-0.000 000 000 742 147 876 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 876(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
-0.000 000 000 742 147 876(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:

|-0.000 000 000 742 147 876| = 0.000 000 000 742 147 876


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: 0.000 000 000 742 147 876.

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) 0.000 000 000 742 147 876 × 2 = 0 + 0.000 000 001 484 295 752;
  • 2) 0.000 000 001 484 295 752 × 2 = 0 + 0.000 000 002 968 591 504;
  • 3) 0.000 000 002 968 591 504 × 2 = 0 + 0.000 000 005 937 183 008;
  • 4) 0.000 000 005 937 183 008 × 2 = 0 + 0.000 000 011 874 366 016;
  • 5) 0.000 000 011 874 366 016 × 2 = 0 + 0.000 000 023 748 732 032;
  • 6) 0.000 000 023 748 732 032 × 2 = 0 + 0.000 000 047 497 464 064;
  • 7) 0.000 000 047 497 464 064 × 2 = 0 + 0.000 000 094 994 928 128;
  • 8) 0.000 000 094 994 928 128 × 2 = 0 + 0.000 000 189 989 856 256;
  • 9) 0.000 000 189 989 856 256 × 2 = 0 + 0.000 000 379 979 712 512;
  • 10) 0.000 000 379 979 712 512 × 2 = 0 + 0.000 000 759 959 425 024;
  • 11) 0.000 000 759 959 425 024 × 2 = 0 + 0.000 001 519 918 850 048;
  • 12) 0.000 001 519 918 850 048 × 2 = 0 + 0.000 003 039 837 700 096;
  • 13) 0.000 003 039 837 700 096 × 2 = 0 + 0.000 006 079 675 400 192;
  • 14) 0.000 006 079 675 400 192 × 2 = 0 + 0.000 012 159 350 800 384;
  • 15) 0.000 012 159 350 800 384 × 2 = 0 + 0.000 024 318 701 600 768;
  • 16) 0.000 024 318 701 600 768 × 2 = 0 + 0.000 048 637 403 201 536;
  • 17) 0.000 048 637 403 201 536 × 2 = 0 + 0.000 097 274 806 403 072;
  • 18) 0.000 097 274 806 403 072 × 2 = 0 + 0.000 194 549 612 806 144;
  • 19) 0.000 194 549 612 806 144 × 2 = 0 + 0.000 389 099 225 612 288;
  • 20) 0.000 389 099 225 612 288 × 2 = 0 + 0.000 778 198 451 224 576;
  • 21) 0.000 778 198 451 224 576 × 2 = 0 + 0.001 556 396 902 449 152;
  • 22) 0.001 556 396 902 449 152 × 2 = 0 + 0.003 112 793 804 898 304;
  • 23) 0.003 112 793 804 898 304 × 2 = 0 + 0.006 225 587 609 796 608;
  • 24) 0.006 225 587 609 796 608 × 2 = 0 + 0.012 451 175 219 593 216;
  • 25) 0.012 451 175 219 593 216 × 2 = 0 + 0.024 902 350 439 186 432;
  • 26) 0.024 902 350 439 186 432 × 2 = 0 + 0.049 804 700 878 372 864;
  • 27) 0.049 804 700 878 372 864 × 2 = 0 + 0.099 609 401 756 745 728;
  • 28) 0.099 609 401 756 745 728 × 2 = 0 + 0.199 218 803 513 491 456;
  • 29) 0.199 218 803 513 491 456 × 2 = 0 + 0.398 437 607 026 982 912;
  • 30) 0.398 437 607 026 982 912 × 2 = 0 + 0.796 875 214 053 965 824;
  • 31) 0.796 875 214 053 965 824 × 2 = 1 + 0.593 750 428 107 931 648;
  • 32) 0.593 750 428 107 931 648 × 2 = 1 + 0.187 500 856 215 863 296;
  • 33) 0.187 500 856 215 863 296 × 2 = 0 + 0.375 001 712 431 726 592;
  • 34) 0.375 001 712 431 726 592 × 2 = 0 + 0.750 003 424 863 453 184;
  • 35) 0.750 003 424 863 453 184 × 2 = 1 + 0.500 006 849 726 906 368;
  • 36) 0.500 006 849 726 906 368 × 2 = 1 + 0.000 013 699 453 812 736;
  • 37) 0.000 013 699 453 812 736 × 2 = 0 + 0.000 027 398 907 625 472;
  • 38) 0.000 027 398 907 625 472 × 2 = 0 + 0.000 054 797 815 250 944;
  • 39) 0.000 054 797 815 250 944 × 2 = 0 + 0.000 109 595 630 501 888;
  • 40) 0.000 109 595 630 501 888 × 2 = 0 + 0.000 219 191 261 003 776;
  • 41) 0.000 219 191 261 003 776 × 2 = 0 + 0.000 438 382 522 007 552;
  • 42) 0.000 438 382 522 007 552 × 2 = 0 + 0.000 876 765 044 015 104;
  • 43) 0.000 876 765 044 015 104 × 2 = 0 + 0.001 753 530 088 030 208;
  • 44) 0.001 753 530 088 030 208 × 2 = 0 + 0.003 507 060 176 060 416;
  • 45) 0.003 507 060 176 060 416 × 2 = 0 + 0.007 014 120 352 120 832;
  • 46) 0.007 014 120 352 120 832 × 2 = 0 + 0.014 028 240 704 241 664;
  • 47) 0.014 028 240 704 241 664 × 2 = 0 + 0.028 056 481 408 483 328;
  • 48) 0.028 056 481 408 483 328 × 2 = 0 + 0.056 112 962 816 966 656;
  • 49) 0.056 112 962 816 966 656 × 2 = 0 + 0.112 225 925 633 933 312;
  • 50) 0.112 225 925 633 933 312 × 2 = 0 + 0.224 451 851 267 866 624;
  • 51) 0.224 451 851 267 866 624 × 2 = 0 + 0.448 903 702 535 733 248;
  • 52) 0.448 903 702 535 733 248 × 2 = 0 + 0.897 807 405 071 466 496;
  • 53) 0.897 807 405 071 466 496 × 2 = 1 + 0.795 614 810 142 932 992;
  • 54) 0.795 614 810 142 932 992 × 2 = 1 + 0.591 229 620 285 865 984;

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:


0.000 000 000 742 147 876(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 876(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 876(10) =

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

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

1.1001 1000 0000 0000 0000 011(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 011


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 0011 =

100 1100 0000 0000 0000 0011


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 0011


Decimal number -0.000 000 000 742 147 876 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0011

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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