-0.000 282 006 006 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 006(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
-0.000 282 006 006(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:

|-0.000 282 006 006| = 0.000 282 006 006


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: 0.000 282 006 006.

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) 0.000 282 006 006 × 2 = 0 + 0.000 564 012 012;
  • 2) 0.000 564 012 012 × 2 = 0 + 0.001 128 024 024;
  • 3) 0.001 128 024 024 × 2 = 0 + 0.002 256 048 048;
  • 4) 0.002 256 048 048 × 2 = 0 + 0.004 512 096 096;
  • 5) 0.004 512 096 096 × 2 = 0 + 0.009 024 192 192;
  • 6) 0.009 024 192 192 × 2 = 0 + 0.018 048 384 384;
  • 7) 0.018 048 384 384 × 2 = 0 + 0.036 096 768 768;
  • 8) 0.036 096 768 768 × 2 = 0 + 0.072 193 537 536;
  • 9) 0.072 193 537 536 × 2 = 0 + 0.144 387 075 072;
  • 10) 0.144 387 075 072 × 2 = 0 + 0.288 774 150 144;
  • 11) 0.288 774 150 144 × 2 = 0 + 0.577 548 300 288;
  • 12) 0.577 548 300 288 × 2 = 1 + 0.155 096 600 576;
  • 13) 0.155 096 600 576 × 2 = 0 + 0.310 193 201 152;
  • 14) 0.310 193 201 152 × 2 = 0 + 0.620 386 402 304;
  • 15) 0.620 386 402 304 × 2 = 1 + 0.240 772 804 608;
  • 16) 0.240 772 804 608 × 2 = 0 + 0.481 545 609 216;
  • 17) 0.481 545 609 216 × 2 = 0 + 0.963 091 218 432;
  • 18) 0.963 091 218 432 × 2 = 1 + 0.926 182 436 864;
  • 19) 0.926 182 436 864 × 2 = 1 + 0.852 364 873 728;
  • 20) 0.852 364 873 728 × 2 = 1 + 0.704 729 747 456;
  • 21) 0.704 729 747 456 × 2 = 1 + 0.409 459 494 912;
  • 22) 0.409 459 494 912 × 2 = 0 + 0.818 918 989 824;
  • 23) 0.818 918 989 824 × 2 = 1 + 0.637 837 979 648;
  • 24) 0.637 837 979 648 × 2 = 1 + 0.275 675 959 296;
  • 25) 0.275 675 959 296 × 2 = 0 + 0.551 351 918 592;
  • 26) 0.551 351 918 592 × 2 = 1 + 0.102 703 837 184;
  • 27) 0.102 703 837 184 × 2 = 0 + 0.205 407 674 368;
  • 28) 0.205 407 674 368 × 2 = 0 + 0.410 815 348 736;
  • 29) 0.410 815 348 736 × 2 = 0 + 0.821 630 697 472;
  • 30) 0.821 630 697 472 × 2 = 1 + 0.643 261 394 944;
  • 31) 0.643 261 394 944 × 2 = 1 + 0.286 522 789 888;
  • 32) 0.286 522 789 888 × 2 = 0 + 0.573 045 579 776;
  • 33) 0.573 045 579 776 × 2 = 1 + 0.146 091 159 552;
  • 34) 0.146 091 159 552 × 2 = 0 + 0.292 182 319 104;
  • 35) 0.292 182 319 104 × 2 = 0 + 0.584 364 638 208;
  • 36) 0.584 364 638 208 × 2 = 1 + 0.168 729 276 416;
  • 37) 0.168 729 276 416 × 2 = 0 + 0.337 458 552 832;
  • 38) 0.337 458 552 832 × 2 = 0 + 0.674 917 105 664;
  • 39) 0.674 917 105 664 × 2 = 1 + 0.349 834 211 328;
  • 40) 0.349 834 211 328 × 2 = 0 + 0.699 668 422 656;
  • 41) 0.699 668 422 656 × 2 = 1 + 0.399 336 845 312;
  • 42) 0.399 336 845 312 × 2 = 0 + 0.798 673 690 624;
  • 43) 0.798 673 690 624 × 2 = 1 + 0.597 347 381 248;
  • 44) 0.597 347 381 248 × 2 = 1 + 0.194 694 762 496;
  • 45) 0.194 694 762 496 × 2 = 0 + 0.389 389 524 992;
  • 46) 0.389 389 524 992 × 2 = 0 + 0.778 779 049 984;
  • 47) 0.778 779 049 984 × 2 = 1 + 0.557 558 099 968;
  • 48) 0.557 558 099 968 × 2 = 1 + 0.115 116 199 936;
  • 49) 0.115 116 199 936 × 2 = 0 + 0.230 232 399 872;
  • 50) 0.230 232 399 872 × 2 = 0 + 0.460 464 799 744;
  • 51) 0.460 464 799 744 × 2 = 0 + 0.920 929 599 488;
  • 52) 0.920 929 599 488 × 2 = 1 + 0.841 859 198 976;
  • 53) 0.841 859 198 976 × 2 = 1 + 0.683 718 397 952;
  • 54) 0.683 718 397 952 × 2 = 1 + 0.367 436 795 904;
  • 55) 0.367 436 795 904 × 2 = 0 + 0.734 873 591 808;
  • 56) 0.734 873 591 808 × 2 = 1 + 0.469 747 183 616;
  • 57) 0.469 747 183 616 × 2 = 0 + 0.939 494 367 232;
  • 58) 0.939 494 367 232 × 2 = 1 + 0.878 988 734 464;
  • 59) 0.878 988 734 464 × 2 = 1 + 0.757 977 468 928;
  • 60) 0.757 977 468 928 × 2 = 1 + 0.515 954 937 856;
  • 61) 0.515 954 937 856 × 2 = 1 + 0.031 909 875 712;
  • 62) 0.031 909 875 712 × 2 = 0 + 0.063 819 751 424;
  • 63) 0.063 819 751 424 × 2 = 0 + 0.127 639 502 848;
  • 64) 0.127 639 502 848 × 2 = 0 + 0.255 279 005 696;

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:


0.000 282 006 006(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1001 0010 1011 0011 0001 1101 0111 1000(2)

6. Positive number before normalization:

0.000 282 006 006(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1001 0010 1011 0011 0001 1101 0111 1000(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:


0.000 282 006 006(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1001 0010 1011 0011 0001 1101 0111 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 1001 0010 1011 0011 0001 1101 0111 1000(2) × 20 =

1.0010 0111 1011 0100 0110 1001 0010 1011 0011 0001 1101 0111 1000(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 1001 0010 1011 0011 0001 1101 0111 1000


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 1001 0010 1011 0011 0001 1101 0111 1000 =

0010 0111 1011 0100 0110 1001 0010 1011 0011 0001 1101 0111 1000


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 1001 0010 1011 0011 0001 1101 0111 1000


Decimal number -0.000 282 006 006 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 1001 0010 1011 0011 0001 1101 0111 1000

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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