-0.000 000 000 742 147 676 646 512 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 512(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
-0.000 000 000 742 147 676 646 512(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:

|-0.000 000 000 742 147 676 646 512| = 0.000 000 000 742 147 676 646 512


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: 0.000 000 000 742 147 676 646 512.

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) 0.000 000 000 742 147 676 646 512 × 2 = 0 + 0.000 000 001 484 295 353 293 024;
  • 2) 0.000 000 001 484 295 353 293 024 × 2 = 0 + 0.000 000 002 968 590 706 586 048;
  • 3) 0.000 000 002 968 590 706 586 048 × 2 = 0 + 0.000 000 005 937 181 413 172 096;
  • 4) 0.000 000 005 937 181 413 172 096 × 2 = 0 + 0.000 000 011 874 362 826 344 192;
  • 5) 0.000 000 011 874 362 826 344 192 × 2 = 0 + 0.000 000 023 748 725 652 688 384;
  • 6) 0.000 000 023 748 725 652 688 384 × 2 = 0 + 0.000 000 047 497 451 305 376 768;
  • 7) 0.000 000 047 497 451 305 376 768 × 2 = 0 + 0.000 000 094 994 902 610 753 536;
  • 8) 0.000 000 094 994 902 610 753 536 × 2 = 0 + 0.000 000 189 989 805 221 507 072;
  • 9) 0.000 000 189 989 805 221 507 072 × 2 = 0 + 0.000 000 379 979 610 443 014 144;
  • 10) 0.000 000 379 979 610 443 014 144 × 2 = 0 + 0.000 000 759 959 220 886 028 288;
  • 11) 0.000 000 759 959 220 886 028 288 × 2 = 0 + 0.000 001 519 918 441 772 056 576;
  • 12) 0.000 001 519 918 441 772 056 576 × 2 = 0 + 0.000 003 039 836 883 544 113 152;
  • 13) 0.000 003 039 836 883 544 113 152 × 2 = 0 + 0.000 006 079 673 767 088 226 304;
  • 14) 0.000 006 079 673 767 088 226 304 × 2 = 0 + 0.000 012 159 347 534 176 452 608;
  • 15) 0.000 012 159 347 534 176 452 608 × 2 = 0 + 0.000 024 318 695 068 352 905 216;
  • 16) 0.000 024 318 695 068 352 905 216 × 2 = 0 + 0.000 048 637 390 136 705 810 432;
  • 17) 0.000 048 637 390 136 705 810 432 × 2 = 0 + 0.000 097 274 780 273 411 620 864;
  • 18) 0.000 097 274 780 273 411 620 864 × 2 = 0 + 0.000 194 549 560 546 823 241 728;
  • 19) 0.000 194 549 560 546 823 241 728 × 2 = 0 + 0.000 389 099 121 093 646 483 456;
  • 20) 0.000 389 099 121 093 646 483 456 × 2 = 0 + 0.000 778 198 242 187 292 966 912;
  • 21) 0.000 778 198 242 187 292 966 912 × 2 = 0 + 0.001 556 396 484 374 585 933 824;
  • 22) 0.001 556 396 484 374 585 933 824 × 2 = 0 + 0.003 112 792 968 749 171 867 648;
  • 23) 0.003 112 792 968 749 171 867 648 × 2 = 0 + 0.006 225 585 937 498 343 735 296;
  • 24) 0.006 225 585 937 498 343 735 296 × 2 = 0 + 0.012 451 171 874 996 687 470 592;
  • 25) 0.012 451 171 874 996 687 470 592 × 2 = 0 + 0.024 902 343 749 993 374 941 184;
  • 26) 0.024 902 343 749 993 374 941 184 × 2 = 0 + 0.049 804 687 499 986 749 882 368;
  • 27) 0.049 804 687 499 986 749 882 368 × 2 = 0 + 0.099 609 374 999 973 499 764 736;
  • 28) 0.099 609 374 999 973 499 764 736 × 2 = 0 + 0.199 218 749 999 946 999 529 472;
  • 29) 0.199 218 749 999 946 999 529 472 × 2 = 0 + 0.398 437 499 999 893 999 058 944;
  • 30) 0.398 437 499 999 893 999 058 944 × 2 = 0 + 0.796 874 999 999 787 998 117 888;
  • 31) 0.796 874 999 999 787 998 117 888 × 2 = 1 + 0.593 749 999 999 575 996 235 776;
  • 32) 0.593 749 999 999 575 996 235 776 × 2 = 1 + 0.187 499 999 999 151 992 471 552;
  • 33) 0.187 499 999 999 151 992 471 552 × 2 = 0 + 0.374 999 999 998 303 984 943 104;
  • 34) 0.374 999 999 998 303 984 943 104 × 2 = 0 + 0.749 999 999 996 607 969 886 208;
  • 35) 0.749 999 999 996 607 969 886 208 × 2 = 1 + 0.499 999 999 993 215 939 772 416;
  • 36) 0.499 999 999 993 215 939 772 416 × 2 = 0 + 0.999 999 999 986 431 879 544 832;
  • 37) 0.999 999 999 986 431 879 544 832 × 2 = 1 + 0.999 999 999 972 863 759 089 664;
  • 38) 0.999 999 999 972 863 759 089 664 × 2 = 1 + 0.999 999 999 945 727 518 179 328;
  • 39) 0.999 999 999 945 727 518 179 328 × 2 = 1 + 0.999 999 999 891 455 036 358 656;
  • 40) 0.999 999 999 891 455 036 358 656 × 2 = 1 + 0.999 999 999 782 910 072 717 312;
  • 41) 0.999 999 999 782 910 072 717 312 × 2 = 1 + 0.999 999 999 565 820 145 434 624;
  • 42) 0.999 999 999 565 820 145 434 624 × 2 = 1 + 0.999 999 999 131 640 290 869 248;
  • 43) 0.999 999 999 131 640 290 869 248 × 2 = 1 + 0.999 999 998 263 280 581 738 496;
  • 44) 0.999 999 998 263 280 581 738 496 × 2 = 1 + 0.999 999 996 526 561 163 476 992;
  • 45) 0.999 999 996 526 561 163 476 992 × 2 = 1 + 0.999 999 993 053 122 326 953 984;
  • 46) 0.999 999 993 053 122 326 953 984 × 2 = 1 + 0.999 999 986 106 244 653 907 968;
  • 47) 0.999 999 986 106 244 653 907 968 × 2 = 1 + 0.999 999 972 212 489 307 815 936;
  • 48) 0.999 999 972 212 489 307 815 936 × 2 = 1 + 0.999 999 944 424 978 615 631 872;
  • 49) 0.999 999 944 424 978 615 631 872 × 2 = 1 + 0.999 999 888 849 957 231 263 744;
  • 50) 0.999 999 888 849 957 231 263 744 × 2 = 1 + 0.999 999 777 699 914 462 527 488;
  • 51) 0.999 999 777 699 914 462 527 488 × 2 = 1 + 0.999 999 555 399 828 925 054 976;
  • 52) 0.999 999 555 399 828 925 054 976 × 2 = 1 + 0.999 999 110 799 657 850 109 952;
  • 53) 0.999 999 110 799 657 850 109 952 × 2 = 1 + 0.999 998 221 599 315 700 219 904;
  • 54) 0.999 998 221 599 315 700 219 904 × 2 = 1 + 0.999 996 443 198 631 400 439 808;

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:


0.000 000 000 742 147 676 646 512(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 512(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:


0.000 000 000 742 147 676 646 512(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 -0.000 000 000 742 147 676 646 512 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