-0.000 000 000 742 147 676 194 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 194(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
-0.000 000 000 742 147 676 194(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:

|-0.000 000 000 742 147 676 194| = 0.000 000 000 742 147 676 194


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: 0.000 000 000 742 147 676 194.

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) 0.000 000 000 742 147 676 194 × 2 = 0 + 0.000 000 001 484 295 352 388;
  • 2) 0.000 000 001 484 295 352 388 × 2 = 0 + 0.000 000 002 968 590 704 776;
  • 3) 0.000 000 002 968 590 704 776 × 2 = 0 + 0.000 000 005 937 181 409 552;
  • 4) 0.000 000 005 937 181 409 552 × 2 = 0 + 0.000 000 011 874 362 819 104;
  • 5) 0.000 000 011 874 362 819 104 × 2 = 0 + 0.000 000 023 748 725 638 208;
  • 6) 0.000 000 023 748 725 638 208 × 2 = 0 + 0.000 000 047 497 451 276 416;
  • 7) 0.000 000 047 497 451 276 416 × 2 = 0 + 0.000 000 094 994 902 552 832;
  • 8) 0.000 000 094 994 902 552 832 × 2 = 0 + 0.000 000 189 989 805 105 664;
  • 9) 0.000 000 189 989 805 105 664 × 2 = 0 + 0.000 000 379 979 610 211 328;
  • 10) 0.000 000 379 979 610 211 328 × 2 = 0 + 0.000 000 759 959 220 422 656;
  • 11) 0.000 000 759 959 220 422 656 × 2 = 0 + 0.000 001 519 918 440 845 312;
  • 12) 0.000 001 519 918 440 845 312 × 2 = 0 + 0.000 003 039 836 881 690 624;
  • 13) 0.000 003 039 836 881 690 624 × 2 = 0 + 0.000 006 079 673 763 381 248;
  • 14) 0.000 006 079 673 763 381 248 × 2 = 0 + 0.000 012 159 347 526 762 496;
  • 15) 0.000 012 159 347 526 762 496 × 2 = 0 + 0.000 024 318 695 053 524 992;
  • 16) 0.000 024 318 695 053 524 992 × 2 = 0 + 0.000 048 637 390 107 049 984;
  • 17) 0.000 048 637 390 107 049 984 × 2 = 0 + 0.000 097 274 780 214 099 968;
  • 18) 0.000 097 274 780 214 099 968 × 2 = 0 + 0.000 194 549 560 428 199 936;
  • 19) 0.000 194 549 560 428 199 936 × 2 = 0 + 0.000 389 099 120 856 399 872;
  • 20) 0.000 389 099 120 856 399 872 × 2 = 0 + 0.000 778 198 241 712 799 744;
  • 21) 0.000 778 198 241 712 799 744 × 2 = 0 + 0.001 556 396 483 425 599 488;
  • 22) 0.001 556 396 483 425 599 488 × 2 = 0 + 0.003 112 792 966 851 198 976;
  • 23) 0.003 112 792 966 851 198 976 × 2 = 0 + 0.006 225 585 933 702 397 952;
  • 24) 0.006 225 585 933 702 397 952 × 2 = 0 + 0.012 451 171 867 404 795 904;
  • 25) 0.012 451 171 867 404 795 904 × 2 = 0 + 0.024 902 343 734 809 591 808;
  • 26) 0.024 902 343 734 809 591 808 × 2 = 0 + 0.049 804 687 469 619 183 616;
  • 27) 0.049 804 687 469 619 183 616 × 2 = 0 + 0.099 609 374 939 238 367 232;
  • 28) 0.099 609 374 939 238 367 232 × 2 = 0 + 0.199 218 749 878 476 734 464;
  • 29) 0.199 218 749 878 476 734 464 × 2 = 0 + 0.398 437 499 756 953 468 928;
  • 30) 0.398 437 499 756 953 468 928 × 2 = 0 + 0.796 874 999 513 906 937 856;
  • 31) 0.796 874 999 513 906 937 856 × 2 = 1 + 0.593 749 999 027 813 875 712;
  • 32) 0.593 749 999 027 813 875 712 × 2 = 1 + 0.187 499 998 055 627 751 424;
  • 33) 0.187 499 998 055 627 751 424 × 2 = 0 + 0.374 999 996 111 255 502 848;
  • 34) 0.374 999 996 111 255 502 848 × 2 = 0 + 0.749 999 992 222 511 005 696;
  • 35) 0.749 999 992 222 511 005 696 × 2 = 1 + 0.499 999 984 445 022 011 392;
  • 36) 0.499 999 984 445 022 011 392 × 2 = 0 + 0.999 999 968 890 044 022 784;
  • 37) 0.999 999 968 890 044 022 784 × 2 = 1 + 0.999 999 937 780 088 045 568;
  • 38) 0.999 999 937 780 088 045 568 × 2 = 1 + 0.999 999 875 560 176 091 136;
  • 39) 0.999 999 875 560 176 091 136 × 2 = 1 + 0.999 999 751 120 352 182 272;
  • 40) 0.999 999 751 120 352 182 272 × 2 = 1 + 0.999 999 502 240 704 364 544;
  • 41) 0.999 999 502 240 704 364 544 × 2 = 1 + 0.999 999 004 481 408 729 088;
  • 42) 0.999 999 004 481 408 729 088 × 2 = 1 + 0.999 998 008 962 817 458 176;
  • 43) 0.999 998 008 962 817 458 176 × 2 = 1 + 0.999 996 017 925 634 916 352;
  • 44) 0.999 996 017 925 634 916 352 × 2 = 1 + 0.999 992 035 851 269 832 704;
  • 45) 0.999 992 035 851 269 832 704 × 2 = 1 + 0.999 984 071 702 539 665 408;
  • 46) 0.999 984 071 702 539 665 408 × 2 = 1 + 0.999 968 143 405 079 330 816;
  • 47) 0.999 968 143 405 079 330 816 × 2 = 1 + 0.999 936 286 810 158 661 632;
  • 48) 0.999 936 286 810 158 661 632 × 2 = 1 + 0.999 872 573 620 317 323 264;
  • 49) 0.999 872 573 620 317 323 264 × 2 = 1 + 0.999 745 147 240 634 646 528;
  • 50) 0.999 745 147 240 634 646 528 × 2 = 1 + 0.999 490 294 481 269 293 056;
  • 51) 0.999 490 294 481 269 293 056 × 2 = 1 + 0.998 980 588 962 538 586 112;
  • 52) 0.998 980 588 962 538 586 112 × 2 = 1 + 0.997 961 177 925 077 172 224;
  • 53) 0.997 961 177 925 077 172 224 × 2 = 1 + 0.995 922 355 850 154 344 448;
  • 54) 0.995 922 355 850 154 344 448 × 2 = 1 + 0.991 844 711 700 308 688 896;

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:


0.000 000 000 742 147 676 194(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 194(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:


0.000 000 000 742 147 676 194(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 -0.000 000 000 742 147 676 194 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