-0.000 000 000 742 147 677 3 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 677 3(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
-0.000 000 000 742 147 677 3(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:

|-0.000 000 000 742 147 677 3| = 0.000 000 000 742 147 677 3


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: 0.000 000 000 742 147 677 3.

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) 0.000 000 000 742 147 677 3 × 2 = 0 + 0.000 000 001 484 295 354 6;
  • 2) 0.000 000 001 484 295 354 6 × 2 = 0 + 0.000 000 002 968 590 709 2;
  • 3) 0.000 000 002 968 590 709 2 × 2 = 0 + 0.000 000 005 937 181 418 4;
  • 4) 0.000 000 005 937 181 418 4 × 2 = 0 + 0.000 000 011 874 362 836 8;
  • 5) 0.000 000 011 874 362 836 8 × 2 = 0 + 0.000 000 023 748 725 673 6;
  • 6) 0.000 000 023 748 725 673 6 × 2 = 0 + 0.000 000 047 497 451 347 2;
  • 7) 0.000 000 047 497 451 347 2 × 2 = 0 + 0.000 000 094 994 902 694 4;
  • 8) 0.000 000 094 994 902 694 4 × 2 = 0 + 0.000 000 189 989 805 388 8;
  • 9) 0.000 000 189 989 805 388 8 × 2 = 0 + 0.000 000 379 979 610 777 6;
  • 10) 0.000 000 379 979 610 777 6 × 2 = 0 + 0.000 000 759 959 221 555 2;
  • 11) 0.000 000 759 959 221 555 2 × 2 = 0 + 0.000 001 519 918 443 110 4;
  • 12) 0.000 001 519 918 443 110 4 × 2 = 0 + 0.000 003 039 836 886 220 8;
  • 13) 0.000 003 039 836 886 220 8 × 2 = 0 + 0.000 006 079 673 772 441 6;
  • 14) 0.000 006 079 673 772 441 6 × 2 = 0 + 0.000 012 159 347 544 883 2;
  • 15) 0.000 012 159 347 544 883 2 × 2 = 0 + 0.000 024 318 695 089 766 4;
  • 16) 0.000 024 318 695 089 766 4 × 2 = 0 + 0.000 048 637 390 179 532 8;
  • 17) 0.000 048 637 390 179 532 8 × 2 = 0 + 0.000 097 274 780 359 065 6;
  • 18) 0.000 097 274 780 359 065 6 × 2 = 0 + 0.000 194 549 560 718 131 2;
  • 19) 0.000 194 549 560 718 131 2 × 2 = 0 + 0.000 389 099 121 436 262 4;
  • 20) 0.000 389 099 121 436 262 4 × 2 = 0 + 0.000 778 198 242 872 524 8;
  • 21) 0.000 778 198 242 872 524 8 × 2 = 0 + 0.001 556 396 485 745 049 6;
  • 22) 0.001 556 396 485 745 049 6 × 2 = 0 + 0.003 112 792 971 490 099 2;
  • 23) 0.003 112 792 971 490 099 2 × 2 = 0 + 0.006 225 585 942 980 198 4;
  • 24) 0.006 225 585 942 980 198 4 × 2 = 0 + 0.012 451 171 885 960 396 8;
  • 25) 0.012 451 171 885 960 396 8 × 2 = 0 + 0.024 902 343 771 920 793 6;
  • 26) 0.024 902 343 771 920 793 6 × 2 = 0 + 0.049 804 687 543 841 587 2;
  • 27) 0.049 804 687 543 841 587 2 × 2 = 0 + 0.099 609 375 087 683 174 4;
  • 28) 0.099 609 375 087 683 174 4 × 2 = 0 + 0.199 218 750 175 366 348 8;
  • 29) 0.199 218 750 175 366 348 8 × 2 = 0 + 0.398 437 500 350 732 697 6;
  • 30) 0.398 437 500 350 732 697 6 × 2 = 0 + 0.796 875 000 701 465 395 2;
  • 31) 0.796 875 000 701 465 395 2 × 2 = 1 + 0.593 750 001 402 930 790 4;
  • 32) 0.593 750 001 402 930 790 4 × 2 = 1 + 0.187 500 002 805 861 580 8;
  • 33) 0.187 500 002 805 861 580 8 × 2 = 0 + 0.375 000 005 611 723 161 6;
  • 34) 0.375 000 005 611 723 161 6 × 2 = 0 + 0.750 000 011 223 446 323 2;
  • 35) 0.750 000 011 223 446 323 2 × 2 = 1 + 0.500 000 022 446 892 646 4;
  • 36) 0.500 000 022 446 892 646 4 × 2 = 1 + 0.000 000 044 893 785 292 8;
  • 37) 0.000 000 044 893 785 292 8 × 2 = 0 + 0.000 000 089 787 570 585 6;
  • 38) 0.000 000 089 787 570 585 6 × 2 = 0 + 0.000 000 179 575 141 171 2;
  • 39) 0.000 000 179 575 141 171 2 × 2 = 0 + 0.000 000 359 150 282 342 4;
  • 40) 0.000 000 359 150 282 342 4 × 2 = 0 + 0.000 000 718 300 564 684 8;
  • 41) 0.000 000 718 300 564 684 8 × 2 = 0 + 0.000 001 436 601 129 369 6;
  • 42) 0.000 001 436 601 129 369 6 × 2 = 0 + 0.000 002 873 202 258 739 2;
  • 43) 0.000 002 873 202 258 739 2 × 2 = 0 + 0.000 005 746 404 517 478 4;
  • 44) 0.000 005 746 404 517 478 4 × 2 = 0 + 0.000 011 492 809 034 956 8;
  • 45) 0.000 011 492 809 034 956 8 × 2 = 0 + 0.000 022 985 618 069 913 6;
  • 46) 0.000 022 985 618 069 913 6 × 2 = 0 + 0.000 045 971 236 139 827 2;
  • 47) 0.000 045 971 236 139 827 2 × 2 = 0 + 0.000 091 942 472 279 654 4;
  • 48) 0.000 091 942 472 279 654 4 × 2 = 0 + 0.000 183 884 944 559 308 8;
  • 49) 0.000 183 884 944 559 308 8 × 2 = 0 + 0.000 367 769 889 118 617 6;
  • 50) 0.000 367 769 889 118 617 6 × 2 = 0 + 0.000 735 539 778 237 235 2;
  • 51) 0.000 735 539 778 237 235 2 × 2 = 0 + 0.001 471 079 556 474 470 4;
  • 52) 0.001 471 079 556 474 470 4 × 2 = 0 + 0.002 942 159 112 948 940 8;
  • 53) 0.002 942 159 112 948 940 8 × 2 = 0 + 0.005 884 318 225 897 881 6;
  • 54) 0.005 884 318 225 897 881 6 × 2 = 0 + 0.011 768 636 451 795 763 2;

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:


0.000 000 000 742 147 677 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 677 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 677 3(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 677 3 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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