-0.000 000 000 742 147 675 52 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 52(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 52(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 52| = 0.000 000 000 742 147 675 52


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 52.

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 52 × 2 = 0 + 0.000 000 001 484 295 351 04;
  • 2) 0.000 000 001 484 295 351 04 × 2 = 0 + 0.000 000 002 968 590 702 08;
  • 3) 0.000 000 002 968 590 702 08 × 2 = 0 + 0.000 000 005 937 181 404 16;
  • 4) 0.000 000 005 937 181 404 16 × 2 = 0 + 0.000 000 011 874 362 808 32;
  • 5) 0.000 000 011 874 362 808 32 × 2 = 0 + 0.000 000 023 748 725 616 64;
  • 6) 0.000 000 023 748 725 616 64 × 2 = 0 + 0.000 000 047 497 451 233 28;
  • 7) 0.000 000 047 497 451 233 28 × 2 = 0 + 0.000 000 094 994 902 466 56;
  • 8) 0.000 000 094 994 902 466 56 × 2 = 0 + 0.000 000 189 989 804 933 12;
  • 9) 0.000 000 189 989 804 933 12 × 2 = 0 + 0.000 000 379 979 609 866 24;
  • 10) 0.000 000 379 979 609 866 24 × 2 = 0 + 0.000 000 759 959 219 732 48;
  • 11) 0.000 000 759 959 219 732 48 × 2 = 0 + 0.000 001 519 918 439 464 96;
  • 12) 0.000 001 519 918 439 464 96 × 2 = 0 + 0.000 003 039 836 878 929 92;
  • 13) 0.000 003 039 836 878 929 92 × 2 = 0 + 0.000 006 079 673 757 859 84;
  • 14) 0.000 006 079 673 757 859 84 × 2 = 0 + 0.000 012 159 347 515 719 68;
  • 15) 0.000 012 159 347 515 719 68 × 2 = 0 + 0.000 024 318 695 031 439 36;
  • 16) 0.000 024 318 695 031 439 36 × 2 = 0 + 0.000 048 637 390 062 878 72;
  • 17) 0.000 048 637 390 062 878 72 × 2 = 0 + 0.000 097 274 780 125 757 44;
  • 18) 0.000 097 274 780 125 757 44 × 2 = 0 + 0.000 194 549 560 251 514 88;
  • 19) 0.000 194 549 560 251 514 88 × 2 = 0 + 0.000 389 099 120 503 029 76;
  • 20) 0.000 389 099 120 503 029 76 × 2 = 0 + 0.000 778 198 241 006 059 52;
  • 21) 0.000 778 198 241 006 059 52 × 2 = 0 + 0.001 556 396 482 012 119 04;
  • 22) 0.001 556 396 482 012 119 04 × 2 = 0 + 0.003 112 792 964 024 238 08;
  • 23) 0.003 112 792 964 024 238 08 × 2 = 0 + 0.006 225 585 928 048 476 16;
  • 24) 0.006 225 585 928 048 476 16 × 2 = 0 + 0.012 451 171 856 096 952 32;
  • 25) 0.012 451 171 856 096 952 32 × 2 = 0 + 0.024 902 343 712 193 904 64;
  • 26) 0.024 902 343 712 193 904 64 × 2 = 0 + 0.049 804 687 424 387 809 28;
  • 27) 0.049 804 687 424 387 809 28 × 2 = 0 + 0.099 609 374 848 775 618 56;
  • 28) 0.099 609 374 848 775 618 56 × 2 = 0 + 0.199 218 749 697 551 237 12;
  • 29) 0.199 218 749 697 551 237 12 × 2 = 0 + 0.398 437 499 395 102 474 24;
  • 30) 0.398 437 499 395 102 474 24 × 2 = 0 + 0.796 874 998 790 204 948 48;
  • 31) 0.796 874 998 790 204 948 48 × 2 = 1 + 0.593 749 997 580 409 896 96;
  • 32) 0.593 749 997 580 409 896 96 × 2 = 1 + 0.187 499 995 160 819 793 92;
  • 33) 0.187 499 995 160 819 793 92 × 2 = 0 + 0.374 999 990 321 639 587 84;
  • 34) 0.374 999 990 321 639 587 84 × 2 = 0 + 0.749 999 980 643 279 175 68;
  • 35) 0.749 999 980 643 279 175 68 × 2 = 1 + 0.499 999 961 286 558 351 36;
  • 36) 0.499 999 961 286 558 351 36 × 2 = 0 + 0.999 999 922 573 116 702 72;
  • 37) 0.999 999 922 573 116 702 72 × 2 = 1 + 0.999 999 845 146 233 405 44;
  • 38) 0.999 999 845 146 233 405 44 × 2 = 1 + 0.999 999 690 292 466 810 88;
  • 39) 0.999 999 690 292 466 810 88 × 2 = 1 + 0.999 999 380 584 933 621 76;
  • 40) 0.999 999 380 584 933 621 76 × 2 = 1 + 0.999 998 761 169 867 243 52;
  • 41) 0.999 998 761 169 867 243 52 × 2 = 1 + 0.999 997 522 339 734 487 04;
  • 42) 0.999 997 522 339 734 487 04 × 2 = 1 + 0.999 995 044 679 468 974 08;
  • 43) 0.999 995 044 679 468 974 08 × 2 = 1 + 0.999 990 089 358 937 948 16;
  • 44) 0.999 990 089 358 937 948 16 × 2 = 1 + 0.999 980 178 717 875 896 32;
  • 45) 0.999 980 178 717 875 896 32 × 2 = 1 + 0.999 960 357 435 751 792 64;
  • 46) 0.999 960 357 435 751 792 64 × 2 = 1 + 0.999 920 714 871 503 585 28;
  • 47) 0.999 920 714 871 503 585 28 × 2 = 1 + 0.999 841 429 743 007 170 56;
  • 48) 0.999 841 429 743 007 170 56 × 2 = 1 + 0.999 682 859 486 014 341 12;
  • 49) 0.999 682 859 486 014 341 12 × 2 = 1 + 0.999 365 718 972 028 682 24;
  • 50) 0.999 365 718 972 028 682 24 × 2 = 1 + 0.998 731 437 944 057 364 48;
  • 51) 0.998 731 437 944 057 364 48 × 2 = 1 + 0.997 462 875 888 114 728 96;
  • 52) 0.997 462 875 888 114 728 96 × 2 = 1 + 0.994 925 751 776 229 457 92;
  • 53) 0.994 925 751 776 229 457 92 × 2 = 1 + 0.989 851 503 552 458 915 84;
  • 54) 0.989 851 503 552 458 915 84 × 2 = 1 + 0.979 703 007 104 917 831 68;

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 52(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 52(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 52(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 52 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