-0.000 000 000 742 147 676 36 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 36(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 36(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 36| = 0.000 000 000 742 147 676 36


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 36.

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 36 × 2 = 0 + 0.000 000 001 484 295 352 72;
  • 2) 0.000 000 001 484 295 352 72 × 2 = 0 + 0.000 000 002 968 590 705 44;
  • 3) 0.000 000 002 968 590 705 44 × 2 = 0 + 0.000 000 005 937 181 410 88;
  • 4) 0.000 000 005 937 181 410 88 × 2 = 0 + 0.000 000 011 874 362 821 76;
  • 5) 0.000 000 011 874 362 821 76 × 2 = 0 + 0.000 000 023 748 725 643 52;
  • 6) 0.000 000 023 748 725 643 52 × 2 = 0 + 0.000 000 047 497 451 287 04;
  • 7) 0.000 000 047 497 451 287 04 × 2 = 0 + 0.000 000 094 994 902 574 08;
  • 8) 0.000 000 094 994 902 574 08 × 2 = 0 + 0.000 000 189 989 805 148 16;
  • 9) 0.000 000 189 989 805 148 16 × 2 = 0 + 0.000 000 379 979 610 296 32;
  • 10) 0.000 000 379 979 610 296 32 × 2 = 0 + 0.000 000 759 959 220 592 64;
  • 11) 0.000 000 759 959 220 592 64 × 2 = 0 + 0.000 001 519 918 441 185 28;
  • 12) 0.000 001 519 918 441 185 28 × 2 = 0 + 0.000 003 039 836 882 370 56;
  • 13) 0.000 003 039 836 882 370 56 × 2 = 0 + 0.000 006 079 673 764 741 12;
  • 14) 0.000 006 079 673 764 741 12 × 2 = 0 + 0.000 012 159 347 529 482 24;
  • 15) 0.000 012 159 347 529 482 24 × 2 = 0 + 0.000 024 318 695 058 964 48;
  • 16) 0.000 024 318 695 058 964 48 × 2 = 0 + 0.000 048 637 390 117 928 96;
  • 17) 0.000 048 637 390 117 928 96 × 2 = 0 + 0.000 097 274 780 235 857 92;
  • 18) 0.000 097 274 780 235 857 92 × 2 = 0 + 0.000 194 549 560 471 715 84;
  • 19) 0.000 194 549 560 471 715 84 × 2 = 0 + 0.000 389 099 120 943 431 68;
  • 20) 0.000 389 099 120 943 431 68 × 2 = 0 + 0.000 778 198 241 886 863 36;
  • 21) 0.000 778 198 241 886 863 36 × 2 = 0 + 0.001 556 396 483 773 726 72;
  • 22) 0.001 556 396 483 773 726 72 × 2 = 0 + 0.003 112 792 967 547 453 44;
  • 23) 0.003 112 792 967 547 453 44 × 2 = 0 + 0.006 225 585 935 094 906 88;
  • 24) 0.006 225 585 935 094 906 88 × 2 = 0 + 0.012 451 171 870 189 813 76;
  • 25) 0.012 451 171 870 189 813 76 × 2 = 0 + 0.024 902 343 740 379 627 52;
  • 26) 0.024 902 343 740 379 627 52 × 2 = 0 + 0.049 804 687 480 759 255 04;
  • 27) 0.049 804 687 480 759 255 04 × 2 = 0 + 0.099 609 374 961 518 510 08;
  • 28) 0.099 609 374 961 518 510 08 × 2 = 0 + 0.199 218 749 923 037 020 16;
  • 29) 0.199 218 749 923 037 020 16 × 2 = 0 + 0.398 437 499 846 074 040 32;
  • 30) 0.398 437 499 846 074 040 32 × 2 = 0 + 0.796 874 999 692 148 080 64;
  • 31) 0.796 874 999 692 148 080 64 × 2 = 1 + 0.593 749 999 384 296 161 28;
  • 32) 0.593 749 999 384 296 161 28 × 2 = 1 + 0.187 499 998 768 592 322 56;
  • 33) 0.187 499 998 768 592 322 56 × 2 = 0 + 0.374 999 997 537 184 645 12;
  • 34) 0.374 999 997 537 184 645 12 × 2 = 0 + 0.749 999 995 074 369 290 24;
  • 35) 0.749 999 995 074 369 290 24 × 2 = 1 + 0.499 999 990 148 738 580 48;
  • 36) 0.499 999 990 148 738 580 48 × 2 = 0 + 0.999 999 980 297 477 160 96;
  • 37) 0.999 999 980 297 477 160 96 × 2 = 1 + 0.999 999 960 594 954 321 92;
  • 38) 0.999 999 960 594 954 321 92 × 2 = 1 + 0.999 999 921 189 908 643 84;
  • 39) 0.999 999 921 189 908 643 84 × 2 = 1 + 0.999 999 842 379 817 287 68;
  • 40) 0.999 999 842 379 817 287 68 × 2 = 1 + 0.999 999 684 759 634 575 36;
  • 41) 0.999 999 684 759 634 575 36 × 2 = 1 + 0.999 999 369 519 269 150 72;
  • 42) 0.999 999 369 519 269 150 72 × 2 = 1 + 0.999 998 739 038 538 301 44;
  • 43) 0.999 998 739 038 538 301 44 × 2 = 1 + 0.999 997 478 077 076 602 88;
  • 44) 0.999 997 478 077 076 602 88 × 2 = 1 + 0.999 994 956 154 153 205 76;
  • 45) 0.999 994 956 154 153 205 76 × 2 = 1 + 0.999 989 912 308 306 411 52;
  • 46) 0.999 989 912 308 306 411 52 × 2 = 1 + 0.999 979 824 616 612 823 04;
  • 47) 0.999 979 824 616 612 823 04 × 2 = 1 + 0.999 959 649 233 225 646 08;
  • 48) 0.999 959 649 233 225 646 08 × 2 = 1 + 0.999 919 298 466 451 292 16;
  • 49) 0.999 919 298 466 451 292 16 × 2 = 1 + 0.999 838 596 932 902 584 32;
  • 50) 0.999 838 596 932 902 584 32 × 2 = 1 + 0.999 677 193 865 805 168 64;
  • 51) 0.999 677 193 865 805 168 64 × 2 = 1 + 0.999 354 387 731 610 337 28;
  • 52) 0.999 354 387 731 610 337 28 × 2 = 1 + 0.998 708 775 463 220 674 56;
  • 53) 0.998 708 775 463 220 674 56 × 2 = 1 + 0.997 417 550 926 441 349 12;
  • 54) 0.997 417 550 926 441 349 12 × 2 = 1 + 0.994 835 101 852 882 698 24;

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 36(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 36(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 36(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 36 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