-0.000 282 005 922 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 922(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
-0.000 282 005 922(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:

|-0.000 282 005 922| = 0.000 282 005 922


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: 0.000 282 005 922.

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) 0.000 282 005 922 × 2 = 0 + 0.000 564 011 844;
  • 2) 0.000 564 011 844 × 2 = 0 + 0.001 128 023 688;
  • 3) 0.001 128 023 688 × 2 = 0 + 0.002 256 047 376;
  • 4) 0.002 256 047 376 × 2 = 0 + 0.004 512 094 752;
  • 5) 0.004 512 094 752 × 2 = 0 + 0.009 024 189 504;
  • 6) 0.009 024 189 504 × 2 = 0 + 0.018 048 379 008;
  • 7) 0.018 048 379 008 × 2 = 0 + 0.036 096 758 016;
  • 8) 0.036 096 758 016 × 2 = 0 + 0.072 193 516 032;
  • 9) 0.072 193 516 032 × 2 = 0 + 0.144 387 032 064;
  • 10) 0.144 387 032 064 × 2 = 0 + 0.288 774 064 128;
  • 11) 0.288 774 064 128 × 2 = 0 + 0.577 548 128 256;
  • 12) 0.577 548 128 256 × 2 = 1 + 0.155 096 256 512;
  • 13) 0.155 096 256 512 × 2 = 0 + 0.310 192 513 024;
  • 14) 0.310 192 513 024 × 2 = 0 + 0.620 385 026 048;
  • 15) 0.620 385 026 048 × 2 = 1 + 0.240 770 052 096;
  • 16) 0.240 770 052 096 × 2 = 0 + 0.481 540 104 192;
  • 17) 0.481 540 104 192 × 2 = 0 + 0.963 080 208 384;
  • 18) 0.963 080 208 384 × 2 = 1 + 0.926 160 416 768;
  • 19) 0.926 160 416 768 × 2 = 1 + 0.852 320 833 536;
  • 20) 0.852 320 833 536 × 2 = 1 + 0.704 641 667 072;
  • 21) 0.704 641 667 072 × 2 = 1 + 0.409 283 334 144;
  • 22) 0.409 283 334 144 × 2 = 0 + 0.818 566 668 288;
  • 23) 0.818 566 668 288 × 2 = 1 + 0.637 133 336 576;
  • 24) 0.637 133 336 576 × 2 = 1 + 0.274 266 673 152;
  • 25) 0.274 266 673 152 × 2 = 0 + 0.548 533 346 304;
  • 26) 0.548 533 346 304 × 2 = 1 + 0.097 066 692 608;
  • 27) 0.097 066 692 608 × 2 = 0 + 0.194 133 385 216;
  • 28) 0.194 133 385 216 × 2 = 0 + 0.388 266 770 432;
  • 29) 0.388 266 770 432 × 2 = 0 + 0.776 533 540 864;
  • 30) 0.776 533 540 864 × 2 = 1 + 0.553 067 081 728;
  • 31) 0.553 067 081 728 × 2 = 1 + 0.106 134 163 456;
  • 32) 0.106 134 163 456 × 2 = 0 + 0.212 268 326 912;
  • 33) 0.212 268 326 912 × 2 = 0 + 0.424 536 653 824;
  • 34) 0.424 536 653 824 × 2 = 0 + 0.849 073 307 648;
  • 35) 0.849 073 307 648 × 2 = 1 + 0.698 146 615 296;
  • 36) 0.698 146 615 296 × 2 = 1 + 0.396 293 230 592;
  • 37) 0.396 293 230 592 × 2 = 0 + 0.792 586 461 184;
  • 38) 0.792 586 461 184 × 2 = 1 + 0.585 172 922 368;
  • 39) 0.585 172 922 368 × 2 = 1 + 0.170 345 844 736;
  • 40) 0.170 345 844 736 × 2 = 0 + 0.340 691 689 472;
  • 41) 0.340 691 689 472 × 2 = 0 + 0.681 383 378 944;
  • 42) 0.681 383 378 944 × 2 = 1 + 0.362 766 757 888;
  • 43) 0.362 766 757 888 × 2 = 0 + 0.725 533 515 776;
  • 44) 0.725 533 515 776 × 2 = 1 + 0.451 067 031 552;
  • 45) 0.451 067 031 552 × 2 = 0 + 0.902 134 063 104;
  • 46) 0.902 134 063 104 × 2 = 1 + 0.804 268 126 208;
  • 47) 0.804 268 126 208 × 2 = 1 + 0.608 536 252 416;
  • 48) 0.608 536 252 416 × 2 = 1 + 0.217 072 504 832;
  • 49) 0.217 072 504 832 × 2 = 0 + 0.434 145 009 664;
  • 50) 0.434 145 009 664 × 2 = 0 + 0.868 290 019 328;
  • 51) 0.868 290 019 328 × 2 = 1 + 0.736 580 038 656;
  • 52) 0.736 580 038 656 × 2 = 1 + 0.473 160 077 312;
  • 53) 0.473 160 077 312 × 2 = 0 + 0.946 320 154 624;
  • 54) 0.946 320 154 624 × 2 = 1 + 0.892 640 309 248;
  • 55) 0.892 640 309 248 × 2 = 1 + 0.785 280 618 496;
  • 56) 0.785 280 618 496 × 2 = 1 + 0.570 561 236 992;
  • 57) 0.570 561 236 992 × 2 = 1 + 0.141 122 473 984;
  • 58) 0.141 122 473 984 × 2 = 0 + 0.282 244 947 968;
  • 59) 0.282 244 947 968 × 2 = 0 + 0.564 489 895 936;
  • 60) 0.564 489 895 936 × 2 = 1 + 0.128 979 791 872;
  • 61) 0.128 979 791 872 × 2 = 0 + 0.257 959 583 744;
  • 62) 0.257 959 583 744 × 2 = 0 + 0.515 919 167 488;
  • 63) 0.515 919 167 488 × 2 = 1 + 0.031 838 334 976;
  • 64) 0.031 838 334 976 × 2 = 0 + 0.063 676 669 952;

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:


0.000 282 005 922(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0110 0101 0111 0011 0111 1001 0010(2)

6. Positive number before normalization:

0.000 282 005 922(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0110 0101 0111 0011 0111 1001 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:


0.000 282 005 922(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0110 0101 0111 0011 0111 1001 0010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0110 0101 0111 0011 0111 1001 0010(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0110 0101 0111 0011 0111 1001 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 0011 0110 0101 0111 0011 0111 1001 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 0011 0110 0101 0111 0011 0111 1001 0010 =

0010 0111 1011 0100 0110 0011 0110 0101 0111 0011 0111 1001 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 0011 0110 0101 0111 0011 0111 1001 0010


Decimal number -0.000 282 005 922 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0110 0101 0111 0011 0111 1001 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