-0.000 282 005 918 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 918(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
-0.000 282 005 918(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:

|-0.000 282 005 918| = 0.000 282 005 918


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: 0.000 282 005 918.

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) 0.000 282 005 918 × 2 = 0 + 0.000 564 011 836;
  • 2) 0.000 564 011 836 × 2 = 0 + 0.001 128 023 672;
  • 3) 0.001 128 023 672 × 2 = 0 + 0.002 256 047 344;
  • 4) 0.002 256 047 344 × 2 = 0 + 0.004 512 094 688;
  • 5) 0.004 512 094 688 × 2 = 0 + 0.009 024 189 376;
  • 6) 0.009 024 189 376 × 2 = 0 + 0.018 048 378 752;
  • 7) 0.018 048 378 752 × 2 = 0 + 0.036 096 757 504;
  • 8) 0.036 096 757 504 × 2 = 0 + 0.072 193 515 008;
  • 9) 0.072 193 515 008 × 2 = 0 + 0.144 387 030 016;
  • 10) 0.144 387 030 016 × 2 = 0 + 0.288 774 060 032;
  • 11) 0.288 774 060 032 × 2 = 0 + 0.577 548 120 064;
  • 12) 0.577 548 120 064 × 2 = 1 + 0.155 096 240 128;
  • 13) 0.155 096 240 128 × 2 = 0 + 0.310 192 480 256;
  • 14) 0.310 192 480 256 × 2 = 0 + 0.620 384 960 512;
  • 15) 0.620 384 960 512 × 2 = 1 + 0.240 769 921 024;
  • 16) 0.240 769 921 024 × 2 = 0 + 0.481 539 842 048;
  • 17) 0.481 539 842 048 × 2 = 0 + 0.963 079 684 096;
  • 18) 0.963 079 684 096 × 2 = 1 + 0.926 159 368 192;
  • 19) 0.926 159 368 192 × 2 = 1 + 0.852 318 736 384;
  • 20) 0.852 318 736 384 × 2 = 1 + 0.704 637 472 768;
  • 21) 0.704 637 472 768 × 2 = 1 + 0.409 274 945 536;
  • 22) 0.409 274 945 536 × 2 = 0 + 0.818 549 891 072;
  • 23) 0.818 549 891 072 × 2 = 1 + 0.637 099 782 144;
  • 24) 0.637 099 782 144 × 2 = 1 + 0.274 199 564 288;
  • 25) 0.274 199 564 288 × 2 = 0 + 0.548 399 128 576;
  • 26) 0.548 399 128 576 × 2 = 1 + 0.096 798 257 152;
  • 27) 0.096 798 257 152 × 2 = 0 + 0.193 596 514 304;
  • 28) 0.193 596 514 304 × 2 = 0 + 0.387 193 028 608;
  • 29) 0.387 193 028 608 × 2 = 0 + 0.774 386 057 216;
  • 30) 0.774 386 057 216 × 2 = 1 + 0.548 772 114 432;
  • 31) 0.548 772 114 432 × 2 = 1 + 0.097 544 228 864;
  • 32) 0.097 544 228 864 × 2 = 0 + 0.195 088 457 728;
  • 33) 0.195 088 457 728 × 2 = 0 + 0.390 176 915 456;
  • 34) 0.390 176 915 456 × 2 = 0 + 0.780 353 830 912;
  • 35) 0.780 353 830 912 × 2 = 1 + 0.560 707 661 824;
  • 36) 0.560 707 661 824 × 2 = 1 + 0.121 415 323 648;
  • 37) 0.121 415 323 648 × 2 = 0 + 0.242 830 647 296;
  • 38) 0.242 830 647 296 × 2 = 0 + 0.485 661 294 592;
  • 39) 0.485 661 294 592 × 2 = 0 + 0.971 322 589 184;
  • 40) 0.971 322 589 184 × 2 = 1 + 0.942 645 178 368;
  • 41) 0.942 645 178 368 × 2 = 1 + 0.885 290 356 736;
  • 42) 0.885 290 356 736 × 2 = 1 + 0.770 580 713 472;
  • 43) 0.770 580 713 472 × 2 = 1 + 0.541 161 426 944;
  • 44) 0.541 161 426 944 × 2 = 1 + 0.082 322 853 888;
  • 45) 0.082 322 853 888 × 2 = 0 + 0.164 645 707 776;
  • 46) 0.164 645 707 776 × 2 = 0 + 0.329 291 415 552;
  • 47) 0.329 291 415 552 × 2 = 0 + 0.658 582 831 104;
  • 48) 0.658 582 831 104 × 2 = 1 + 0.317 165 662 208;
  • 49) 0.317 165 662 208 × 2 = 0 + 0.634 331 324 416;
  • 50) 0.634 331 324 416 × 2 = 1 + 0.268 662 648 832;
  • 51) 0.268 662 648 832 × 2 = 0 + 0.537 325 297 664;
  • 52) 0.537 325 297 664 × 2 = 1 + 0.074 650 595 328;
  • 53) 0.074 650 595 328 × 2 = 0 + 0.149 301 190 656;
  • 54) 0.149 301 190 656 × 2 = 0 + 0.298 602 381 312;
  • 55) 0.298 602 381 312 × 2 = 0 + 0.597 204 762 624;
  • 56) 0.597 204 762 624 × 2 = 1 + 0.194 409 525 248;
  • 57) 0.194 409 525 248 × 2 = 0 + 0.388 819 050 496;
  • 58) 0.388 819 050 496 × 2 = 0 + 0.777 638 100 992;
  • 59) 0.777 638 100 992 × 2 = 1 + 0.555 276 201 984;
  • 60) 0.555 276 201 984 × 2 = 1 + 0.110 552 403 968;
  • 61) 0.110 552 403 968 × 2 = 0 + 0.221 104 807 936;
  • 62) 0.221 104 807 936 × 2 = 0 + 0.442 209 615 872;
  • 63) 0.442 209 615 872 × 2 = 0 + 0.884 419 231 744;
  • 64) 0.884 419 231 744 × 2 = 1 + 0.768 838 463 488;

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:


0.000 282 005 918(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 1111 0001 0101 0001 0011 0001(2)

6. Positive number before normalization:

0.000 282 005 918(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 1111 0001 0101 0001 0011 0001(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:


0.000 282 005 918(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 1111 0001 0101 0001 0011 0001(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0011 0001 1111 0001 0101 0001 0011 0001(2) × 20 =

1.0010 0111 1011 0100 0110 0011 0001 1111 0001 0101 0001 0011 0001(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 0001 1111 0001 0101 0001 0011 0001


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 0001 1111 0001 0101 0001 0011 0001 =

0010 0111 1011 0100 0110 0011 0001 1111 0001 0101 0001 0011 0001


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 0001 1111 0001 0101 0001 0011 0001


Decimal number -0.000 282 005 918 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0011 0001 1111 0001 0101 0001 0011 0001

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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