-0.000 000 000 742 147 676 621 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 621(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
-0.000 000 000 742 147 676 621(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:

|-0.000 000 000 742 147 676 621| = 0.000 000 000 742 147 676 621


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: 0.000 000 000 742 147 676 621.

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) 0.000 000 000 742 147 676 621 × 2 = 0 + 0.000 000 001 484 295 353 242;
  • 2) 0.000 000 001 484 295 353 242 × 2 = 0 + 0.000 000 002 968 590 706 484;
  • 3) 0.000 000 002 968 590 706 484 × 2 = 0 + 0.000 000 005 937 181 412 968;
  • 4) 0.000 000 005 937 181 412 968 × 2 = 0 + 0.000 000 011 874 362 825 936;
  • 5) 0.000 000 011 874 362 825 936 × 2 = 0 + 0.000 000 023 748 725 651 872;
  • 6) 0.000 000 023 748 725 651 872 × 2 = 0 + 0.000 000 047 497 451 303 744;
  • 7) 0.000 000 047 497 451 303 744 × 2 = 0 + 0.000 000 094 994 902 607 488;
  • 8) 0.000 000 094 994 902 607 488 × 2 = 0 + 0.000 000 189 989 805 214 976;
  • 9) 0.000 000 189 989 805 214 976 × 2 = 0 + 0.000 000 379 979 610 429 952;
  • 10) 0.000 000 379 979 610 429 952 × 2 = 0 + 0.000 000 759 959 220 859 904;
  • 11) 0.000 000 759 959 220 859 904 × 2 = 0 + 0.000 001 519 918 441 719 808;
  • 12) 0.000 001 519 918 441 719 808 × 2 = 0 + 0.000 003 039 836 883 439 616;
  • 13) 0.000 003 039 836 883 439 616 × 2 = 0 + 0.000 006 079 673 766 879 232;
  • 14) 0.000 006 079 673 766 879 232 × 2 = 0 + 0.000 012 159 347 533 758 464;
  • 15) 0.000 012 159 347 533 758 464 × 2 = 0 + 0.000 024 318 695 067 516 928;
  • 16) 0.000 024 318 695 067 516 928 × 2 = 0 + 0.000 048 637 390 135 033 856;
  • 17) 0.000 048 637 390 135 033 856 × 2 = 0 + 0.000 097 274 780 270 067 712;
  • 18) 0.000 097 274 780 270 067 712 × 2 = 0 + 0.000 194 549 560 540 135 424;
  • 19) 0.000 194 549 560 540 135 424 × 2 = 0 + 0.000 389 099 121 080 270 848;
  • 20) 0.000 389 099 121 080 270 848 × 2 = 0 + 0.000 778 198 242 160 541 696;
  • 21) 0.000 778 198 242 160 541 696 × 2 = 0 + 0.001 556 396 484 321 083 392;
  • 22) 0.001 556 396 484 321 083 392 × 2 = 0 + 0.003 112 792 968 642 166 784;
  • 23) 0.003 112 792 968 642 166 784 × 2 = 0 + 0.006 225 585 937 284 333 568;
  • 24) 0.006 225 585 937 284 333 568 × 2 = 0 + 0.012 451 171 874 568 667 136;
  • 25) 0.012 451 171 874 568 667 136 × 2 = 0 + 0.024 902 343 749 137 334 272;
  • 26) 0.024 902 343 749 137 334 272 × 2 = 0 + 0.049 804 687 498 274 668 544;
  • 27) 0.049 804 687 498 274 668 544 × 2 = 0 + 0.099 609 374 996 549 337 088;
  • 28) 0.099 609 374 996 549 337 088 × 2 = 0 + 0.199 218 749 993 098 674 176;
  • 29) 0.199 218 749 993 098 674 176 × 2 = 0 + 0.398 437 499 986 197 348 352;
  • 30) 0.398 437 499 986 197 348 352 × 2 = 0 + 0.796 874 999 972 394 696 704;
  • 31) 0.796 874 999 972 394 696 704 × 2 = 1 + 0.593 749 999 944 789 393 408;
  • 32) 0.593 749 999 944 789 393 408 × 2 = 1 + 0.187 499 999 889 578 786 816;
  • 33) 0.187 499 999 889 578 786 816 × 2 = 0 + 0.374 999 999 779 157 573 632;
  • 34) 0.374 999 999 779 157 573 632 × 2 = 0 + 0.749 999 999 558 315 147 264;
  • 35) 0.749 999 999 558 315 147 264 × 2 = 1 + 0.499 999 999 116 630 294 528;
  • 36) 0.499 999 999 116 630 294 528 × 2 = 0 + 0.999 999 998 233 260 589 056;
  • 37) 0.999 999 998 233 260 589 056 × 2 = 1 + 0.999 999 996 466 521 178 112;
  • 38) 0.999 999 996 466 521 178 112 × 2 = 1 + 0.999 999 992 933 042 356 224;
  • 39) 0.999 999 992 933 042 356 224 × 2 = 1 + 0.999 999 985 866 084 712 448;
  • 40) 0.999 999 985 866 084 712 448 × 2 = 1 + 0.999 999 971 732 169 424 896;
  • 41) 0.999 999 971 732 169 424 896 × 2 = 1 + 0.999 999 943 464 338 849 792;
  • 42) 0.999 999 943 464 338 849 792 × 2 = 1 + 0.999 999 886 928 677 699 584;
  • 43) 0.999 999 886 928 677 699 584 × 2 = 1 + 0.999 999 773 857 355 399 168;
  • 44) 0.999 999 773 857 355 399 168 × 2 = 1 + 0.999 999 547 714 710 798 336;
  • 45) 0.999 999 547 714 710 798 336 × 2 = 1 + 0.999 999 095 429 421 596 672;
  • 46) 0.999 999 095 429 421 596 672 × 2 = 1 + 0.999 998 190 858 843 193 344;
  • 47) 0.999 998 190 858 843 193 344 × 2 = 1 + 0.999 996 381 717 686 386 688;
  • 48) 0.999 996 381 717 686 386 688 × 2 = 1 + 0.999 992 763 435 372 773 376;
  • 49) 0.999 992 763 435 372 773 376 × 2 = 1 + 0.999 985 526 870 745 546 752;
  • 50) 0.999 985 526 870 745 546 752 × 2 = 1 + 0.999 971 053 741 491 093 504;
  • 51) 0.999 971 053 741 491 093 504 × 2 = 1 + 0.999 942 107 482 982 187 008;
  • 52) 0.999 942 107 482 982 187 008 × 2 = 1 + 0.999 884 214 965 964 374 016;
  • 53) 0.999 884 214 965 964 374 016 × 2 = 1 + 0.999 768 429 931 928 748 032;
  • 54) 0.999 768 429 931 928 748 032 × 2 = 1 + 0.999 536 859 863 857 496 064;

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:


0.000 000 000 742 147 676 621(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 621(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:


0.000 000 000 742 147 676 621(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 -0.000 000 000 742 147 676 621 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