-0.000 282 005 913 26 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 913 26(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
-0.000 282 005 913 26(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:

|-0.000 282 005 913 26| = 0.000 282 005 913 26


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: 0.000 282 005 913 26.

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) 0.000 282 005 913 26 × 2 = 0 + 0.000 564 011 826 52;
  • 2) 0.000 564 011 826 52 × 2 = 0 + 0.001 128 023 653 04;
  • 3) 0.001 128 023 653 04 × 2 = 0 + 0.002 256 047 306 08;
  • 4) 0.002 256 047 306 08 × 2 = 0 + 0.004 512 094 612 16;
  • 5) 0.004 512 094 612 16 × 2 = 0 + 0.009 024 189 224 32;
  • 6) 0.009 024 189 224 32 × 2 = 0 + 0.018 048 378 448 64;
  • 7) 0.018 048 378 448 64 × 2 = 0 + 0.036 096 756 897 28;
  • 8) 0.036 096 756 897 28 × 2 = 0 + 0.072 193 513 794 56;
  • 9) 0.072 193 513 794 56 × 2 = 0 + 0.144 387 027 589 12;
  • 10) 0.144 387 027 589 12 × 2 = 0 + 0.288 774 055 178 24;
  • 11) 0.288 774 055 178 24 × 2 = 0 + 0.577 548 110 356 48;
  • 12) 0.577 548 110 356 48 × 2 = 1 + 0.155 096 220 712 96;
  • 13) 0.155 096 220 712 96 × 2 = 0 + 0.310 192 441 425 92;
  • 14) 0.310 192 441 425 92 × 2 = 0 + 0.620 384 882 851 84;
  • 15) 0.620 384 882 851 84 × 2 = 1 + 0.240 769 765 703 68;
  • 16) 0.240 769 765 703 68 × 2 = 0 + 0.481 539 531 407 36;
  • 17) 0.481 539 531 407 36 × 2 = 0 + 0.963 079 062 814 72;
  • 18) 0.963 079 062 814 72 × 2 = 1 + 0.926 158 125 629 44;
  • 19) 0.926 158 125 629 44 × 2 = 1 + 0.852 316 251 258 88;
  • 20) 0.852 316 251 258 88 × 2 = 1 + 0.704 632 502 517 76;
  • 21) 0.704 632 502 517 76 × 2 = 1 + 0.409 265 005 035 52;
  • 22) 0.409 265 005 035 52 × 2 = 0 + 0.818 530 010 071 04;
  • 23) 0.818 530 010 071 04 × 2 = 1 + 0.637 060 020 142 08;
  • 24) 0.637 060 020 142 08 × 2 = 1 + 0.274 120 040 284 16;
  • 25) 0.274 120 040 284 16 × 2 = 0 + 0.548 240 080 568 32;
  • 26) 0.548 240 080 568 32 × 2 = 1 + 0.096 480 161 136 64;
  • 27) 0.096 480 161 136 64 × 2 = 0 + 0.192 960 322 273 28;
  • 28) 0.192 960 322 273 28 × 2 = 0 + 0.385 920 644 546 56;
  • 29) 0.385 920 644 546 56 × 2 = 0 + 0.771 841 289 093 12;
  • 30) 0.771 841 289 093 12 × 2 = 1 + 0.543 682 578 186 24;
  • 31) 0.543 682 578 186 24 × 2 = 1 + 0.087 365 156 372 48;
  • 32) 0.087 365 156 372 48 × 2 = 0 + 0.174 730 312 744 96;
  • 33) 0.174 730 312 744 96 × 2 = 0 + 0.349 460 625 489 92;
  • 34) 0.349 460 625 489 92 × 2 = 0 + 0.698 921 250 979 84;
  • 35) 0.698 921 250 979 84 × 2 = 1 + 0.397 842 501 959 68;
  • 36) 0.397 842 501 959 68 × 2 = 0 + 0.795 685 003 919 36;
  • 37) 0.795 685 003 919 36 × 2 = 1 + 0.591 370 007 838 72;
  • 38) 0.591 370 007 838 72 × 2 = 1 + 0.182 740 015 677 44;
  • 39) 0.182 740 015 677 44 × 2 = 0 + 0.365 480 031 354 88;
  • 40) 0.365 480 031 354 88 × 2 = 0 + 0.730 960 062 709 76;
  • 41) 0.730 960 062 709 76 × 2 = 1 + 0.461 920 125 419 52;
  • 42) 0.461 920 125 419 52 × 2 = 0 + 0.923 840 250 839 04;
  • 43) 0.923 840 250 839 04 × 2 = 1 + 0.847 680 501 678 08;
  • 44) 0.847 680 501 678 08 × 2 = 1 + 0.695 361 003 356 16;
  • 45) 0.695 361 003 356 16 × 2 = 1 + 0.390 722 006 712 32;
  • 46) 0.390 722 006 712 32 × 2 = 0 + 0.781 444 013 424 64;
  • 47) 0.781 444 013 424 64 × 2 = 1 + 0.562 888 026 849 28;
  • 48) 0.562 888 026 849 28 × 2 = 1 + 0.125 776 053 698 56;
  • 49) 0.125 776 053 698 56 × 2 = 0 + 0.251 552 107 397 12;
  • 50) 0.251 552 107 397 12 × 2 = 0 + 0.503 104 214 794 24;
  • 51) 0.503 104 214 794 24 × 2 = 1 + 0.006 208 429 588 48;
  • 52) 0.006 208 429 588 48 × 2 = 0 + 0.012 416 859 176 96;
  • 53) 0.012 416 859 176 96 × 2 = 0 + 0.024 833 718 353 92;
  • 54) 0.024 833 718 353 92 × 2 = 0 + 0.049 667 436 707 84;
  • 55) 0.049 667 436 707 84 × 2 = 0 + 0.099 334 873 415 68;
  • 56) 0.099 334 873 415 68 × 2 = 0 + 0.198 669 746 831 36;
  • 57) 0.198 669 746 831 36 × 2 = 0 + 0.397 339 493 662 72;
  • 58) 0.397 339 493 662 72 × 2 = 0 + 0.794 678 987 325 44;
  • 59) 0.794 678 987 325 44 × 2 = 1 + 0.589 357 974 650 88;
  • 60) 0.589 357 974 650 88 × 2 = 1 + 0.178 715 949 301 76;
  • 61) 0.178 715 949 301 76 × 2 = 0 + 0.357 431 898 603 52;
  • 62) 0.357 431 898 603 52 × 2 = 0 + 0.714 863 797 207 04;
  • 63) 0.714 863 797 207 04 × 2 = 1 + 0.429 727 594 414 08;
  • 64) 0.429 727 594 414 08 × 2 = 0 + 0.859 455 188 828 16;

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:


0.000 282 005 913 26(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1011 1011 0010 0000 0011 0010(2)

6. Positive number before normalization:

0.000 282 005 913 26(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1011 1011 0010 0000 0011 0010(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:


0.000 282 005 913 26(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1011 1011 0010 0000 0011 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1100 1011 1011 0010 0000 0011 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1100 1011 1011 0010 0000 0011 0010(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 0010 1100 1011 1011 0010 0000 0011 0010


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 0010 1100 1011 1011 0010 0000 0011 0010 =

0010 0111 1011 0100 0110 0010 1100 1011 1011 0010 0000 0011 0010


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 0010 1100 1011 1011 0010 0000 0011 0010


Decimal number -0.000 282 005 913 26 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1100 1011 1011 0010 0000 0011 0010

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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