-0.000 000 000 742 147 676 659 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 659 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
-0.000 000 000 742 147 676 659 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:

|-0.000 000 000 742 147 676 659 5| = 0.000 000 000 742 147 676 659 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: 0.000 000 000 742 147 676 659 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) 0.000 000 000 742 147 676 659 5 × 2 = 0 + 0.000 000 001 484 295 353 319;
  • 2) 0.000 000 001 484 295 353 319 × 2 = 0 + 0.000 000 002 968 590 706 638;
  • 3) 0.000 000 002 968 590 706 638 × 2 = 0 + 0.000 000 005 937 181 413 276;
  • 4) 0.000 000 005 937 181 413 276 × 2 = 0 + 0.000 000 011 874 362 826 552;
  • 5) 0.000 000 011 874 362 826 552 × 2 = 0 + 0.000 000 023 748 725 653 104;
  • 6) 0.000 000 023 748 725 653 104 × 2 = 0 + 0.000 000 047 497 451 306 208;
  • 7) 0.000 000 047 497 451 306 208 × 2 = 0 + 0.000 000 094 994 902 612 416;
  • 8) 0.000 000 094 994 902 612 416 × 2 = 0 + 0.000 000 189 989 805 224 832;
  • 9) 0.000 000 189 989 805 224 832 × 2 = 0 + 0.000 000 379 979 610 449 664;
  • 10) 0.000 000 379 979 610 449 664 × 2 = 0 + 0.000 000 759 959 220 899 328;
  • 11) 0.000 000 759 959 220 899 328 × 2 = 0 + 0.000 001 519 918 441 798 656;
  • 12) 0.000 001 519 918 441 798 656 × 2 = 0 + 0.000 003 039 836 883 597 312;
  • 13) 0.000 003 039 836 883 597 312 × 2 = 0 + 0.000 006 079 673 767 194 624;
  • 14) 0.000 006 079 673 767 194 624 × 2 = 0 + 0.000 012 159 347 534 389 248;
  • 15) 0.000 012 159 347 534 389 248 × 2 = 0 + 0.000 024 318 695 068 778 496;
  • 16) 0.000 024 318 695 068 778 496 × 2 = 0 + 0.000 048 637 390 137 556 992;
  • 17) 0.000 048 637 390 137 556 992 × 2 = 0 + 0.000 097 274 780 275 113 984;
  • 18) 0.000 097 274 780 275 113 984 × 2 = 0 + 0.000 194 549 560 550 227 968;
  • 19) 0.000 194 549 560 550 227 968 × 2 = 0 + 0.000 389 099 121 100 455 936;
  • 20) 0.000 389 099 121 100 455 936 × 2 = 0 + 0.000 778 198 242 200 911 872;
  • 21) 0.000 778 198 242 200 911 872 × 2 = 0 + 0.001 556 396 484 401 823 744;
  • 22) 0.001 556 396 484 401 823 744 × 2 = 0 + 0.003 112 792 968 803 647 488;
  • 23) 0.003 112 792 968 803 647 488 × 2 = 0 + 0.006 225 585 937 607 294 976;
  • 24) 0.006 225 585 937 607 294 976 × 2 = 0 + 0.012 451 171 875 214 589 952;
  • 25) 0.012 451 171 875 214 589 952 × 2 = 0 + 0.024 902 343 750 429 179 904;
  • 26) 0.024 902 343 750 429 179 904 × 2 = 0 + 0.049 804 687 500 858 359 808;
  • 27) 0.049 804 687 500 858 359 808 × 2 = 0 + 0.099 609 375 001 716 719 616;
  • 28) 0.099 609 375 001 716 719 616 × 2 = 0 + 0.199 218 750 003 433 439 232;
  • 29) 0.199 218 750 003 433 439 232 × 2 = 0 + 0.398 437 500 006 866 878 464;
  • 30) 0.398 437 500 006 866 878 464 × 2 = 0 + 0.796 875 000 013 733 756 928;
  • 31) 0.796 875 000 013 733 756 928 × 2 = 1 + 0.593 750 000 027 467 513 856;
  • 32) 0.593 750 000 027 467 513 856 × 2 = 1 + 0.187 500 000 054 935 027 712;
  • 33) 0.187 500 000 054 935 027 712 × 2 = 0 + 0.375 000 000 109 870 055 424;
  • 34) 0.375 000 000 109 870 055 424 × 2 = 0 + 0.750 000 000 219 740 110 848;
  • 35) 0.750 000 000 219 740 110 848 × 2 = 1 + 0.500 000 000 439 480 221 696;
  • 36) 0.500 000 000 439 480 221 696 × 2 = 1 + 0.000 000 000 878 960 443 392;
  • 37) 0.000 000 000 878 960 443 392 × 2 = 0 + 0.000 000 001 757 920 886 784;
  • 38) 0.000 000 001 757 920 886 784 × 2 = 0 + 0.000 000 003 515 841 773 568;
  • 39) 0.000 000 003 515 841 773 568 × 2 = 0 + 0.000 000 007 031 683 547 136;
  • 40) 0.000 000 007 031 683 547 136 × 2 = 0 + 0.000 000 014 063 367 094 272;
  • 41) 0.000 000 014 063 367 094 272 × 2 = 0 + 0.000 000 028 126 734 188 544;
  • 42) 0.000 000 028 126 734 188 544 × 2 = 0 + 0.000 000 056 253 468 377 088;
  • 43) 0.000 000 056 253 468 377 088 × 2 = 0 + 0.000 000 112 506 936 754 176;
  • 44) 0.000 000 112 506 936 754 176 × 2 = 0 + 0.000 000 225 013 873 508 352;
  • 45) 0.000 000 225 013 873 508 352 × 2 = 0 + 0.000 000 450 027 747 016 704;
  • 46) 0.000 000 450 027 747 016 704 × 2 = 0 + 0.000 000 900 055 494 033 408;
  • 47) 0.000 000 900 055 494 033 408 × 2 = 0 + 0.000 001 800 110 988 066 816;
  • 48) 0.000 001 800 110 988 066 816 × 2 = 0 + 0.000 003 600 221 976 133 632;
  • 49) 0.000 003 600 221 976 133 632 × 2 = 0 + 0.000 007 200 443 952 267 264;
  • 50) 0.000 007 200 443 952 267 264 × 2 = 0 + 0.000 014 400 887 904 534 528;
  • 51) 0.000 014 400 887 904 534 528 × 2 = 0 + 0.000 028 801 775 809 069 056;
  • 52) 0.000 028 801 775 809 069 056 × 2 = 0 + 0.000 057 603 551 618 138 112;
  • 53) 0.000 057 603 551 618 138 112 × 2 = 0 + 0.000 115 207 103 236 276 224;
  • 54) 0.000 115 207 103 236 276 224 × 2 = 0 + 0.000 230 414 206 472 552 448;

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:


0.000 000 000 742 147 676 659 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 659 5(10) =

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


0.000 000 000 742 147 676 659 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 000


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 0000 =

100 1100 0000 0000 0000 0000


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 0000


Decimal number -0.000 000 000 742 147 676 659 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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