-0.000 000 000 742 147 894 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 894(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
-0.000 000 000 742 147 894(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:

|-0.000 000 000 742 147 894| = 0.000 000 000 742 147 894


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: 0.000 000 000 742 147 894.

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) 0.000 000 000 742 147 894 × 2 = 0 + 0.000 000 001 484 295 788;
  • 2) 0.000 000 001 484 295 788 × 2 = 0 + 0.000 000 002 968 591 576;
  • 3) 0.000 000 002 968 591 576 × 2 = 0 + 0.000 000 005 937 183 152;
  • 4) 0.000 000 005 937 183 152 × 2 = 0 + 0.000 000 011 874 366 304;
  • 5) 0.000 000 011 874 366 304 × 2 = 0 + 0.000 000 023 748 732 608;
  • 6) 0.000 000 023 748 732 608 × 2 = 0 + 0.000 000 047 497 465 216;
  • 7) 0.000 000 047 497 465 216 × 2 = 0 + 0.000 000 094 994 930 432;
  • 8) 0.000 000 094 994 930 432 × 2 = 0 + 0.000 000 189 989 860 864;
  • 9) 0.000 000 189 989 860 864 × 2 = 0 + 0.000 000 379 979 721 728;
  • 10) 0.000 000 379 979 721 728 × 2 = 0 + 0.000 000 759 959 443 456;
  • 11) 0.000 000 759 959 443 456 × 2 = 0 + 0.000 001 519 918 886 912;
  • 12) 0.000 001 519 918 886 912 × 2 = 0 + 0.000 003 039 837 773 824;
  • 13) 0.000 003 039 837 773 824 × 2 = 0 + 0.000 006 079 675 547 648;
  • 14) 0.000 006 079 675 547 648 × 2 = 0 + 0.000 012 159 351 095 296;
  • 15) 0.000 012 159 351 095 296 × 2 = 0 + 0.000 024 318 702 190 592;
  • 16) 0.000 024 318 702 190 592 × 2 = 0 + 0.000 048 637 404 381 184;
  • 17) 0.000 048 637 404 381 184 × 2 = 0 + 0.000 097 274 808 762 368;
  • 18) 0.000 097 274 808 762 368 × 2 = 0 + 0.000 194 549 617 524 736;
  • 19) 0.000 194 549 617 524 736 × 2 = 0 + 0.000 389 099 235 049 472;
  • 20) 0.000 389 099 235 049 472 × 2 = 0 + 0.000 778 198 470 098 944;
  • 21) 0.000 778 198 470 098 944 × 2 = 0 + 0.001 556 396 940 197 888;
  • 22) 0.001 556 396 940 197 888 × 2 = 0 + 0.003 112 793 880 395 776;
  • 23) 0.003 112 793 880 395 776 × 2 = 0 + 0.006 225 587 760 791 552;
  • 24) 0.006 225 587 760 791 552 × 2 = 0 + 0.012 451 175 521 583 104;
  • 25) 0.012 451 175 521 583 104 × 2 = 0 + 0.024 902 351 043 166 208;
  • 26) 0.024 902 351 043 166 208 × 2 = 0 + 0.049 804 702 086 332 416;
  • 27) 0.049 804 702 086 332 416 × 2 = 0 + 0.099 609 404 172 664 832;
  • 28) 0.099 609 404 172 664 832 × 2 = 0 + 0.199 218 808 345 329 664;
  • 29) 0.199 218 808 345 329 664 × 2 = 0 + 0.398 437 616 690 659 328;
  • 30) 0.398 437 616 690 659 328 × 2 = 0 + 0.796 875 233 381 318 656;
  • 31) 0.796 875 233 381 318 656 × 2 = 1 + 0.593 750 466 762 637 312;
  • 32) 0.593 750 466 762 637 312 × 2 = 1 + 0.187 500 933 525 274 624;
  • 33) 0.187 500 933 525 274 624 × 2 = 0 + 0.375 001 867 050 549 248;
  • 34) 0.375 001 867 050 549 248 × 2 = 0 + 0.750 003 734 101 098 496;
  • 35) 0.750 003 734 101 098 496 × 2 = 1 + 0.500 007 468 202 196 992;
  • 36) 0.500 007 468 202 196 992 × 2 = 1 + 0.000 014 936 404 393 984;
  • 37) 0.000 014 936 404 393 984 × 2 = 0 + 0.000 029 872 808 787 968;
  • 38) 0.000 029 872 808 787 968 × 2 = 0 + 0.000 059 745 617 575 936;
  • 39) 0.000 059 745 617 575 936 × 2 = 0 + 0.000 119 491 235 151 872;
  • 40) 0.000 119 491 235 151 872 × 2 = 0 + 0.000 238 982 470 303 744;
  • 41) 0.000 238 982 470 303 744 × 2 = 0 + 0.000 477 964 940 607 488;
  • 42) 0.000 477 964 940 607 488 × 2 = 0 + 0.000 955 929 881 214 976;
  • 43) 0.000 955 929 881 214 976 × 2 = 0 + 0.001 911 859 762 429 952;
  • 44) 0.001 911 859 762 429 952 × 2 = 0 + 0.003 823 719 524 859 904;
  • 45) 0.003 823 719 524 859 904 × 2 = 0 + 0.007 647 439 049 719 808;
  • 46) 0.007 647 439 049 719 808 × 2 = 0 + 0.015 294 878 099 439 616;
  • 47) 0.015 294 878 099 439 616 × 2 = 0 + 0.030 589 756 198 879 232;
  • 48) 0.030 589 756 198 879 232 × 2 = 0 + 0.061 179 512 397 758 464;
  • 49) 0.061 179 512 397 758 464 × 2 = 0 + 0.122 359 024 795 516 928;
  • 50) 0.122 359 024 795 516 928 × 2 = 0 + 0.244 718 049 591 033 856;
  • 51) 0.244 718 049 591 033 856 × 2 = 0 + 0.489 436 099 182 067 712;
  • 52) 0.489 436 099 182 067 712 × 2 = 0 + 0.978 872 198 364 135 424;
  • 53) 0.978 872 198 364 135 424 × 2 = 1 + 0.957 744 396 728 270 848;
  • 54) 0.957 744 396 728 270 848 × 2 = 1 + 0.915 488 793 456 541 696;

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:


0.000 000 000 742 147 894(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 894(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:


0.000 000 000 742 147 894(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 -0.000 000 000 742 147 894 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