-0.000 000 000 742 147 676 005 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 005(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
-0.000 000 000 742 147 676 005(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:

|-0.000 000 000 742 147 676 005| = 0.000 000 000 742 147 676 005


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: 0.000 000 000 742 147 676 005.

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) 0.000 000 000 742 147 676 005 × 2 = 0 + 0.000 000 001 484 295 352 01;
  • 2) 0.000 000 001 484 295 352 01 × 2 = 0 + 0.000 000 002 968 590 704 02;
  • 3) 0.000 000 002 968 590 704 02 × 2 = 0 + 0.000 000 005 937 181 408 04;
  • 4) 0.000 000 005 937 181 408 04 × 2 = 0 + 0.000 000 011 874 362 816 08;
  • 5) 0.000 000 011 874 362 816 08 × 2 = 0 + 0.000 000 023 748 725 632 16;
  • 6) 0.000 000 023 748 725 632 16 × 2 = 0 + 0.000 000 047 497 451 264 32;
  • 7) 0.000 000 047 497 451 264 32 × 2 = 0 + 0.000 000 094 994 902 528 64;
  • 8) 0.000 000 094 994 902 528 64 × 2 = 0 + 0.000 000 189 989 805 057 28;
  • 9) 0.000 000 189 989 805 057 28 × 2 = 0 + 0.000 000 379 979 610 114 56;
  • 10) 0.000 000 379 979 610 114 56 × 2 = 0 + 0.000 000 759 959 220 229 12;
  • 11) 0.000 000 759 959 220 229 12 × 2 = 0 + 0.000 001 519 918 440 458 24;
  • 12) 0.000 001 519 918 440 458 24 × 2 = 0 + 0.000 003 039 836 880 916 48;
  • 13) 0.000 003 039 836 880 916 48 × 2 = 0 + 0.000 006 079 673 761 832 96;
  • 14) 0.000 006 079 673 761 832 96 × 2 = 0 + 0.000 012 159 347 523 665 92;
  • 15) 0.000 012 159 347 523 665 92 × 2 = 0 + 0.000 024 318 695 047 331 84;
  • 16) 0.000 024 318 695 047 331 84 × 2 = 0 + 0.000 048 637 390 094 663 68;
  • 17) 0.000 048 637 390 094 663 68 × 2 = 0 + 0.000 097 274 780 189 327 36;
  • 18) 0.000 097 274 780 189 327 36 × 2 = 0 + 0.000 194 549 560 378 654 72;
  • 19) 0.000 194 549 560 378 654 72 × 2 = 0 + 0.000 389 099 120 757 309 44;
  • 20) 0.000 389 099 120 757 309 44 × 2 = 0 + 0.000 778 198 241 514 618 88;
  • 21) 0.000 778 198 241 514 618 88 × 2 = 0 + 0.001 556 396 483 029 237 76;
  • 22) 0.001 556 396 483 029 237 76 × 2 = 0 + 0.003 112 792 966 058 475 52;
  • 23) 0.003 112 792 966 058 475 52 × 2 = 0 + 0.006 225 585 932 116 951 04;
  • 24) 0.006 225 585 932 116 951 04 × 2 = 0 + 0.012 451 171 864 233 902 08;
  • 25) 0.012 451 171 864 233 902 08 × 2 = 0 + 0.024 902 343 728 467 804 16;
  • 26) 0.024 902 343 728 467 804 16 × 2 = 0 + 0.049 804 687 456 935 608 32;
  • 27) 0.049 804 687 456 935 608 32 × 2 = 0 + 0.099 609 374 913 871 216 64;
  • 28) 0.099 609 374 913 871 216 64 × 2 = 0 + 0.199 218 749 827 742 433 28;
  • 29) 0.199 218 749 827 742 433 28 × 2 = 0 + 0.398 437 499 655 484 866 56;
  • 30) 0.398 437 499 655 484 866 56 × 2 = 0 + 0.796 874 999 310 969 733 12;
  • 31) 0.796 874 999 310 969 733 12 × 2 = 1 + 0.593 749 998 621 939 466 24;
  • 32) 0.593 749 998 621 939 466 24 × 2 = 1 + 0.187 499 997 243 878 932 48;
  • 33) 0.187 499 997 243 878 932 48 × 2 = 0 + 0.374 999 994 487 757 864 96;
  • 34) 0.374 999 994 487 757 864 96 × 2 = 0 + 0.749 999 988 975 515 729 92;
  • 35) 0.749 999 988 975 515 729 92 × 2 = 1 + 0.499 999 977 951 031 459 84;
  • 36) 0.499 999 977 951 031 459 84 × 2 = 0 + 0.999 999 955 902 062 919 68;
  • 37) 0.999 999 955 902 062 919 68 × 2 = 1 + 0.999 999 911 804 125 839 36;
  • 38) 0.999 999 911 804 125 839 36 × 2 = 1 + 0.999 999 823 608 251 678 72;
  • 39) 0.999 999 823 608 251 678 72 × 2 = 1 + 0.999 999 647 216 503 357 44;
  • 40) 0.999 999 647 216 503 357 44 × 2 = 1 + 0.999 999 294 433 006 714 88;
  • 41) 0.999 999 294 433 006 714 88 × 2 = 1 + 0.999 998 588 866 013 429 76;
  • 42) 0.999 998 588 866 013 429 76 × 2 = 1 + 0.999 997 177 732 026 859 52;
  • 43) 0.999 997 177 732 026 859 52 × 2 = 1 + 0.999 994 355 464 053 719 04;
  • 44) 0.999 994 355 464 053 719 04 × 2 = 1 + 0.999 988 710 928 107 438 08;
  • 45) 0.999 988 710 928 107 438 08 × 2 = 1 + 0.999 977 421 856 214 876 16;
  • 46) 0.999 977 421 856 214 876 16 × 2 = 1 + 0.999 954 843 712 429 752 32;
  • 47) 0.999 954 843 712 429 752 32 × 2 = 1 + 0.999 909 687 424 859 504 64;
  • 48) 0.999 909 687 424 859 504 64 × 2 = 1 + 0.999 819 374 849 719 009 28;
  • 49) 0.999 819 374 849 719 009 28 × 2 = 1 + 0.999 638 749 699 438 018 56;
  • 50) 0.999 638 749 699 438 018 56 × 2 = 1 + 0.999 277 499 398 876 037 12;
  • 51) 0.999 277 499 398 876 037 12 × 2 = 1 + 0.998 554 998 797 752 074 24;
  • 52) 0.998 554 998 797 752 074 24 × 2 = 1 + 0.997 109 997 595 504 148 48;
  • 53) 0.997 109 997 595 504 148 48 × 2 = 1 + 0.994 219 995 191 008 296 96;
  • 54) 0.994 219 995 191 008 296 96 × 2 = 1 + 0.988 439 990 382 016 593 92;

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:


0.000 000 000 742 147 676 005(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 005(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:


0.000 000 000 742 147 676 005(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 -0.000 000 000 742 147 676 005 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