-0.000 000 000 742 147 687 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 687(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
-0.000 000 000 742 147 687(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:

|-0.000 000 000 742 147 687| = 0.000 000 000 742 147 687


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: 0.000 000 000 742 147 687.

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) 0.000 000 000 742 147 687 × 2 = 0 + 0.000 000 001 484 295 374;
  • 2) 0.000 000 001 484 295 374 × 2 = 0 + 0.000 000 002 968 590 748;
  • 3) 0.000 000 002 968 590 748 × 2 = 0 + 0.000 000 005 937 181 496;
  • 4) 0.000 000 005 937 181 496 × 2 = 0 + 0.000 000 011 874 362 992;
  • 5) 0.000 000 011 874 362 992 × 2 = 0 + 0.000 000 023 748 725 984;
  • 6) 0.000 000 023 748 725 984 × 2 = 0 + 0.000 000 047 497 451 968;
  • 7) 0.000 000 047 497 451 968 × 2 = 0 + 0.000 000 094 994 903 936;
  • 8) 0.000 000 094 994 903 936 × 2 = 0 + 0.000 000 189 989 807 872;
  • 9) 0.000 000 189 989 807 872 × 2 = 0 + 0.000 000 379 979 615 744;
  • 10) 0.000 000 379 979 615 744 × 2 = 0 + 0.000 000 759 959 231 488;
  • 11) 0.000 000 759 959 231 488 × 2 = 0 + 0.000 001 519 918 462 976;
  • 12) 0.000 001 519 918 462 976 × 2 = 0 + 0.000 003 039 836 925 952;
  • 13) 0.000 003 039 836 925 952 × 2 = 0 + 0.000 006 079 673 851 904;
  • 14) 0.000 006 079 673 851 904 × 2 = 0 + 0.000 012 159 347 703 808;
  • 15) 0.000 012 159 347 703 808 × 2 = 0 + 0.000 024 318 695 407 616;
  • 16) 0.000 024 318 695 407 616 × 2 = 0 + 0.000 048 637 390 815 232;
  • 17) 0.000 048 637 390 815 232 × 2 = 0 + 0.000 097 274 781 630 464;
  • 18) 0.000 097 274 781 630 464 × 2 = 0 + 0.000 194 549 563 260 928;
  • 19) 0.000 194 549 563 260 928 × 2 = 0 + 0.000 389 099 126 521 856;
  • 20) 0.000 389 099 126 521 856 × 2 = 0 + 0.000 778 198 253 043 712;
  • 21) 0.000 778 198 253 043 712 × 2 = 0 + 0.001 556 396 506 087 424;
  • 22) 0.001 556 396 506 087 424 × 2 = 0 + 0.003 112 793 012 174 848;
  • 23) 0.003 112 793 012 174 848 × 2 = 0 + 0.006 225 586 024 349 696;
  • 24) 0.006 225 586 024 349 696 × 2 = 0 + 0.012 451 172 048 699 392;
  • 25) 0.012 451 172 048 699 392 × 2 = 0 + 0.024 902 344 097 398 784;
  • 26) 0.024 902 344 097 398 784 × 2 = 0 + 0.049 804 688 194 797 568;
  • 27) 0.049 804 688 194 797 568 × 2 = 0 + 0.099 609 376 389 595 136;
  • 28) 0.099 609 376 389 595 136 × 2 = 0 + 0.199 218 752 779 190 272;
  • 29) 0.199 218 752 779 190 272 × 2 = 0 + 0.398 437 505 558 380 544;
  • 30) 0.398 437 505 558 380 544 × 2 = 0 + 0.796 875 011 116 761 088;
  • 31) 0.796 875 011 116 761 088 × 2 = 1 + 0.593 750 022 233 522 176;
  • 32) 0.593 750 022 233 522 176 × 2 = 1 + 0.187 500 044 467 044 352;
  • 33) 0.187 500 044 467 044 352 × 2 = 0 + 0.375 000 088 934 088 704;
  • 34) 0.375 000 088 934 088 704 × 2 = 0 + 0.750 000 177 868 177 408;
  • 35) 0.750 000 177 868 177 408 × 2 = 1 + 0.500 000 355 736 354 816;
  • 36) 0.500 000 355 736 354 816 × 2 = 1 + 0.000 000 711 472 709 632;
  • 37) 0.000 000 711 472 709 632 × 2 = 0 + 0.000 001 422 945 419 264;
  • 38) 0.000 001 422 945 419 264 × 2 = 0 + 0.000 002 845 890 838 528;
  • 39) 0.000 002 845 890 838 528 × 2 = 0 + 0.000 005 691 781 677 056;
  • 40) 0.000 005 691 781 677 056 × 2 = 0 + 0.000 011 383 563 354 112;
  • 41) 0.000 011 383 563 354 112 × 2 = 0 + 0.000 022 767 126 708 224;
  • 42) 0.000 022 767 126 708 224 × 2 = 0 + 0.000 045 534 253 416 448;
  • 43) 0.000 045 534 253 416 448 × 2 = 0 + 0.000 091 068 506 832 896;
  • 44) 0.000 091 068 506 832 896 × 2 = 0 + 0.000 182 137 013 665 792;
  • 45) 0.000 182 137 013 665 792 × 2 = 0 + 0.000 364 274 027 331 584;
  • 46) 0.000 364 274 027 331 584 × 2 = 0 + 0.000 728 548 054 663 168;
  • 47) 0.000 728 548 054 663 168 × 2 = 0 + 0.001 457 096 109 326 336;
  • 48) 0.001 457 096 109 326 336 × 2 = 0 + 0.002 914 192 218 652 672;
  • 49) 0.002 914 192 218 652 672 × 2 = 0 + 0.005 828 384 437 305 344;
  • 50) 0.005 828 384 437 305 344 × 2 = 0 + 0.011 656 768 874 610 688;
  • 51) 0.011 656 768 874 610 688 × 2 = 0 + 0.023 313 537 749 221 376;
  • 52) 0.023 313 537 749 221 376 × 2 = 0 + 0.046 627 075 498 442 752;
  • 53) 0.046 627 075 498 442 752 × 2 = 0 + 0.093 254 150 996 885 504;
  • 54) 0.093 254 150 996 885 504 × 2 = 0 + 0.186 508 301 993 771 008;

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:


0.000 000 000 742 147 687(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 687(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:


0.000 000 000 742 147 687(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 -0.000 000 000 742 147 687 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