-0.000 000 000 742 147 676 112 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 112(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
-0.000 000 000 742 147 676 112(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:

|-0.000 000 000 742 147 676 112| = 0.000 000 000 742 147 676 112


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: 0.000 000 000 742 147 676 112.

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) 0.000 000 000 742 147 676 112 × 2 = 0 + 0.000 000 001 484 295 352 224;
  • 2) 0.000 000 001 484 295 352 224 × 2 = 0 + 0.000 000 002 968 590 704 448;
  • 3) 0.000 000 002 968 590 704 448 × 2 = 0 + 0.000 000 005 937 181 408 896;
  • 4) 0.000 000 005 937 181 408 896 × 2 = 0 + 0.000 000 011 874 362 817 792;
  • 5) 0.000 000 011 874 362 817 792 × 2 = 0 + 0.000 000 023 748 725 635 584;
  • 6) 0.000 000 023 748 725 635 584 × 2 = 0 + 0.000 000 047 497 451 271 168;
  • 7) 0.000 000 047 497 451 271 168 × 2 = 0 + 0.000 000 094 994 902 542 336;
  • 8) 0.000 000 094 994 902 542 336 × 2 = 0 + 0.000 000 189 989 805 084 672;
  • 9) 0.000 000 189 989 805 084 672 × 2 = 0 + 0.000 000 379 979 610 169 344;
  • 10) 0.000 000 379 979 610 169 344 × 2 = 0 + 0.000 000 759 959 220 338 688;
  • 11) 0.000 000 759 959 220 338 688 × 2 = 0 + 0.000 001 519 918 440 677 376;
  • 12) 0.000 001 519 918 440 677 376 × 2 = 0 + 0.000 003 039 836 881 354 752;
  • 13) 0.000 003 039 836 881 354 752 × 2 = 0 + 0.000 006 079 673 762 709 504;
  • 14) 0.000 006 079 673 762 709 504 × 2 = 0 + 0.000 012 159 347 525 419 008;
  • 15) 0.000 012 159 347 525 419 008 × 2 = 0 + 0.000 024 318 695 050 838 016;
  • 16) 0.000 024 318 695 050 838 016 × 2 = 0 + 0.000 048 637 390 101 676 032;
  • 17) 0.000 048 637 390 101 676 032 × 2 = 0 + 0.000 097 274 780 203 352 064;
  • 18) 0.000 097 274 780 203 352 064 × 2 = 0 + 0.000 194 549 560 406 704 128;
  • 19) 0.000 194 549 560 406 704 128 × 2 = 0 + 0.000 389 099 120 813 408 256;
  • 20) 0.000 389 099 120 813 408 256 × 2 = 0 + 0.000 778 198 241 626 816 512;
  • 21) 0.000 778 198 241 626 816 512 × 2 = 0 + 0.001 556 396 483 253 633 024;
  • 22) 0.001 556 396 483 253 633 024 × 2 = 0 + 0.003 112 792 966 507 266 048;
  • 23) 0.003 112 792 966 507 266 048 × 2 = 0 + 0.006 225 585 933 014 532 096;
  • 24) 0.006 225 585 933 014 532 096 × 2 = 0 + 0.012 451 171 866 029 064 192;
  • 25) 0.012 451 171 866 029 064 192 × 2 = 0 + 0.024 902 343 732 058 128 384;
  • 26) 0.024 902 343 732 058 128 384 × 2 = 0 + 0.049 804 687 464 116 256 768;
  • 27) 0.049 804 687 464 116 256 768 × 2 = 0 + 0.099 609 374 928 232 513 536;
  • 28) 0.099 609 374 928 232 513 536 × 2 = 0 + 0.199 218 749 856 465 027 072;
  • 29) 0.199 218 749 856 465 027 072 × 2 = 0 + 0.398 437 499 712 930 054 144;
  • 30) 0.398 437 499 712 930 054 144 × 2 = 0 + 0.796 874 999 425 860 108 288;
  • 31) 0.796 874 999 425 860 108 288 × 2 = 1 + 0.593 749 998 851 720 216 576;
  • 32) 0.593 749 998 851 720 216 576 × 2 = 1 + 0.187 499 997 703 440 433 152;
  • 33) 0.187 499 997 703 440 433 152 × 2 = 0 + 0.374 999 995 406 880 866 304;
  • 34) 0.374 999 995 406 880 866 304 × 2 = 0 + 0.749 999 990 813 761 732 608;
  • 35) 0.749 999 990 813 761 732 608 × 2 = 1 + 0.499 999 981 627 523 465 216;
  • 36) 0.499 999 981 627 523 465 216 × 2 = 0 + 0.999 999 963 255 046 930 432;
  • 37) 0.999 999 963 255 046 930 432 × 2 = 1 + 0.999 999 926 510 093 860 864;
  • 38) 0.999 999 926 510 093 860 864 × 2 = 1 + 0.999 999 853 020 187 721 728;
  • 39) 0.999 999 853 020 187 721 728 × 2 = 1 + 0.999 999 706 040 375 443 456;
  • 40) 0.999 999 706 040 375 443 456 × 2 = 1 + 0.999 999 412 080 750 886 912;
  • 41) 0.999 999 412 080 750 886 912 × 2 = 1 + 0.999 998 824 161 501 773 824;
  • 42) 0.999 998 824 161 501 773 824 × 2 = 1 + 0.999 997 648 323 003 547 648;
  • 43) 0.999 997 648 323 003 547 648 × 2 = 1 + 0.999 995 296 646 007 095 296;
  • 44) 0.999 995 296 646 007 095 296 × 2 = 1 + 0.999 990 593 292 014 190 592;
  • 45) 0.999 990 593 292 014 190 592 × 2 = 1 + 0.999 981 186 584 028 381 184;
  • 46) 0.999 981 186 584 028 381 184 × 2 = 1 + 0.999 962 373 168 056 762 368;
  • 47) 0.999 962 373 168 056 762 368 × 2 = 1 + 0.999 924 746 336 113 524 736;
  • 48) 0.999 924 746 336 113 524 736 × 2 = 1 + 0.999 849 492 672 227 049 472;
  • 49) 0.999 849 492 672 227 049 472 × 2 = 1 + 0.999 698 985 344 454 098 944;
  • 50) 0.999 698 985 344 454 098 944 × 2 = 1 + 0.999 397 970 688 908 197 888;
  • 51) 0.999 397 970 688 908 197 888 × 2 = 1 + 0.998 795 941 377 816 395 776;
  • 52) 0.998 795 941 377 816 395 776 × 2 = 1 + 0.997 591 882 755 632 791 552;
  • 53) 0.997 591 882 755 632 791 552 × 2 = 1 + 0.995 183 765 511 265 583 104;
  • 54) 0.995 183 765 511 265 583 104 × 2 = 1 + 0.990 367 531 022 531 166 208;

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:


0.000 000 000 742 147 676 112(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 112(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:


0.000 000 000 742 147 676 112(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 -0.000 000 000 742 147 676 112 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