-0.000 000 000 742 147 676 726 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 726(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
-0.000 000 000 742 147 676 726(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:

|-0.000 000 000 742 147 676 726| = 0.000 000 000 742 147 676 726


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: 0.000 000 000 742 147 676 726.

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) 0.000 000 000 742 147 676 726 × 2 = 0 + 0.000 000 001 484 295 353 452;
  • 2) 0.000 000 001 484 295 353 452 × 2 = 0 + 0.000 000 002 968 590 706 904;
  • 3) 0.000 000 002 968 590 706 904 × 2 = 0 + 0.000 000 005 937 181 413 808;
  • 4) 0.000 000 005 937 181 413 808 × 2 = 0 + 0.000 000 011 874 362 827 616;
  • 5) 0.000 000 011 874 362 827 616 × 2 = 0 + 0.000 000 023 748 725 655 232;
  • 6) 0.000 000 023 748 725 655 232 × 2 = 0 + 0.000 000 047 497 451 310 464;
  • 7) 0.000 000 047 497 451 310 464 × 2 = 0 + 0.000 000 094 994 902 620 928;
  • 8) 0.000 000 094 994 902 620 928 × 2 = 0 + 0.000 000 189 989 805 241 856;
  • 9) 0.000 000 189 989 805 241 856 × 2 = 0 + 0.000 000 379 979 610 483 712;
  • 10) 0.000 000 379 979 610 483 712 × 2 = 0 + 0.000 000 759 959 220 967 424;
  • 11) 0.000 000 759 959 220 967 424 × 2 = 0 + 0.000 001 519 918 441 934 848;
  • 12) 0.000 001 519 918 441 934 848 × 2 = 0 + 0.000 003 039 836 883 869 696;
  • 13) 0.000 003 039 836 883 869 696 × 2 = 0 + 0.000 006 079 673 767 739 392;
  • 14) 0.000 006 079 673 767 739 392 × 2 = 0 + 0.000 012 159 347 535 478 784;
  • 15) 0.000 012 159 347 535 478 784 × 2 = 0 + 0.000 024 318 695 070 957 568;
  • 16) 0.000 024 318 695 070 957 568 × 2 = 0 + 0.000 048 637 390 141 915 136;
  • 17) 0.000 048 637 390 141 915 136 × 2 = 0 + 0.000 097 274 780 283 830 272;
  • 18) 0.000 097 274 780 283 830 272 × 2 = 0 + 0.000 194 549 560 567 660 544;
  • 19) 0.000 194 549 560 567 660 544 × 2 = 0 + 0.000 389 099 121 135 321 088;
  • 20) 0.000 389 099 121 135 321 088 × 2 = 0 + 0.000 778 198 242 270 642 176;
  • 21) 0.000 778 198 242 270 642 176 × 2 = 0 + 0.001 556 396 484 541 284 352;
  • 22) 0.001 556 396 484 541 284 352 × 2 = 0 + 0.003 112 792 969 082 568 704;
  • 23) 0.003 112 792 969 082 568 704 × 2 = 0 + 0.006 225 585 938 165 137 408;
  • 24) 0.006 225 585 938 165 137 408 × 2 = 0 + 0.012 451 171 876 330 274 816;
  • 25) 0.012 451 171 876 330 274 816 × 2 = 0 + 0.024 902 343 752 660 549 632;
  • 26) 0.024 902 343 752 660 549 632 × 2 = 0 + 0.049 804 687 505 321 099 264;
  • 27) 0.049 804 687 505 321 099 264 × 2 = 0 + 0.099 609 375 010 642 198 528;
  • 28) 0.099 609 375 010 642 198 528 × 2 = 0 + 0.199 218 750 021 284 397 056;
  • 29) 0.199 218 750 021 284 397 056 × 2 = 0 + 0.398 437 500 042 568 794 112;
  • 30) 0.398 437 500 042 568 794 112 × 2 = 0 + 0.796 875 000 085 137 588 224;
  • 31) 0.796 875 000 085 137 588 224 × 2 = 1 + 0.593 750 000 170 275 176 448;
  • 32) 0.593 750 000 170 275 176 448 × 2 = 1 + 0.187 500 000 340 550 352 896;
  • 33) 0.187 500 000 340 550 352 896 × 2 = 0 + 0.375 000 000 681 100 705 792;
  • 34) 0.375 000 000 681 100 705 792 × 2 = 0 + 0.750 000 001 362 201 411 584;
  • 35) 0.750 000 001 362 201 411 584 × 2 = 1 + 0.500 000 002 724 402 823 168;
  • 36) 0.500 000 002 724 402 823 168 × 2 = 1 + 0.000 000 005 448 805 646 336;
  • 37) 0.000 000 005 448 805 646 336 × 2 = 0 + 0.000 000 010 897 611 292 672;
  • 38) 0.000 000 010 897 611 292 672 × 2 = 0 + 0.000 000 021 795 222 585 344;
  • 39) 0.000 000 021 795 222 585 344 × 2 = 0 + 0.000 000 043 590 445 170 688;
  • 40) 0.000 000 043 590 445 170 688 × 2 = 0 + 0.000 000 087 180 890 341 376;
  • 41) 0.000 000 087 180 890 341 376 × 2 = 0 + 0.000 000 174 361 780 682 752;
  • 42) 0.000 000 174 361 780 682 752 × 2 = 0 + 0.000 000 348 723 561 365 504;
  • 43) 0.000 000 348 723 561 365 504 × 2 = 0 + 0.000 000 697 447 122 731 008;
  • 44) 0.000 000 697 447 122 731 008 × 2 = 0 + 0.000 001 394 894 245 462 016;
  • 45) 0.000 001 394 894 245 462 016 × 2 = 0 + 0.000 002 789 788 490 924 032;
  • 46) 0.000 002 789 788 490 924 032 × 2 = 0 + 0.000 005 579 576 981 848 064;
  • 47) 0.000 005 579 576 981 848 064 × 2 = 0 + 0.000 011 159 153 963 696 128;
  • 48) 0.000 011 159 153 963 696 128 × 2 = 0 + 0.000 022 318 307 927 392 256;
  • 49) 0.000 022 318 307 927 392 256 × 2 = 0 + 0.000 044 636 615 854 784 512;
  • 50) 0.000 044 636 615 854 784 512 × 2 = 0 + 0.000 089 273 231 709 569 024;
  • 51) 0.000 089 273 231 709 569 024 × 2 = 0 + 0.000 178 546 463 419 138 048;
  • 52) 0.000 178 546 463 419 138 048 × 2 = 0 + 0.000 357 092 926 838 276 096;
  • 53) 0.000 357 092 926 838 276 096 × 2 = 0 + 0.000 714 185 853 676 552 192;
  • 54) 0.000 714 185 853 676 552 192 × 2 = 0 + 0.001 428 371 707 353 104 384;

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:


0.000 000 000 742 147 676 726(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 726(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:


0.000 000 000 742 147 676 726(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 -0.000 000 000 742 147 676 726 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