.000 000 001 484 295 358 4 × 2 = 0 + 0.000 000 002 968 590 716 8;
  • 3) 0.000 000 002 968 590 716 8 × 2 = 0 + 0.000 000 005 937 181 433 6;
  • 4) 0.000 000 005 937 181 433 6 × 2 = 0 + 0.000 000 011 874 362 867 2;
  • 5) 0.000 000 011 874 362 867 2 × 2 = 0 + 0.000 000 023 748 725 734 4;
  • 6) 0.000 000 023 748 725 734 4 × 2 = 0 + 0.000 000 047 497 451 468 8;
  • 7) 0.000 000 047 497 451 468 8 × 2 = 0 + 0.000 000 094 994 902 937 6;
  • 8) 0.000 000 094 994 902 937 6 × 2 = 0 + 0.000 000 189 989 805 875 2;
  • 9) 0.000 000 189 989 805 875 2 × 2 = 0 + 0.000 000 379 979 611 750 4;
  • 10) 0.000 000 379 979 611 750 4 × 2 = 0 + 0.000 000 759 959 223 500 8;
  • 11) 0.000 000 759 959 223 500 8 × 2 = 0 + 0.000 001 519 918 447 001 6;
  • 12) 0.000 001 519 918 447 001 6 × 2 = 0 + 0.000 003 039 836 894 003 2;
  • 13) 0.000 003 039 836 894 003 2 × 2 = 0 + 0.000 006 079 673 788 006 4;
  • 14) 0.000 006 079 673 788 006 4 × 2 = 0 + 0.000 012 159 347 576 012 8;
  • 15) 0.000 012 159 347 576 012 8 × 2 = 0 + 0.000 024 318 695 152 025 6;
  • 16) 0.000 024 318 695 152 025 6 × 2 = 0 + 0.000 048 637 390 304 051 2;
  • 17) 0.000 048 637 390 304 051 2 × 2 = 0 + 0.000 097 274 780 608 102 4;
  • 18) 0.000 097 274 780 608 102 4 × 2 = 0 + 0.000 194 549 561 216 204 8;
  • 19) 0.000 194 549 561 216 204 8 × 2 = 0 + 0.000 389 099 122 432 409 6;
  • 20) 0.000 389 099 122 432 409 6 × 2 = 0 + 0.000 778 198 244 864 819 2;
  • 21) 0.000 778 198 244 864 819 2 × 2 = 0 + 0.001 556 396 489 729 638 4;
  • 22) 0.001 556 396 489 729 638 4 × 2 = 0 + 0.003 112 792 979 459 276 8;
  • 23) 0.003 112 792 979 459 276 8 × 2 = 0 + 0.006 225 585 958 918 553 6;
  • 24) 0.006 225 585 958 918 553 6 × 2 = 0 + 0.012 451 171 917 837 107 2;
  • 25) 0.012 451 171 917 837 107 2 × 2 = 0 + 0.024 902 343 835 674 214 4;
  • 26) 0.024 902 343 835 674 214 4 × 2 = 0 + 0.049 804 687 671 348 428 8;
  • 27) 0.049 804 687 671 348 428 8 × 2 = 0 + 0.099 609 375 342 696 857 6;
  • 28) 0.099 609 375 342 696 857 6 × 2 = 0 + 0.199 218 750 685 393 715 2;
  • 29) 0.199 218 750 685 393 715 2 × 2 = 0 + 0.398 437 501 370 787 430 4;
  • 30) 0.398 437 501 370 787 430 4 × 2 = 0 + 0.796 875 002 741 574 860 8;
  • 31) 0.796 875 002 741 574 860 8 × 2 = 1 + 0.593 750 005 483 149 721 6;
  • 32) 0.593 750 005 483 149 721 6 × 2 = 1 + 0.187 500 010 966 299 443 2;
  • 33) 0.187 500 010 966 299 443 2 × 2 = 0 + 0.375 000 021 932 598 886 4;
  • 34) 0.375 000 021 932 598 886 4 × 2 = 0 + 0.750 000 043 865 197 772 8;
  • 35) 0.750 000 043 865 197 772 8 × 2 = 1 + 0.500 000 087 730 395 545 6;
  • 36) 0.500 000 087 730 395 545 6 × 2 = 1 + 0.000 000 175 460 791 091 2;
  • 37) 0.000 000 175 460 791 091 2 × 2 = 0 + 0.000 000 350 921 582 182 4;
  • 38) 0.000 000 350 921 582 182 4 × 2 = 0 + 0.000 000 701 843 164 364 8;
  • 39) 0.000 000 701 843 164 364 8 × 2 = 0 + 0.000 001 403 686 328 729 6;
  • 40) 0.000 001 403 686 328 729 6 × 2 = 0 + 0.000 002 807 372 657 459 2;
  • 41) 0.000 002 807 372 657 459 2 × 2 = 0 + 0.000 005 614 745 314 918 4;
  • 42) 0.000 005 614 745 314 918 4 × 2 = 0 + 0.000 011 229 490 629 836 8;
  • 43) 0.000 011 229 490 629 836 8 × 2 = 0 + 0.000 022 458 981 259 673 6;
  • 44) 0.000 022 458 981 259 673 6 × 2 = 0 + 0.000 044 917 962 519 347 2;
  • 45) 0.000 044 917 962 519 347 2 × 2 = 0 + 0.000 089 835 925 038 694 4;
  • 46) 0.000 089 835 925 038 694 4 × 2 = 0 + 0.000 179 671 850 077 388 8;
  • 47) 0.000 179 671 850 077 388 8 × 2 = 0 + 0.000 359 343 700 154 777 6;
  • 48) 0.000 359 343 700 154 777 6 × 2 = 0 + 0.000 718 687 400 309 555 2;
  • 49) 0.000 718 687 400 309 555 2 × 2 = 0 + 0.001 437 374 800 619 110 4;
  • 50) 0.001 437 374 800 619 110 4 × 2 = 0 + 0.002 874 749 601 238 220 8;
  • 51) 0.002 874 749 601 238 220 8 × 2 = 0 + 0.005 749 499 202 476 441 6;
  • 52) 0.005 749 499 202 476 441 6 × 2 = 0 + 0.011 498 998 404 952 883 2;
  • 53) 0.011 498 998 404 952 883 2 × 2 = 0 + 0.022 997 996 809 905 766 4;
  • 54) 0.022 997 996 809 905 766 4 × 2 = 0 + 0.045 995 993 619 811 532 8;

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:


0.000 000 000 742 147 679 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 679 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 679 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 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 0000

Mantissa (23 bits) =
100 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 679 2 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

Share

» Convert decimal -0.000 000 000 742 147 672 7 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