-0.000 282 006 322 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 006 322(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
-0.000 282 006 322(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:

|-0.000 282 006 322| = 0.000 282 006 322


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: 0.000 282 006 322.

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) 0.000 282 006 322 × 2 = 0 + 0.000 564 012 644;
  • 2) 0.000 564 012 644 × 2 = 0 + 0.001 128 025 288;
  • 3) 0.001 128 025 288 × 2 = 0 + 0.002 256 050 576;
  • 4) 0.002 256 050 576 × 2 = 0 + 0.004 512 101 152;
  • 5) 0.004 512 101 152 × 2 = 0 + 0.009 024 202 304;
  • 6) 0.009 024 202 304 × 2 = 0 + 0.018 048 404 608;
  • 7) 0.018 048 404 608 × 2 = 0 + 0.036 096 809 216;
  • 8) 0.036 096 809 216 × 2 = 0 + 0.072 193 618 432;
  • 9) 0.072 193 618 432 × 2 = 0 + 0.144 387 236 864;
  • 10) 0.144 387 236 864 × 2 = 0 + 0.288 774 473 728;
  • 11) 0.288 774 473 728 × 2 = 0 + 0.577 548 947 456;
  • 12) 0.577 548 947 456 × 2 = 1 + 0.155 097 894 912;
  • 13) 0.155 097 894 912 × 2 = 0 + 0.310 195 789 824;
  • 14) 0.310 195 789 824 × 2 = 0 + 0.620 391 579 648;
  • 15) 0.620 391 579 648 × 2 = 1 + 0.240 783 159 296;
  • 16) 0.240 783 159 296 × 2 = 0 + 0.481 566 318 592;
  • 17) 0.481 566 318 592 × 2 = 0 + 0.963 132 637 184;
  • 18) 0.963 132 637 184 × 2 = 1 + 0.926 265 274 368;
  • 19) 0.926 265 274 368 × 2 = 1 + 0.852 530 548 736;
  • 20) 0.852 530 548 736 × 2 = 1 + 0.705 061 097 472;
  • 21) 0.705 061 097 472 × 2 = 1 + 0.410 122 194 944;
  • 22) 0.410 122 194 944 × 2 = 0 + 0.820 244 389 888;
  • 23) 0.820 244 389 888 × 2 = 1 + 0.640 488 779 776;
  • 24) 0.640 488 779 776 × 2 = 1 + 0.280 977 559 552;
  • 25) 0.280 977 559 552 × 2 = 0 + 0.561 955 119 104;
  • 26) 0.561 955 119 104 × 2 = 1 + 0.123 910 238 208;
  • 27) 0.123 910 238 208 × 2 = 0 + 0.247 820 476 416;
  • 28) 0.247 820 476 416 × 2 = 0 + 0.495 640 952 832;
  • 29) 0.495 640 952 832 × 2 = 0 + 0.991 281 905 664;
  • 30) 0.991 281 905 664 × 2 = 1 + 0.982 563 811 328;
  • 31) 0.982 563 811 328 × 2 = 1 + 0.965 127 622 656;
  • 32) 0.965 127 622 656 × 2 = 1 + 0.930 255 245 312;
  • 33) 0.930 255 245 312 × 2 = 1 + 0.860 510 490 624;
  • 34) 0.860 510 490 624 × 2 = 1 + 0.721 020 981 248;
  • 35) 0.721 020 981 248 × 2 = 1 + 0.442 041 962 496;
  • 36) 0.442 041 962 496 × 2 = 0 + 0.884 083 924 992;
  • 37) 0.884 083 924 992 × 2 = 1 + 0.768 167 849 984;
  • 38) 0.768 167 849 984 × 2 = 1 + 0.536 335 699 968;
  • 39) 0.536 335 699 968 × 2 = 1 + 0.072 671 399 936;
  • 40) 0.072 671 399 936 × 2 = 0 + 0.145 342 799 872;
  • 41) 0.145 342 799 872 × 2 = 0 + 0.290 685 599 744;
  • 42) 0.290 685 599 744 × 2 = 0 + 0.581 371 199 488;
  • 43) 0.581 371 199 488 × 2 = 1 + 0.162 742 398 976;
  • 44) 0.162 742 398 976 × 2 = 0 + 0.325 484 797 952;
  • 45) 0.325 484 797 952 × 2 = 0 + 0.650 969 595 904;
  • 46) 0.650 969 595 904 × 2 = 1 + 0.301 939 191 808;
  • 47) 0.301 939 191 808 × 2 = 0 + 0.603 878 383 616;
  • 48) 0.603 878 383 616 × 2 = 1 + 0.207 756 767 232;
  • 49) 0.207 756 767 232 × 2 = 0 + 0.415 513 534 464;
  • 50) 0.415 513 534 464 × 2 = 0 + 0.831 027 068 928;
  • 51) 0.831 027 068 928 × 2 = 1 + 0.662 054 137 856;
  • 52) 0.662 054 137 856 × 2 = 1 + 0.324 108 275 712;
  • 53) 0.324 108 275 712 × 2 = 0 + 0.648 216 551 424;
  • 54) 0.648 216 551 424 × 2 = 1 + 0.296 433 102 848;
  • 55) 0.296 433 102 848 × 2 = 0 + 0.592 866 205 696;
  • 56) 0.592 866 205 696 × 2 = 1 + 0.185 732 411 392;
  • 57) 0.185 732 411 392 × 2 = 0 + 0.371 464 822 784;
  • 58) 0.371 464 822 784 × 2 = 0 + 0.742 929 645 568;
  • 59) 0.742 929 645 568 × 2 = 1 + 0.485 859 291 136;
  • 60) 0.485 859 291 136 × 2 = 0 + 0.971 718 582 272;
  • 61) 0.971 718 582 272 × 2 = 1 + 0.943 437 164 544;
  • 62) 0.943 437 164 544 × 2 = 1 + 0.886 874 329 088;
  • 63) 0.886 874 329 088 × 2 = 1 + 0.773 748 658 176;
  • 64) 0.773 748 658 176 × 2 = 1 + 0.547 497 316 352;

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:


0.000 282 006 322(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1110 1110 0010 0101 0011 0101 0010 1111(2)

6. Positive number before normalization:

0.000 282 006 322(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1110 1110 0010 0101 0011 0101 0010 1111(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:


0.000 282 006 322(10) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1110 1110 0010 0101 0011 0101 0010 1111(2) =

0.0000 0000 0001 0010 0111 1011 0100 0111 1110 1110 0010 0101 0011 0101 0010 1111(2) × 20 =

1.0010 0111 1011 0100 0111 1110 1110 0010 0101 0011 0101 0010 1111(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 0111 1110 1110 0010 0101 0011 0101 0010 1111


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 0111 1110 1110 0010 0101 0011 0101 0010 1111 =

0010 0111 1011 0100 0111 1110 1110 0010 0101 0011 0101 0010 1111


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 0111 1110 1110 0010 0101 0011 0101 0010 1111


Decimal number -0.000 282 006 322 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0111 1110 1110 0010 0101 0011 0101 0010 1111

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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