-0.000 000 000 742 147 821 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 821(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
-0.000 000 000 742 147 821(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:

|-0.000 000 000 742 147 821| = 0.000 000 000 742 147 821


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: 0.000 000 000 742 147 821.

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) 0.000 000 000 742 147 821 × 2 = 0 + 0.000 000 001 484 295 642;
  • 2) 0.000 000 001 484 295 642 × 2 = 0 + 0.000 000 002 968 591 284;
  • 3) 0.000 000 002 968 591 284 × 2 = 0 + 0.000 000 005 937 182 568;
  • 4) 0.000 000 005 937 182 568 × 2 = 0 + 0.000 000 011 874 365 136;
  • 5) 0.000 000 011 874 365 136 × 2 = 0 + 0.000 000 023 748 730 272;
  • 6) 0.000 000 023 748 730 272 × 2 = 0 + 0.000 000 047 497 460 544;
  • 7) 0.000 000 047 497 460 544 × 2 = 0 + 0.000 000 094 994 921 088;
  • 8) 0.000 000 094 994 921 088 × 2 = 0 + 0.000 000 189 989 842 176;
  • 9) 0.000 000 189 989 842 176 × 2 = 0 + 0.000 000 379 979 684 352;
  • 10) 0.000 000 379 979 684 352 × 2 = 0 + 0.000 000 759 959 368 704;
  • 11) 0.000 000 759 959 368 704 × 2 = 0 + 0.000 001 519 918 737 408;
  • 12) 0.000 001 519 918 737 408 × 2 = 0 + 0.000 003 039 837 474 816;
  • 13) 0.000 003 039 837 474 816 × 2 = 0 + 0.000 006 079 674 949 632;
  • 14) 0.000 006 079 674 949 632 × 2 = 0 + 0.000 012 159 349 899 264;
  • 15) 0.000 012 159 349 899 264 × 2 = 0 + 0.000 024 318 699 798 528;
  • 16) 0.000 024 318 699 798 528 × 2 = 0 + 0.000 048 637 399 597 056;
  • 17) 0.000 048 637 399 597 056 × 2 = 0 + 0.000 097 274 799 194 112;
  • 18) 0.000 097 274 799 194 112 × 2 = 0 + 0.000 194 549 598 388 224;
  • 19) 0.000 194 549 598 388 224 × 2 = 0 + 0.000 389 099 196 776 448;
  • 20) 0.000 389 099 196 776 448 × 2 = 0 + 0.000 778 198 393 552 896;
  • 21) 0.000 778 198 393 552 896 × 2 = 0 + 0.001 556 396 787 105 792;
  • 22) 0.001 556 396 787 105 792 × 2 = 0 + 0.003 112 793 574 211 584;
  • 23) 0.003 112 793 574 211 584 × 2 = 0 + 0.006 225 587 148 423 168;
  • 24) 0.006 225 587 148 423 168 × 2 = 0 + 0.012 451 174 296 846 336;
  • 25) 0.012 451 174 296 846 336 × 2 = 0 + 0.024 902 348 593 692 672;
  • 26) 0.024 902 348 593 692 672 × 2 = 0 + 0.049 804 697 187 385 344;
  • 27) 0.049 804 697 187 385 344 × 2 = 0 + 0.099 609 394 374 770 688;
  • 28) 0.099 609 394 374 770 688 × 2 = 0 + 0.199 218 788 749 541 376;
  • 29) 0.199 218 788 749 541 376 × 2 = 0 + 0.398 437 577 499 082 752;
  • 30) 0.398 437 577 499 082 752 × 2 = 0 + 0.796 875 154 998 165 504;
  • 31) 0.796 875 154 998 165 504 × 2 = 1 + 0.593 750 309 996 331 008;
  • 32) 0.593 750 309 996 331 008 × 2 = 1 + 0.187 500 619 992 662 016;
  • 33) 0.187 500 619 992 662 016 × 2 = 0 + 0.375 001 239 985 324 032;
  • 34) 0.375 001 239 985 324 032 × 2 = 0 + 0.750 002 479 970 648 064;
  • 35) 0.750 002 479 970 648 064 × 2 = 1 + 0.500 004 959 941 296 128;
  • 36) 0.500 004 959 941 296 128 × 2 = 1 + 0.000 009 919 882 592 256;
  • 37) 0.000 009 919 882 592 256 × 2 = 0 + 0.000 019 839 765 184 512;
  • 38) 0.000 019 839 765 184 512 × 2 = 0 + 0.000 039 679 530 369 024;
  • 39) 0.000 039 679 530 369 024 × 2 = 0 + 0.000 079 359 060 738 048;
  • 40) 0.000 079 359 060 738 048 × 2 = 0 + 0.000 158 718 121 476 096;
  • 41) 0.000 158 718 121 476 096 × 2 = 0 + 0.000 317 436 242 952 192;
  • 42) 0.000 317 436 242 952 192 × 2 = 0 + 0.000 634 872 485 904 384;
  • 43) 0.000 634 872 485 904 384 × 2 = 0 + 0.001 269 744 971 808 768;
  • 44) 0.001 269 744 971 808 768 × 2 = 0 + 0.002 539 489 943 617 536;
  • 45) 0.002 539 489 943 617 536 × 2 = 0 + 0.005 078 979 887 235 072;
  • 46) 0.005 078 979 887 235 072 × 2 = 0 + 0.010 157 959 774 470 144;
  • 47) 0.010 157 959 774 470 144 × 2 = 0 + 0.020 315 919 548 940 288;
  • 48) 0.020 315 919 548 940 288 × 2 = 0 + 0.040 631 839 097 880 576;
  • 49) 0.040 631 839 097 880 576 × 2 = 0 + 0.081 263 678 195 761 152;
  • 50) 0.081 263 678 195 761 152 × 2 = 0 + 0.162 527 356 391 522 304;
  • 51) 0.162 527 356 391 522 304 × 2 = 0 + 0.325 054 712 783 044 608;
  • 52) 0.325 054 712 783 044 608 × 2 = 0 + 0.650 109 425 566 089 216;
  • 53) 0.650 109 425 566 089 216 × 2 = 1 + 0.300 218 851 132 178 432;
  • 54) 0.300 218 851 132 178 432 × 2 = 0 + 0.600 437 702 264 356 864;

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:


0.000 000 000 742 147 821(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 821(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(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:


0.000 000 000 742 147 821(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) × 20 =

1.1001 1000 0000 0000 0000 010(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 010


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 0010 =

100 1100 0000 0000 0000 0010


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 0010


Decimal number -0.000 000 000 742 147 821 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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