-0.000 000 000 742 147 676 362 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 362(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
-0.000 000 000 742 147 676 362(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:

|-0.000 000 000 742 147 676 362| = 0.000 000 000 742 147 676 362


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: 0.000 000 000 742 147 676 362.

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) 0.000 000 000 742 147 676 362 × 2 = 0 + 0.000 000 001 484 295 352 724;
  • 2) 0.000 000 001 484 295 352 724 × 2 = 0 + 0.000 000 002 968 590 705 448;
  • 3) 0.000 000 002 968 590 705 448 × 2 = 0 + 0.000 000 005 937 181 410 896;
  • 4) 0.000 000 005 937 181 410 896 × 2 = 0 + 0.000 000 011 874 362 821 792;
  • 5) 0.000 000 011 874 362 821 792 × 2 = 0 + 0.000 000 023 748 725 643 584;
  • 6) 0.000 000 023 748 725 643 584 × 2 = 0 + 0.000 000 047 497 451 287 168;
  • 7) 0.000 000 047 497 451 287 168 × 2 = 0 + 0.000 000 094 994 902 574 336;
  • 8) 0.000 000 094 994 902 574 336 × 2 = 0 + 0.000 000 189 989 805 148 672;
  • 9) 0.000 000 189 989 805 148 672 × 2 = 0 + 0.000 000 379 979 610 297 344;
  • 10) 0.000 000 379 979 610 297 344 × 2 = 0 + 0.000 000 759 959 220 594 688;
  • 11) 0.000 000 759 959 220 594 688 × 2 = 0 + 0.000 001 519 918 441 189 376;
  • 12) 0.000 001 519 918 441 189 376 × 2 = 0 + 0.000 003 039 836 882 378 752;
  • 13) 0.000 003 039 836 882 378 752 × 2 = 0 + 0.000 006 079 673 764 757 504;
  • 14) 0.000 006 079 673 764 757 504 × 2 = 0 + 0.000 012 159 347 529 515 008;
  • 15) 0.000 012 159 347 529 515 008 × 2 = 0 + 0.000 024 318 695 059 030 016;
  • 16) 0.000 024 318 695 059 030 016 × 2 = 0 + 0.000 048 637 390 118 060 032;
  • 17) 0.000 048 637 390 118 060 032 × 2 = 0 + 0.000 097 274 780 236 120 064;
  • 18) 0.000 097 274 780 236 120 064 × 2 = 0 + 0.000 194 549 560 472 240 128;
  • 19) 0.000 194 549 560 472 240 128 × 2 = 0 + 0.000 389 099 120 944 480 256;
  • 20) 0.000 389 099 120 944 480 256 × 2 = 0 + 0.000 778 198 241 888 960 512;
  • 21) 0.000 778 198 241 888 960 512 × 2 = 0 + 0.001 556 396 483 777 921 024;
  • 22) 0.001 556 396 483 777 921 024 × 2 = 0 + 0.003 112 792 967 555 842 048;
  • 23) 0.003 112 792 967 555 842 048 × 2 = 0 + 0.006 225 585 935 111 684 096;
  • 24) 0.006 225 585 935 111 684 096 × 2 = 0 + 0.012 451 171 870 223 368 192;
  • 25) 0.012 451 171 870 223 368 192 × 2 = 0 + 0.024 902 343 740 446 736 384;
  • 26) 0.024 902 343 740 446 736 384 × 2 = 0 + 0.049 804 687 480 893 472 768;
  • 27) 0.049 804 687 480 893 472 768 × 2 = 0 + 0.099 609 374 961 786 945 536;
  • 28) 0.099 609 374 961 786 945 536 × 2 = 0 + 0.199 218 749 923 573 891 072;
  • 29) 0.199 218 749 923 573 891 072 × 2 = 0 + 0.398 437 499 847 147 782 144;
  • 30) 0.398 437 499 847 147 782 144 × 2 = 0 + 0.796 874 999 694 295 564 288;
  • 31) 0.796 874 999 694 295 564 288 × 2 = 1 + 0.593 749 999 388 591 128 576;
  • 32) 0.593 749 999 388 591 128 576 × 2 = 1 + 0.187 499 998 777 182 257 152;
  • 33) 0.187 499 998 777 182 257 152 × 2 = 0 + 0.374 999 997 554 364 514 304;
  • 34) 0.374 999 997 554 364 514 304 × 2 = 0 + 0.749 999 995 108 729 028 608;
  • 35) 0.749 999 995 108 729 028 608 × 2 = 1 + 0.499 999 990 217 458 057 216;
  • 36) 0.499 999 990 217 458 057 216 × 2 = 0 + 0.999 999 980 434 916 114 432;
  • 37) 0.999 999 980 434 916 114 432 × 2 = 1 + 0.999 999 960 869 832 228 864;
  • 38) 0.999 999 960 869 832 228 864 × 2 = 1 + 0.999 999 921 739 664 457 728;
  • 39) 0.999 999 921 739 664 457 728 × 2 = 1 + 0.999 999 843 479 328 915 456;
  • 40) 0.999 999 843 479 328 915 456 × 2 = 1 + 0.999 999 686 958 657 830 912;
  • 41) 0.999 999 686 958 657 830 912 × 2 = 1 + 0.999 999 373 917 315 661 824;
  • 42) 0.999 999 373 917 315 661 824 × 2 = 1 + 0.999 998 747 834 631 323 648;
  • 43) 0.999 998 747 834 631 323 648 × 2 = 1 + 0.999 997 495 669 262 647 296;
  • 44) 0.999 997 495 669 262 647 296 × 2 = 1 + 0.999 994 991 338 525 294 592;
  • 45) 0.999 994 991 338 525 294 592 × 2 = 1 + 0.999 989 982 677 050 589 184;
  • 46) 0.999 989 982 677 050 589 184 × 2 = 1 + 0.999 979 965 354 101 178 368;
  • 47) 0.999 979 965 354 101 178 368 × 2 = 1 + 0.999 959 930 708 202 356 736;
  • 48) 0.999 959 930 708 202 356 736 × 2 = 1 + 0.999 919 861 416 404 713 472;
  • 49) 0.999 919 861 416 404 713 472 × 2 = 1 + 0.999 839 722 832 809 426 944;
  • 50) 0.999 839 722 832 809 426 944 × 2 = 1 + 0.999 679 445 665 618 853 888;
  • 51) 0.999 679 445 665 618 853 888 × 2 = 1 + 0.999 358 891 331 237 707 776;
  • 52) 0.999 358 891 331 237 707 776 × 2 = 1 + 0.998 717 782 662 475 415 552;
  • 53) 0.998 717 782 662 475 415 552 × 2 = 1 + 0.997 435 565 324 950 831 104;
  • 54) 0.997 435 565 324 950 831 104 × 2 = 1 + 0.994 871 130 649 901 662 208;

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:


0.000 000 000 742 147 676 362(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 362(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:


0.000 000 000 742 147 676 362(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 -0.000 000 000 742 147 676 362 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