-0.000 000 000 742 147 676 263 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 263(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
-0.000 000 000 742 147 676 263(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:

|-0.000 000 000 742 147 676 263| = 0.000 000 000 742 147 676 263


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: 0.000 000 000 742 147 676 263.

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) 0.000 000 000 742 147 676 263 × 2 = 0 + 0.000 000 001 484 295 352 526;
  • 2) 0.000 000 001 484 295 352 526 × 2 = 0 + 0.000 000 002 968 590 705 052;
  • 3) 0.000 000 002 968 590 705 052 × 2 = 0 + 0.000 000 005 937 181 410 104;
  • 4) 0.000 000 005 937 181 410 104 × 2 = 0 + 0.000 000 011 874 362 820 208;
  • 5) 0.000 000 011 874 362 820 208 × 2 = 0 + 0.000 000 023 748 725 640 416;
  • 6) 0.000 000 023 748 725 640 416 × 2 = 0 + 0.000 000 047 497 451 280 832;
  • 7) 0.000 000 047 497 451 280 832 × 2 = 0 + 0.000 000 094 994 902 561 664;
  • 8) 0.000 000 094 994 902 561 664 × 2 = 0 + 0.000 000 189 989 805 123 328;
  • 9) 0.000 000 189 989 805 123 328 × 2 = 0 + 0.000 000 379 979 610 246 656;
  • 10) 0.000 000 379 979 610 246 656 × 2 = 0 + 0.000 000 759 959 220 493 312;
  • 11) 0.000 000 759 959 220 493 312 × 2 = 0 + 0.000 001 519 918 440 986 624;
  • 12) 0.000 001 519 918 440 986 624 × 2 = 0 + 0.000 003 039 836 881 973 248;
  • 13) 0.000 003 039 836 881 973 248 × 2 = 0 + 0.000 006 079 673 763 946 496;
  • 14) 0.000 006 079 673 763 946 496 × 2 = 0 + 0.000 012 159 347 527 892 992;
  • 15) 0.000 012 159 347 527 892 992 × 2 = 0 + 0.000 024 318 695 055 785 984;
  • 16) 0.000 024 318 695 055 785 984 × 2 = 0 + 0.000 048 637 390 111 571 968;
  • 17) 0.000 048 637 390 111 571 968 × 2 = 0 + 0.000 097 274 780 223 143 936;
  • 18) 0.000 097 274 780 223 143 936 × 2 = 0 + 0.000 194 549 560 446 287 872;
  • 19) 0.000 194 549 560 446 287 872 × 2 = 0 + 0.000 389 099 120 892 575 744;
  • 20) 0.000 389 099 120 892 575 744 × 2 = 0 + 0.000 778 198 241 785 151 488;
  • 21) 0.000 778 198 241 785 151 488 × 2 = 0 + 0.001 556 396 483 570 302 976;
  • 22) 0.001 556 396 483 570 302 976 × 2 = 0 + 0.003 112 792 967 140 605 952;
  • 23) 0.003 112 792 967 140 605 952 × 2 = 0 + 0.006 225 585 934 281 211 904;
  • 24) 0.006 225 585 934 281 211 904 × 2 = 0 + 0.012 451 171 868 562 423 808;
  • 25) 0.012 451 171 868 562 423 808 × 2 = 0 + 0.024 902 343 737 124 847 616;
  • 26) 0.024 902 343 737 124 847 616 × 2 = 0 + 0.049 804 687 474 249 695 232;
  • 27) 0.049 804 687 474 249 695 232 × 2 = 0 + 0.099 609 374 948 499 390 464;
  • 28) 0.099 609 374 948 499 390 464 × 2 = 0 + 0.199 218 749 896 998 780 928;
  • 29) 0.199 218 749 896 998 780 928 × 2 = 0 + 0.398 437 499 793 997 561 856;
  • 30) 0.398 437 499 793 997 561 856 × 2 = 0 + 0.796 874 999 587 995 123 712;
  • 31) 0.796 874 999 587 995 123 712 × 2 = 1 + 0.593 749 999 175 990 247 424;
  • 32) 0.593 749 999 175 990 247 424 × 2 = 1 + 0.187 499 998 351 980 494 848;
  • 33) 0.187 499 998 351 980 494 848 × 2 = 0 + 0.374 999 996 703 960 989 696;
  • 34) 0.374 999 996 703 960 989 696 × 2 = 0 + 0.749 999 993 407 921 979 392;
  • 35) 0.749 999 993 407 921 979 392 × 2 = 1 + 0.499 999 986 815 843 958 784;
  • 36) 0.499 999 986 815 843 958 784 × 2 = 0 + 0.999 999 973 631 687 917 568;
  • 37) 0.999 999 973 631 687 917 568 × 2 = 1 + 0.999 999 947 263 375 835 136;
  • 38) 0.999 999 947 263 375 835 136 × 2 = 1 + 0.999 999 894 526 751 670 272;
  • 39) 0.999 999 894 526 751 670 272 × 2 = 1 + 0.999 999 789 053 503 340 544;
  • 40) 0.999 999 789 053 503 340 544 × 2 = 1 + 0.999 999 578 107 006 681 088;
  • 41) 0.999 999 578 107 006 681 088 × 2 = 1 + 0.999 999 156 214 013 362 176;
  • 42) 0.999 999 156 214 013 362 176 × 2 = 1 + 0.999 998 312 428 026 724 352;
  • 43) 0.999 998 312 428 026 724 352 × 2 = 1 + 0.999 996 624 856 053 448 704;
  • 44) 0.999 996 624 856 053 448 704 × 2 = 1 + 0.999 993 249 712 106 897 408;
  • 45) 0.999 993 249 712 106 897 408 × 2 = 1 + 0.999 986 499 424 213 794 816;
  • 46) 0.999 986 499 424 213 794 816 × 2 = 1 + 0.999 972 998 848 427 589 632;
  • 47) 0.999 972 998 848 427 589 632 × 2 = 1 + 0.999 945 997 696 855 179 264;
  • 48) 0.999 945 997 696 855 179 264 × 2 = 1 + 0.999 891 995 393 710 358 528;
  • 49) 0.999 891 995 393 710 358 528 × 2 = 1 + 0.999 783 990 787 420 717 056;
  • 50) 0.999 783 990 787 420 717 056 × 2 = 1 + 0.999 567 981 574 841 434 112;
  • 51) 0.999 567 981 574 841 434 112 × 2 = 1 + 0.999 135 963 149 682 868 224;
  • 52) 0.999 135 963 149 682 868 224 × 2 = 1 + 0.998 271 926 299 365 736 448;
  • 53) 0.998 271 926 299 365 736 448 × 2 = 1 + 0.996 543 852 598 731 472 896;
  • 54) 0.996 543 852 598 731 472 896 × 2 = 1 + 0.993 087 705 197 462 945 792;

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:


0.000 000 000 742 147 676 263(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 263(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:


0.000 000 000 742 147 676 263(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 -0.000 000 000 742 147 676 263 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