-0.000 000 000 742 147 676 243 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 243(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
-0.000 000 000 742 147 676 243(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:

|-0.000 000 000 742 147 676 243| = 0.000 000 000 742 147 676 243


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: 0.000 000 000 742 147 676 243.

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) 0.000 000 000 742 147 676 243 × 2 = 0 + 0.000 000 001 484 295 352 486;
  • 2) 0.000 000 001 484 295 352 486 × 2 = 0 + 0.000 000 002 968 590 704 972;
  • 3) 0.000 000 002 968 590 704 972 × 2 = 0 + 0.000 000 005 937 181 409 944;
  • 4) 0.000 000 005 937 181 409 944 × 2 = 0 + 0.000 000 011 874 362 819 888;
  • 5) 0.000 000 011 874 362 819 888 × 2 = 0 + 0.000 000 023 748 725 639 776;
  • 6) 0.000 000 023 748 725 639 776 × 2 = 0 + 0.000 000 047 497 451 279 552;
  • 7) 0.000 000 047 497 451 279 552 × 2 = 0 + 0.000 000 094 994 902 559 104;
  • 8) 0.000 000 094 994 902 559 104 × 2 = 0 + 0.000 000 189 989 805 118 208;
  • 9) 0.000 000 189 989 805 118 208 × 2 = 0 + 0.000 000 379 979 610 236 416;
  • 10) 0.000 000 379 979 610 236 416 × 2 = 0 + 0.000 000 759 959 220 472 832;
  • 11) 0.000 000 759 959 220 472 832 × 2 = 0 + 0.000 001 519 918 440 945 664;
  • 12) 0.000 001 519 918 440 945 664 × 2 = 0 + 0.000 003 039 836 881 891 328;
  • 13) 0.000 003 039 836 881 891 328 × 2 = 0 + 0.000 006 079 673 763 782 656;
  • 14) 0.000 006 079 673 763 782 656 × 2 = 0 + 0.000 012 159 347 527 565 312;
  • 15) 0.000 012 159 347 527 565 312 × 2 = 0 + 0.000 024 318 695 055 130 624;
  • 16) 0.000 024 318 695 055 130 624 × 2 = 0 + 0.000 048 637 390 110 261 248;
  • 17) 0.000 048 637 390 110 261 248 × 2 = 0 + 0.000 097 274 780 220 522 496;
  • 18) 0.000 097 274 780 220 522 496 × 2 = 0 + 0.000 194 549 560 441 044 992;
  • 19) 0.000 194 549 560 441 044 992 × 2 = 0 + 0.000 389 099 120 882 089 984;
  • 20) 0.000 389 099 120 882 089 984 × 2 = 0 + 0.000 778 198 241 764 179 968;
  • 21) 0.000 778 198 241 764 179 968 × 2 = 0 + 0.001 556 396 483 528 359 936;
  • 22) 0.001 556 396 483 528 359 936 × 2 = 0 + 0.003 112 792 967 056 719 872;
  • 23) 0.003 112 792 967 056 719 872 × 2 = 0 + 0.006 225 585 934 113 439 744;
  • 24) 0.006 225 585 934 113 439 744 × 2 = 0 + 0.012 451 171 868 226 879 488;
  • 25) 0.012 451 171 868 226 879 488 × 2 = 0 + 0.024 902 343 736 453 758 976;
  • 26) 0.024 902 343 736 453 758 976 × 2 = 0 + 0.049 804 687 472 907 517 952;
  • 27) 0.049 804 687 472 907 517 952 × 2 = 0 + 0.099 609 374 945 815 035 904;
  • 28) 0.099 609 374 945 815 035 904 × 2 = 0 + 0.199 218 749 891 630 071 808;
  • 29) 0.199 218 749 891 630 071 808 × 2 = 0 + 0.398 437 499 783 260 143 616;
  • 30) 0.398 437 499 783 260 143 616 × 2 = 0 + 0.796 874 999 566 520 287 232;
  • 31) 0.796 874 999 566 520 287 232 × 2 = 1 + 0.593 749 999 133 040 574 464;
  • 32) 0.593 749 999 133 040 574 464 × 2 = 1 + 0.187 499 998 266 081 148 928;
  • 33) 0.187 499 998 266 081 148 928 × 2 = 0 + 0.374 999 996 532 162 297 856;
  • 34) 0.374 999 996 532 162 297 856 × 2 = 0 + 0.749 999 993 064 324 595 712;
  • 35) 0.749 999 993 064 324 595 712 × 2 = 1 + 0.499 999 986 128 649 191 424;
  • 36) 0.499 999 986 128 649 191 424 × 2 = 0 + 0.999 999 972 257 298 382 848;
  • 37) 0.999 999 972 257 298 382 848 × 2 = 1 + 0.999 999 944 514 596 765 696;
  • 38) 0.999 999 944 514 596 765 696 × 2 = 1 + 0.999 999 889 029 193 531 392;
  • 39) 0.999 999 889 029 193 531 392 × 2 = 1 + 0.999 999 778 058 387 062 784;
  • 40) 0.999 999 778 058 387 062 784 × 2 = 1 + 0.999 999 556 116 774 125 568;
  • 41) 0.999 999 556 116 774 125 568 × 2 = 1 + 0.999 999 112 233 548 251 136;
  • 42) 0.999 999 112 233 548 251 136 × 2 = 1 + 0.999 998 224 467 096 502 272;
  • 43) 0.999 998 224 467 096 502 272 × 2 = 1 + 0.999 996 448 934 193 004 544;
  • 44) 0.999 996 448 934 193 004 544 × 2 = 1 + 0.999 992 897 868 386 009 088;
  • 45) 0.999 992 897 868 386 009 088 × 2 = 1 + 0.999 985 795 736 772 018 176;
  • 46) 0.999 985 795 736 772 018 176 × 2 = 1 + 0.999 971 591 473 544 036 352;
  • 47) 0.999 971 591 473 544 036 352 × 2 = 1 + 0.999 943 182 947 088 072 704;
  • 48) 0.999 943 182 947 088 072 704 × 2 = 1 + 0.999 886 365 894 176 145 408;
  • 49) 0.999 886 365 894 176 145 408 × 2 = 1 + 0.999 772 731 788 352 290 816;
  • 50) 0.999 772 731 788 352 290 816 × 2 = 1 + 0.999 545 463 576 704 581 632;
  • 51) 0.999 545 463 576 704 581 632 × 2 = 1 + 0.999 090 927 153 409 163 264;
  • 52) 0.999 090 927 153 409 163 264 × 2 = 1 + 0.998 181 854 306 818 326 528;
  • 53) 0.998 181 854 306 818 326 528 × 2 = 1 + 0.996 363 708 613 636 653 056;
  • 54) 0.996 363 708 613 636 653 056 × 2 = 1 + 0.992 727 417 227 273 306 112;

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:


0.000 000 000 742 147 676 243(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 243(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 243(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 0111 1111 1111 1111 111


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 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 243 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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