-0.000 000 000 742 147 541 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 541(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
-0.000 000 000 742 147 541(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:

|-0.000 000 000 742 147 541| = 0.000 000 000 742 147 541


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: 0.000 000 000 742 147 541.

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) 0.000 000 000 742 147 541 × 2 = 0 + 0.000 000 001 484 295 082;
  • 2) 0.000 000 001 484 295 082 × 2 = 0 + 0.000 000 002 968 590 164;
  • 3) 0.000 000 002 968 590 164 × 2 = 0 + 0.000 000 005 937 180 328;
  • 4) 0.000 000 005 937 180 328 × 2 = 0 + 0.000 000 011 874 360 656;
  • 5) 0.000 000 011 874 360 656 × 2 = 0 + 0.000 000 023 748 721 312;
  • 6) 0.000 000 023 748 721 312 × 2 = 0 + 0.000 000 047 497 442 624;
  • 7) 0.000 000 047 497 442 624 × 2 = 0 + 0.000 000 094 994 885 248;
  • 8) 0.000 000 094 994 885 248 × 2 = 0 + 0.000 000 189 989 770 496;
  • 9) 0.000 000 189 989 770 496 × 2 = 0 + 0.000 000 379 979 540 992;
  • 10) 0.000 000 379 979 540 992 × 2 = 0 + 0.000 000 759 959 081 984;
  • 11) 0.000 000 759 959 081 984 × 2 = 0 + 0.000 001 519 918 163 968;
  • 12) 0.000 001 519 918 163 968 × 2 = 0 + 0.000 003 039 836 327 936;
  • 13) 0.000 003 039 836 327 936 × 2 = 0 + 0.000 006 079 672 655 872;
  • 14) 0.000 006 079 672 655 872 × 2 = 0 + 0.000 012 159 345 311 744;
  • 15) 0.000 012 159 345 311 744 × 2 = 0 + 0.000 024 318 690 623 488;
  • 16) 0.000 024 318 690 623 488 × 2 = 0 + 0.000 048 637 381 246 976;
  • 17) 0.000 048 637 381 246 976 × 2 = 0 + 0.000 097 274 762 493 952;
  • 18) 0.000 097 274 762 493 952 × 2 = 0 + 0.000 194 549 524 987 904;
  • 19) 0.000 194 549 524 987 904 × 2 = 0 + 0.000 389 099 049 975 808;
  • 20) 0.000 389 099 049 975 808 × 2 = 0 + 0.000 778 198 099 951 616;
  • 21) 0.000 778 198 099 951 616 × 2 = 0 + 0.001 556 396 199 903 232;
  • 22) 0.001 556 396 199 903 232 × 2 = 0 + 0.003 112 792 399 806 464;
  • 23) 0.003 112 792 399 806 464 × 2 = 0 + 0.006 225 584 799 612 928;
  • 24) 0.006 225 584 799 612 928 × 2 = 0 + 0.012 451 169 599 225 856;
  • 25) 0.012 451 169 599 225 856 × 2 = 0 + 0.024 902 339 198 451 712;
  • 26) 0.024 902 339 198 451 712 × 2 = 0 + 0.049 804 678 396 903 424;
  • 27) 0.049 804 678 396 903 424 × 2 = 0 + 0.099 609 356 793 806 848;
  • 28) 0.099 609 356 793 806 848 × 2 = 0 + 0.199 218 713 587 613 696;
  • 29) 0.199 218 713 587 613 696 × 2 = 0 + 0.398 437 427 175 227 392;
  • 30) 0.398 437 427 175 227 392 × 2 = 0 + 0.796 874 854 350 454 784;
  • 31) 0.796 874 854 350 454 784 × 2 = 1 + 0.593 749 708 700 909 568;
  • 32) 0.593 749 708 700 909 568 × 2 = 1 + 0.187 499 417 401 819 136;
  • 33) 0.187 499 417 401 819 136 × 2 = 0 + 0.374 998 834 803 638 272;
  • 34) 0.374 998 834 803 638 272 × 2 = 0 + 0.749 997 669 607 276 544;
  • 35) 0.749 997 669 607 276 544 × 2 = 1 + 0.499 995 339 214 553 088;
  • 36) 0.499 995 339 214 553 088 × 2 = 0 + 0.999 990 678 429 106 176;
  • 37) 0.999 990 678 429 106 176 × 2 = 1 + 0.999 981 356 858 212 352;
  • 38) 0.999 981 356 858 212 352 × 2 = 1 + 0.999 962 713 716 424 704;
  • 39) 0.999 962 713 716 424 704 × 2 = 1 + 0.999 925 427 432 849 408;
  • 40) 0.999 925 427 432 849 408 × 2 = 1 + 0.999 850 854 865 698 816;
  • 41) 0.999 850 854 865 698 816 × 2 = 1 + 0.999 701 709 731 397 632;
  • 42) 0.999 701 709 731 397 632 × 2 = 1 + 0.999 403 419 462 795 264;
  • 43) 0.999 403 419 462 795 264 × 2 = 1 + 0.998 806 838 925 590 528;
  • 44) 0.998 806 838 925 590 528 × 2 = 1 + 0.997 613 677 851 181 056;
  • 45) 0.997 613 677 851 181 056 × 2 = 1 + 0.995 227 355 702 362 112;
  • 46) 0.995 227 355 702 362 112 × 2 = 1 + 0.990 454 711 404 724 224;
  • 47) 0.990 454 711 404 724 224 × 2 = 1 + 0.980 909 422 809 448 448;
  • 48) 0.980 909 422 809 448 448 × 2 = 1 + 0.961 818 845 618 896 896;
  • 49) 0.961 818 845 618 896 896 × 2 = 1 + 0.923 637 691 237 793 792;
  • 50) 0.923 637 691 237 793 792 × 2 = 1 + 0.847 275 382 475 587 584;
  • 51) 0.847 275 382 475 587 584 × 2 = 1 + 0.694 550 764 951 175 168;
  • 52) 0.694 550 764 951 175 168 × 2 = 1 + 0.389 101 529 902 350 336;
  • 53) 0.389 101 529 902 350 336 × 2 = 0 + 0.778 203 059 804 700 672;
  • 54) 0.778 203 059 804 700 672 × 2 = 1 + 0.556 406 119 609 401 344;

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:


0.000 000 000 742 147 541(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 541(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(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:


0.000 000 000 742 147 541(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) × 20 =

1.1001 0111 1111 1111 1111 101(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 101


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 1101 =

100 1011 1111 1111 1111 1101


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 1101


Decimal number -0.000 000 000 742 147 541 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1101

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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