-0.000 000 000 742 147 676 211 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 211(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
-0.000 000 000 742 147 676 211(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:

|-0.000 000 000 742 147 676 211| = 0.000 000 000 742 147 676 211


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: 0.000 000 000 742 147 676 211.

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) 0.000 000 000 742 147 676 211 × 2 = 0 + 0.000 000 001 484 295 352 422;
  • 2) 0.000 000 001 484 295 352 422 × 2 = 0 + 0.000 000 002 968 590 704 844;
  • 3) 0.000 000 002 968 590 704 844 × 2 = 0 + 0.000 000 005 937 181 409 688;
  • 4) 0.000 000 005 937 181 409 688 × 2 = 0 + 0.000 000 011 874 362 819 376;
  • 5) 0.000 000 011 874 362 819 376 × 2 = 0 + 0.000 000 023 748 725 638 752;
  • 6) 0.000 000 023 748 725 638 752 × 2 = 0 + 0.000 000 047 497 451 277 504;
  • 7) 0.000 000 047 497 451 277 504 × 2 = 0 + 0.000 000 094 994 902 555 008;
  • 8) 0.000 000 094 994 902 555 008 × 2 = 0 + 0.000 000 189 989 805 110 016;
  • 9) 0.000 000 189 989 805 110 016 × 2 = 0 + 0.000 000 379 979 610 220 032;
  • 10) 0.000 000 379 979 610 220 032 × 2 = 0 + 0.000 000 759 959 220 440 064;
  • 11) 0.000 000 759 959 220 440 064 × 2 = 0 + 0.000 001 519 918 440 880 128;
  • 12) 0.000 001 519 918 440 880 128 × 2 = 0 + 0.000 003 039 836 881 760 256;
  • 13) 0.000 003 039 836 881 760 256 × 2 = 0 + 0.000 006 079 673 763 520 512;
  • 14) 0.000 006 079 673 763 520 512 × 2 = 0 + 0.000 012 159 347 527 041 024;
  • 15) 0.000 012 159 347 527 041 024 × 2 = 0 + 0.000 024 318 695 054 082 048;
  • 16) 0.000 024 318 695 054 082 048 × 2 = 0 + 0.000 048 637 390 108 164 096;
  • 17) 0.000 048 637 390 108 164 096 × 2 = 0 + 0.000 097 274 780 216 328 192;
  • 18) 0.000 097 274 780 216 328 192 × 2 = 0 + 0.000 194 549 560 432 656 384;
  • 19) 0.000 194 549 560 432 656 384 × 2 = 0 + 0.000 389 099 120 865 312 768;
  • 20) 0.000 389 099 120 865 312 768 × 2 = 0 + 0.000 778 198 241 730 625 536;
  • 21) 0.000 778 198 241 730 625 536 × 2 = 0 + 0.001 556 396 483 461 251 072;
  • 22) 0.001 556 396 483 461 251 072 × 2 = 0 + 0.003 112 792 966 922 502 144;
  • 23) 0.003 112 792 966 922 502 144 × 2 = 0 + 0.006 225 585 933 845 004 288;
  • 24) 0.006 225 585 933 845 004 288 × 2 = 0 + 0.012 451 171 867 690 008 576;
  • 25) 0.012 451 171 867 690 008 576 × 2 = 0 + 0.024 902 343 735 380 017 152;
  • 26) 0.024 902 343 735 380 017 152 × 2 = 0 + 0.049 804 687 470 760 034 304;
  • 27) 0.049 804 687 470 760 034 304 × 2 = 0 + 0.099 609 374 941 520 068 608;
  • 28) 0.099 609 374 941 520 068 608 × 2 = 0 + 0.199 218 749 883 040 137 216;
  • 29) 0.199 218 749 883 040 137 216 × 2 = 0 + 0.398 437 499 766 080 274 432;
  • 30) 0.398 437 499 766 080 274 432 × 2 = 0 + 0.796 874 999 532 160 548 864;
  • 31) 0.796 874 999 532 160 548 864 × 2 = 1 + 0.593 749 999 064 321 097 728;
  • 32) 0.593 749 999 064 321 097 728 × 2 = 1 + 0.187 499 998 128 642 195 456;
  • 33) 0.187 499 998 128 642 195 456 × 2 = 0 + 0.374 999 996 257 284 390 912;
  • 34) 0.374 999 996 257 284 390 912 × 2 = 0 + 0.749 999 992 514 568 781 824;
  • 35) 0.749 999 992 514 568 781 824 × 2 = 1 + 0.499 999 985 029 137 563 648;
  • 36) 0.499 999 985 029 137 563 648 × 2 = 0 + 0.999 999 970 058 275 127 296;
  • 37) 0.999 999 970 058 275 127 296 × 2 = 1 + 0.999 999 940 116 550 254 592;
  • 38) 0.999 999 940 116 550 254 592 × 2 = 1 + 0.999 999 880 233 100 509 184;
  • 39) 0.999 999 880 233 100 509 184 × 2 = 1 + 0.999 999 760 466 201 018 368;
  • 40) 0.999 999 760 466 201 018 368 × 2 = 1 + 0.999 999 520 932 402 036 736;
  • 41) 0.999 999 520 932 402 036 736 × 2 = 1 + 0.999 999 041 864 804 073 472;
  • 42) 0.999 999 041 864 804 073 472 × 2 = 1 + 0.999 998 083 729 608 146 944;
  • 43) 0.999 998 083 729 608 146 944 × 2 = 1 + 0.999 996 167 459 216 293 888;
  • 44) 0.999 996 167 459 216 293 888 × 2 = 1 + 0.999 992 334 918 432 587 776;
  • 45) 0.999 992 334 918 432 587 776 × 2 = 1 + 0.999 984 669 836 865 175 552;
  • 46) 0.999 984 669 836 865 175 552 × 2 = 1 + 0.999 969 339 673 730 351 104;
  • 47) 0.999 969 339 673 730 351 104 × 2 = 1 + 0.999 938 679 347 460 702 208;
  • 48) 0.999 938 679 347 460 702 208 × 2 = 1 + 0.999 877 358 694 921 404 416;
  • 49) 0.999 877 358 694 921 404 416 × 2 = 1 + 0.999 754 717 389 842 808 832;
  • 50) 0.999 754 717 389 842 808 832 × 2 = 1 + 0.999 509 434 779 685 617 664;
  • 51) 0.999 509 434 779 685 617 664 × 2 = 1 + 0.999 018 869 559 371 235 328;
  • 52) 0.999 018 869 559 371 235 328 × 2 = 1 + 0.998 037 739 118 742 470 656;
  • 53) 0.998 037 739 118 742 470 656 × 2 = 1 + 0.996 075 478 237 484 941 312;
  • 54) 0.996 075 478 237 484 941 312 × 2 = 1 + 0.992 150 956 474 969 882 624;

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:


0.000 000 000 742 147 676 211(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 211(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:


0.000 000 000 742 147 676 211(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 -0.000 000 000 742 147 676 211 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