-0.000 000 000 742 147 675 47 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 47(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
-0.000 000 000 742 147 675 47(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:

|-0.000 000 000 742 147 675 47| = 0.000 000 000 742 147 675 47


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: 0.000 000 000 742 147 675 47.

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) 0.000 000 000 742 147 675 47 × 2 = 0 + 0.000 000 001 484 295 350 94;
  • 2) 0.000 000 001 484 295 350 94 × 2 = 0 + 0.000 000 002 968 590 701 88;
  • 3) 0.000 000 002 968 590 701 88 × 2 = 0 + 0.000 000 005 937 181 403 76;
  • 4) 0.000 000 005 937 181 403 76 × 2 = 0 + 0.000 000 011 874 362 807 52;
  • 5) 0.000 000 011 874 362 807 52 × 2 = 0 + 0.000 000 023 748 725 615 04;
  • 6) 0.000 000 023 748 725 615 04 × 2 = 0 + 0.000 000 047 497 451 230 08;
  • 7) 0.000 000 047 497 451 230 08 × 2 = 0 + 0.000 000 094 994 902 460 16;
  • 8) 0.000 000 094 994 902 460 16 × 2 = 0 + 0.000 000 189 989 804 920 32;
  • 9) 0.000 000 189 989 804 920 32 × 2 = 0 + 0.000 000 379 979 609 840 64;
  • 10) 0.000 000 379 979 609 840 64 × 2 = 0 + 0.000 000 759 959 219 681 28;
  • 11) 0.000 000 759 959 219 681 28 × 2 = 0 + 0.000 001 519 918 439 362 56;
  • 12) 0.000 001 519 918 439 362 56 × 2 = 0 + 0.000 003 039 836 878 725 12;
  • 13) 0.000 003 039 836 878 725 12 × 2 = 0 + 0.000 006 079 673 757 450 24;
  • 14) 0.000 006 079 673 757 450 24 × 2 = 0 + 0.000 012 159 347 514 900 48;
  • 15) 0.000 012 159 347 514 900 48 × 2 = 0 + 0.000 024 318 695 029 800 96;
  • 16) 0.000 024 318 695 029 800 96 × 2 = 0 + 0.000 048 637 390 059 601 92;
  • 17) 0.000 048 637 390 059 601 92 × 2 = 0 + 0.000 097 274 780 119 203 84;
  • 18) 0.000 097 274 780 119 203 84 × 2 = 0 + 0.000 194 549 560 238 407 68;
  • 19) 0.000 194 549 560 238 407 68 × 2 = 0 + 0.000 389 099 120 476 815 36;
  • 20) 0.000 389 099 120 476 815 36 × 2 = 0 + 0.000 778 198 240 953 630 72;
  • 21) 0.000 778 198 240 953 630 72 × 2 = 0 + 0.001 556 396 481 907 261 44;
  • 22) 0.001 556 396 481 907 261 44 × 2 = 0 + 0.003 112 792 963 814 522 88;
  • 23) 0.003 112 792 963 814 522 88 × 2 = 0 + 0.006 225 585 927 629 045 76;
  • 24) 0.006 225 585 927 629 045 76 × 2 = 0 + 0.012 451 171 855 258 091 52;
  • 25) 0.012 451 171 855 258 091 52 × 2 = 0 + 0.024 902 343 710 516 183 04;
  • 26) 0.024 902 343 710 516 183 04 × 2 = 0 + 0.049 804 687 421 032 366 08;
  • 27) 0.049 804 687 421 032 366 08 × 2 = 0 + 0.099 609 374 842 064 732 16;
  • 28) 0.099 609 374 842 064 732 16 × 2 = 0 + 0.199 218 749 684 129 464 32;
  • 29) 0.199 218 749 684 129 464 32 × 2 = 0 + 0.398 437 499 368 258 928 64;
  • 30) 0.398 437 499 368 258 928 64 × 2 = 0 + 0.796 874 998 736 517 857 28;
  • 31) 0.796 874 998 736 517 857 28 × 2 = 1 + 0.593 749 997 473 035 714 56;
  • 32) 0.593 749 997 473 035 714 56 × 2 = 1 + 0.187 499 994 946 071 429 12;
  • 33) 0.187 499 994 946 071 429 12 × 2 = 0 + 0.374 999 989 892 142 858 24;
  • 34) 0.374 999 989 892 142 858 24 × 2 = 0 + 0.749 999 979 784 285 716 48;
  • 35) 0.749 999 979 784 285 716 48 × 2 = 1 + 0.499 999 959 568 571 432 96;
  • 36) 0.499 999 959 568 571 432 96 × 2 = 0 + 0.999 999 919 137 142 865 92;
  • 37) 0.999 999 919 137 142 865 92 × 2 = 1 + 0.999 999 838 274 285 731 84;
  • 38) 0.999 999 838 274 285 731 84 × 2 = 1 + 0.999 999 676 548 571 463 68;
  • 39) 0.999 999 676 548 571 463 68 × 2 = 1 + 0.999 999 353 097 142 927 36;
  • 40) 0.999 999 353 097 142 927 36 × 2 = 1 + 0.999 998 706 194 285 854 72;
  • 41) 0.999 998 706 194 285 854 72 × 2 = 1 + 0.999 997 412 388 571 709 44;
  • 42) 0.999 997 412 388 571 709 44 × 2 = 1 + 0.999 994 824 777 143 418 88;
  • 43) 0.999 994 824 777 143 418 88 × 2 = 1 + 0.999 989 649 554 286 837 76;
  • 44) 0.999 989 649 554 286 837 76 × 2 = 1 + 0.999 979 299 108 573 675 52;
  • 45) 0.999 979 299 108 573 675 52 × 2 = 1 + 0.999 958 598 217 147 351 04;
  • 46) 0.999 958 598 217 147 351 04 × 2 = 1 + 0.999 917 196 434 294 702 08;
  • 47) 0.999 917 196 434 294 702 08 × 2 = 1 + 0.999 834 392 868 589 404 16;
  • 48) 0.999 834 392 868 589 404 16 × 2 = 1 + 0.999 668 785 737 178 808 32;
  • 49) 0.999 668 785 737 178 808 32 × 2 = 1 + 0.999 337 571 474 357 616 64;
  • 50) 0.999 337 571 474 357 616 64 × 2 = 1 + 0.998 675 142 948 715 233 28;
  • 51) 0.998 675 142 948 715 233 28 × 2 = 1 + 0.997 350 285 897 430 466 56;
  • 52) 0.997 350 285 897 430 466 56 × 2 = 1 + 0.994 700 571 794 860 933 12;
  • 53) 0.994 700 571 794 860 933 12 × 2 = 1 + 0.989 401 143 589 721 866 24;
  • 54) 0.989 401 143 589 721 866 24 × 2 = 1 + 0.978 802 287 179 443 732 48;

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:


0.000 000 000 742 147 675 47(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 47(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:


0.000 000 000 742 147 675 47(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 -0.000 000 000 742 147 675 47 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 675 52 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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