-0.000 282 005 899 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 899 1(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
-0.000 282 005 899 1(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 282 005 899 1| = 0.000 282 005 899 1


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: 0.000 282 005 899 1.

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) 0.000 282 005 899 1 × 2 = 0 + 0.000 564 011 798 2;
  • 2) 0.000 564 011 798 2 × 2 = 0 + 0.001 128 023 596 4;
  • 3) 0.001 128 023 596 4 × 2 = 0 + 0.002 256 047 192 8;
  • 4) 0.002 256 047 192 8 × 2 = 0 + 0.004 512 094 385 6;
  • 5) 0.004 512 094 385 6 × 2 = 0 + 0.009 024 188 771 2;
  • 6) 0.009 024 188 771 2 × 2 = 0 + 0.018 048 377 542 4;
  • 7) 0.018 048 377 542 4 × 2 = 0 + 0.036 096 755 084 8;
  • 8) 0.036 096 755 084 8 × 2 = 0 + 0.072 193 510 169 6;
  • 9) 0.072 193 510 169 6 × 2 = 0 + 0.144 387 020 339 2;
  • 10) 0.144 387 020 339 2 × 2 = 0 + 0.288 774 040 678 4;
  • 11) 0.288 774 040 678 4 × 2 = 0 + 0.577 548 081 356 8;
  • 12) 0.577 548 081 356 8 × 2 = 1 + 0.155 096 162 713 6;
  • 13) 0.155 096 162 713 6 × 2 = 0 + 0.310 192 325 427 2;
  • 14) 0.310 192 325 427 2 × 2 = 0 + 0.620 384 650 854 4;
  • 15) 0.620 384 650 854 4 × 2 = 1 + 0.240 769 301 708 8;
  • 16) 0.240 769 301 708 8 × 2 = 0 + 0.481 538 603 417 6;
  • 17) 0.481 538 603 417 6 × 2 = 0 + 0.963 077 206 835 2;
  • 18) 0.963 077 206 835 2 × 2 = 1 + 0.926 154 413 670 4;
  • 19) 0.926 154 413 670 4 × 2 = 1 + 0.852 308 827 340 8;
  • 20) 0.852 308 827 340 8 × 2 = 1 + 0.704 617 654 681 6;
  • 21) 0.704 617 654 681 6 × 2 = 1 + 0.409 235 309 363 2;
  • 22) 0.409 235 309 363 2 × 2 = 0 + 0.818 470 618 726 4;
  • 23) 0.818 470 618 726 4 × 2 = 1 + 0.636 941 237 452 8;
  • 24) 0.636 941 237 452 8 × 2 = 1 + 0.273 882 474 905 6;
  • 25) 0.273 882 474 905 6 × 2 = 0 + 0.547 764 949 811 2;
  • 26) 0.547 764 949 811 2 × 2 = 1 + 0.095 529 899 622 4;
  • 27) 0.095 529 899 622 4 × 2 = 0 + 0.191 059 799 244 8;
  • 28) 0.191 059 799 244 8 × 2 = 0 + 0.382 119 598 489 6;
  • 29) 0.382 119 598 489 6 × 2 = 0 + 0.764 239 196 979 2;
  • 30) 0.764 239 196 979 2 × 2 = 1 + 0.528 478 393 958 4;
  • 31) 0.528 478 393 958 4 × 2 = 1 + 0.056 956 787 916 8;
  • 32) 0.056 956 787 916 8 × 2 = 0 + 0.113 913 575 833 6;
  • 33) 0.113 913 575 833 6 × 2 = 0 + 0.227 827 151 667 2;
  • 34) 0.227 827 151 667 2 × 2 = 0 + 0.455 654 303 334 4;
  • 35) 0.455 654 303 334 4 × 2 = 0 + 0.911 308 606 668 8;
  • 36) 0.911 308 606 668 8 × 2 = 1 + 0.822 617 213 337 6;
  • 37) 0.822 617 213 337 6 × 2 = 1 + 0.645 234 426 675 2;
  • 38) 0.645 234 426 675 2 × 2 = 1 + 0.290 468 853 350 4;
  • 39) 0.290 468 853 350 4 × 2 = 0 + 0.580 937 706 700 8;
  • 40) 0.580 937 706 700 8 × 2 = 1 + 0.161 875 413 401 6;
  • 41) 0.161 875 413 401 6 × 2 = 0 + 0.323 750 826 803 2;
  • 42) 0.323 750 826 803 2 × 2 = 0 + 0.647 501 653 606 4;
  • 43) 0.647 501 653 606 4 × 2 = 1 + 0.295 003 307 212 8;
  • 44) 0.295 003 307 212 8 × 2 = 0 + 0.590 006 614 425 6;
  • 45) 0.590 006 614 425 6 × 2 = 1 + 0.180 013 228 851 2;
  • 46) 0.180 013 228 851 2 × 2 = 0 + 0.360 026 457 702 4;
  • 47) 0.360 026 457 702 4 × 2 = 0 + 0.720 052 915 404 8;
  • 48) 0.720 052 915 404 8 × 2 = 1 + 0.440 105 830 809 6;
  • 49) 0.440 105 830 809 6 × 2 = 0 + 0.880 211 661 619 2;
  • 50) 0.880 211 661 619 2 × 2 = 1 + 0.760 423 323 238 4;
  • 51) 0.760 423 323 238 4 × 2 = 1 + 0.520 846 646 476 8;
  • 52) 0.520 846 646 476 8 × 2 = 1 + 0.041 693 292 953 6;
  • 53) 0.041 693 292 953 6 × 2 = 0 + 0.083 386 585 907 2;
  • 54) 0.083 386 585 907 2 × 2 = 0 + 0.166 773 171 814 4;
  • 55) 0.166 773 171 814 4 × 2 = 0 + 0.333 546 343 628 8;
  • 56) 0.333 546 343 628 8 × 2 = 0 + 0.667 092 687 257 6;
  • 57) 0.667 092 687 257 6 × 2 = 1 + 0.334 185 374 515 2;
  • 58) 0.334 185 374 515 2 × 2 = 0 + 0.668 370 749 030 4;
  • 59) 0.668 370 749 030 4 × 2 = 1 + 0.336 741 498 060 8;
  • 60) 0.336 741 498 060 8 × 2 = 0 + 0.673 482 996 121 6;
  • 61) 0.673 482 996 121 6 × 2 = 1 + 0.346 965 992 243 2;
  • 62) 0.346 965 992 243 2 × 2 = 0 + 0.693 931 984 486 4;
  • 63) 0.693 931 984 486 4 × 2 = 1 + 0.387 863 968 972 8;
  • 64) 0.387 863 968 972 8 × 2 = 0 + 0.775 727 937 945 6;

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:


0.000 282 005 899 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010(2)

6. Positive number before normalization:

0.000 282 005 899 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 282 005 899 1(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010(2) × 2-12


8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -12

Mantissa (not normalized):
1.0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(10)


10. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 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) =

1011(10) =

011 1111 0011(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 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010 =

0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 0011

Mantissa (52 bits) =
0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010


Decimal number -0.000 282 005 899 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0001 1101 0010 1001 0111 0000 1010 1010

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double 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 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 from the previous 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 multiplying operations, starting from the top of the list constructed 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, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' 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 -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 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:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. 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) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) 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:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100