-0.000 000 000 742 131 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 131 5(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
-0.000 000 000 742 131 5(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:

|-0.000 000 000 742 131 5| = 0.000 000 000 742 131 5


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: 0.000 000 000 742 131 5.

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) 0.000 000 000 742 131 5 × 2 = 0 + 0.000 000 001 484 263;
  • 2) 0.000 000 001 484 263 × 2 = 0 + 0.000 000 002 968 526;
  • 3) 0.000 000 002 968 526 × 2 = 0 + 0.000 000 005 937 052;
  • 4) 0.000 000 005 937 052 × 2 = 0 + 0.000 000 011 874 104;
  • 5) 0.000 000 011 874 104 × 2 = 0 + 0.000 000 023 748 208;
  • 6) 0.000 000 023 748 208 × 2 = 0 + 0.000 000 047 496 416;
  • 7) 0.000 000 047 496 416 × 2 = 0 + 0.000 000 094 992 832;
  • 8) 0.000 000 094 992 832 × 2 = 0 + 0.000 000 189 985 664;
  • 9) 0.000 000 189 985 664 × 2 = 0 + 0.000 000 379 971 328;
  • 10) 0.000 000 379 971 328 × 2 = 0 + 0.000 000 759 942 656;
  • 11) 0.000 000 759 942 656 × 2 = 0 + 0.000 001 519 885 312;
  • 12) 0.000 001 519 885 312 × 2 = 0 + 0.000 003 039 770 624;
  • 13) 0.000 003 039 770 624 × 2 = 0 + 0.000 006 079 541 248;
  • 14) 0.000 006 079 541 248 × 2 = 0 + 0.000 012 159 082 496;
  • 15) 0.000 012 159 082 496 × 2 = 0 + 0.000 024 318 164 992;
  • 16) 0.000 024 318 164 992 × 2 = 0 + 0.000 048 636 329 984;
  • 17) 0.000 048 636 329 984 × 2 = 0 + 0.000 097 272 659 968;
  • 18) 0.000 097 272 659 968 × 2 = 0 + 0.000 194 545 319 936;
  • 19) 0.000 194 545 319 936 × 2 = 0 + 0.000 389 090 639 872;
  • 20) 0.000 389 090 639 872 × 2 = 0 + 0.000 778 181 279 744;
  • 21) 0.000 778 181 279 744 × 2 = 0 + 0.001 556 362 559 488;
  • 22) 0.001 556 362 559 488 × 2 = 0 + 0.003 112 725 118 976;
  • 23) 0.003 112 725 118 976 × 2 = 0 + 0.006 225 450 237 952;
  • 24) 0.006 225 450 237 952 × 2 = 0 + 0.012 450 900 475 904;
  • 25) 0.012 450 900 475 904 × 2 = 0 + 0.024 901 800 951 808;
  • 26) 0.024 901 800 951 808 × 2 = 0 + 0.049 803 601 903 616;
  • 27) 0.049 803 601 903 616 × 2 = 0 + 0.099 607 203 807 232;
  • 28) 0.099 607 203 807 232 × 2 = 0 + 0.199 214 407 614 464;
  • 29) 0.199 214 407 614 464 × 2 = 0 + 0.398 428 815 228 928;
  • 30) 0.398 428 815 228 928 × 2 = 0 + 0.796 857 630 457 856;
  • 31) 0.796 857 630 457 856 × 2 = 1 + 0.593 715 260 915 712;
  • 32) 0.593 715 260 915 712 × 2 = 1 + 0.187 430 521 831 424;
  • 33) 0.187 430 521 831 424 × 2 = 0 + 0.374 861 043 662 848;
  • 34) 0.374 861 043 662 848 × 2 = 0 + 0.749 722 087 325 696;
  • 35) 0.749 722 087 325 696 × 2 = 1 + 0.499 444 174 651 392;
  • 36) 0.499 444 174 651 392 × 2 = 0 + 0.998 888 349 302 784;
  • 37) 0.998 888 349 302 784 × 2 = 1 + 0.997 776 698 605 568;
  • 38) 0.997 776 698 605 568 × 2 = 1 + 0.995 553 397 211 136;
  • 39) 0.995 553 397 211 136 × 2 = 1 + 0.991 106 794 422 272;
  • 40) 0.991 106 794 422 272 × 2 = 1 + 0.982 213 588 844 544;
  • 41) 0.982 213 588 844 544 × 2 = 1 + 0.964 427 177 689 088;
  • 42) 0.964 427 177 689 088 × 2 = 1 + 0.928 854 355 378 176;
  • 43) 0.928 854 355 378 176 × 2 = 1 + 0.857 708 710 756 352;
  • 44) 0.857 708 710 756 352 × 2 = 1 + 0.715 417 421 512 704;
  • 45) 0.715 417 421 512 704 × 2 = 1 + 0.430 834 843 025 408;
  • 46) 0.430 834 843 025 408 × 2 = 0 + 0.861 669 686 050 816;
  • 47) 0.861 669 686 050 816 × 2 = 1 + 0.723 339 372 101 632;
  • 48) 0.723 339 372 101 632 × 2 = 1 + 0.446 678 744 203 264;
  • 49) 0.446 678 744 203 264 × 2 = 0 + 0.893 357 488 406 528;
  • 50) 0.893 357 488 406 528 × 2 = 1 + 0.786 714 976 813 056;
  • 51) 0.786 714 976 813 056 × 2 = 1 + 0.573 429 953 626 112;
  • 52) 0.573 429 953 626 112 × 2 = 1 + 0.146 859 907 252 224;
  • 53) 0.146 859 907 252 224 × 2 = 0 + 0.293 719 814 504 448;
  • 54) 0.293 719 814 504 448 × 2 = 0 + 0.587 439 629 008 896;

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:


0.000 000 000 742 131 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 0111 00(2)

6. Positive number before normalization:

0.000 000 000 742 131 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 0111 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:


0.000 000 000 742 131 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 0111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1011 0111 00(2) × 20 =

1.1001 0111 1111 1101 1011 100(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 1101 1011 100


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 1110 1101 1100 =

100 1011 1111 1110 1101 1100


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 1110 1101 1100


Decimal number -0.000 000 000 742 131 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1110 1101 1100

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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