-0.000 000 000 742 147 723 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 723(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
-0.000 000 000 742 147 723(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:

|-0.000 000 000 742 147 723| = 0.000 000 000 742 147 723


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: 0.000 000 000 742 147 723.

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) 0.000 000 000 742 147 723 × 2 = 0 + 0.000 000 001 484 295 446;
  • 2) 0.000 000 001 484 295 446 × 2 = 0 + 0.000 000 002 968 590 892;
  • 3) 0.000 000 002 968 590 892 × 2 = 0 + 0.000 000 005 937 181 784;
  • 4) 0.000 000 005 937 181 784 × 2 = 0 + 0.000 000 011 874 363 568;
  • 5) 0.000 000 011 874 363 568 × 2 = 0 + 0.000 000 023 748 727 136;
  • 6) 0.000 000 023 748 727 136 × 2 = 0 + 0.000 000 047 497 454 272;
  • 7) 0.000 000 047 497 454 272 × 2 = 0 + 0.000 000 094 994 908 544;
  • 8) 0.000 000 094 994 908 544 × 2 = 0 + 0.000 000 189 989 817 088;
  • 9) 0.000 000 189 989 817 088 × 2 = 0 + 0.000 000 379 979 634 176;
  • 10) 0.000 000 379 979 634 176 × 2 = 0 + 0.000 000 759 959 268 352;
  • 11) 0.000 000 759 959 268 352 × 2 = 0 + 0.000 001 519 918 536 704;
  • 12) 0.000 001 519 918 536 704 × 2 = 0 + 0.000 003 039 837 073 408;
  • 13) 0.000 003 039 837 073 408 × 2 = 0 + 0.000 006 079 674 146 816;
  • 14) 0.000 006 079 674 146 816 × 2 = 0 + 0.000 012 159 348 293 632;
  • 15) 0.000 012 159 348 293 632 × 2 = 0 + 0.000 024 318 696 587 264;
  • 16) 0.000 024 318 696 587 264 × 2 = 0 + 0.000 048 637 393 174 528;
  • 17) 0.000 048 637 393 174 528 × 2 = 0 + 0.000 097 274 786 349 056;
  • 18) 0.000 097 274 786 349 056 × 2 = 0 + 0.000 194 549 572 698 112;
  • 19) 0.000 194 549 572 698 112 × 2 = 0 + 0.000 389 099 145 396 224;
  • 20) 0.000 389 099 145 396 224 × 2 = 0 + 0.000 778 198 290 792 448;
  • 21) 0.000 778 198 290 792 448 × 2 = 0 + 0.001 556 396 581 584 896;
  • 22) 0.001 556 396 581 584 896 × 2 = 0 + 0.003 112 793 163 169 792;
  • 23) 0.003 112 793 163 169 792 × 2 = 0 + 0.006 225 586 326 339 584;
  • 24) 0.006 225 586 326 339 584 × 2 = 0 + 0.012 451 172 652 679 168;
  • 25) 0.012 451 172 652 679 168 × 2 = 0 + 0.024 902 345 305 358 336;
  • 26) 0.024 902 345 305 358 336 × 2 = 0 + 0.049 804 690 610 716 672;
  • 27) 0.049 804 690 610 716 672 × 2 = 0 + 0.099 609 381 221 433 344;
  • 28) 0.099 609 381 221 433 344 × 2 = 0 + 0.199 218 762 442 866 688;
  • 29) 0.199 218 762 442 866 688 × 2 = 0 + 0.398 437 524 885 733 376;
  • 30) 0.398 437 524 885 733 376 × 2 = 0 + 0.796 875 049 771 466 752;
  • 31) 0.796 875 049 771 466 752 × 2 = 1 + 0.593 750 099 542 933 504;
  • 32) 0.593 750 099 542 933 504 × 2 = 1 + 0.187 500 199 085 867 008;
  • 33) 0.187 500 199 085 867 008 × 2 = 0 + 0.375 000 398 171 734 016;
  • 34) 0.375 000 398 171 734 016 × 2 = 0 + 0.750 000 796 343 468 032;
  • 35) 0.750 000 796 343 468 032 × 2 = 1 + 0.500 001 592 686 936 064;
  • 36) 0.500 001 592 686 936 064 × 2 = 1 + 0.000 003 185 373 872 128;
  • 37) 0.000 003 185 373 872 128 × 2 = 0 + 0.000 006 370 747 744 256;
  • 38) 0.000 006 370 747 744 256 × 2 = 0 + 0.000 012 741 495 488 512;
  • 39) 0.000 012 741 495 488 512 × 2 = 0 + 0.000 025 482 990 977 024;
  • 40) 0.000 025 482 990 977 024 × 2 = 0 + 0.000 050 965 981 954 048;
  • 41) 0.000 050 965 981 954 048 × 2 = 0 + 0.000 101 931 963 908 096;
  • 42) 0.000 101 931 963 908 096 × 2 = 0 + 0.000 203 863 927 816 192;
  • 43) 0.000 203 863 927 816 192 × 2 = 0 + 0.000 407 727 855 632 384;
  • 44) 0.000 407 727 855 632 384 × 2 = 0 + 0.000 815 455 711 264 768;
  • 45) 0.000 815 455 711 264 768 × 2 = 0 + 0.001 630 911 422 529 536;
  • 46) 0.001 630 911 422 529 536 × 2 = 0 + 0.003 261 822 845 059 072;
  • 47) 0.003 261 822 845 059 072 × 2 = 0 + 0.006 523 645 690 118 144;
  • 48) 0.006 523 645 690 118 144 × 2 = 0 + 0.013 047 291 380 236 288;
  • 49) 0.013 047 291 380 236 288 × 2 = 0 + 0.026 094 582 760 472 576;
  • 50) 0.026 094 582 760 472 576 × 2 = 0 + 0.052 189 165 520 945 152;
  • 51) 0.052 189 165 520 945 152 × 2 = 0 + 0.104 378 331 041 890 304;
  • 52) 0.104 378 331 041 890 304 × 2 = 0 + 0.208 756 662 083 780 608;
  • 53) 0.208 756 662 083 780 608 × 2 = 0 + 0.417 513 324 167 561 216;
  • 54) 0.417 513 324 167 561 216 × 2 = 0 + 0.835 026 648 335 122 432;

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:


0.000 000 000 742 147 723(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 723(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:


0.000 000 000 742 147 723(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 -0.000 000 000 742 147 723 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