-0.000 282 005 914 402 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 402(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
-0.000 282 005 914 402(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:

|-0.000 282 005 914 402| = 0.000 282 005 914 402


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: 0.000 282 005 914 402.

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) 0.000 282 005 914 402 × 2 = 0 + 0.000 564 011 828 804;
  • 2) 0.000 564 011 828 804 × 2 = 0 + 0.001 128 023 657 608;
  • 3) 0.001 128 023 657 608 × 2 = 0 + 0.002 256 047 315 216;
  • 4) 0.002 256 047 315 216 × 2 = 0 + 0.004 512 094 630 432;
  • 5) 0.004 512 094 630 432 × 2 = 0 + 0.009 024 189 260 864;
  • 6) 0.009 024 189 260 864 × 2 = 0 + 0.018 048 378 521 728;
  • 7) 0.018 048 378 521 728 × 2 = 0 + 0.036 096 757 043 456;
  • 8) 0.036 096 757 043 456 × 2 = 0 + 0.072 193 514 086 912;
  • 9) 0.072 193 514 086 912 × 2 = 0 + 0.144 387 028 173 824;
  • 10) 0.144 387 028 173 824 × 2 = 0 + 0.288 774 056 347 648;
  • 11) 0.288 774 056 347 648 × 2 = 0 + 0.577 548 112 695 296;
  • 12) 0.577 548 112 695 296 × 2 = 1 + 0.155 096 225 390 592;
  • 13) 0.155 096 225 390 592 × 2 = 0 + 0.310 192 450 781 184;
  • 14) 0.310 192 450 781 184 × 2 = 0 + 0.620 384 901 562 368;
  • 15) 0.620 384 901 562 368 × 2 = 1 + 0.240 769 803 124 736;
  • 16) 0.240 769 803 124 736 × 2 = 0 + 0.481 539 606 249 472;
  • 17) 0.481 539 606 249 472 × 2 = 0 + 0.963 079 212 498 944;
  • 18) 0.963 079 212 498 944 × 2 = 1 + 0.926 158 424 997 888;
  • 19) 0.926 158 424 997 888 × 2 = 1 + 0.852 316 849 995 776;
  • 20) 0.852 316 849 995 776 × 2 = 1 + 0.704 633 699 991 552;
  • 21) 0.704 633 699 991 552 × 2 = 1 + 0.409 267 399 983 104;
  • 22) 0.409 267 399 983 104 × 2 = 0 + 0.818 534 799 966 208;
  • 23) 0.818 534 799 966 208 × 2 = 1 + 0.637 069 599 932 416;
  • 24) 0.637 069 599 932 416 × 2 = 1 + 0.274 139 199 864 832;
  • 25) 0.274 139 199 864 832 × 2 = 0 + 0.548 278 399 729 664;
  • 26) 0.548 278 399 729 664 × 2 = 1 + 0.096 556 799 459 328;
  • 27) 0.096 556 799 459 328 × 2 = 0 + 0.193 113 598 918 656;
  • 28) 0.193 113 598 918 656 × 2 = 0 + 0.386 227 197 837 312;
  • 29) 0.386 227 197 837 312 × 2 = 0 + 0.772 454 395 674 624;
  • 30) 0.772 454 395 674 624 × 2 = 1 + 0.544 908 791 349 248;
  • 31) 0.544 908 791 349 248 × 2 = 1 + 0.089 817 582 698 496;
  • 32) 0.089 817 582 698 496 × 2 = 0 + 0.179 635 165 396 992;
  • 33) 0.179 635 165 396 992 × 2 = 0 + 0.359 270 330 793 984;
  • 34) 0.359 270 330 793 984 × 2 = 0 + 0.718 540 661 587 968;
  • 35) 0.718 540 661 587 968 × 2 = 1 + 0.437 081 323 175 936;
  • 36) 0.437 081 323 175 936 × 2 = 0 + 0.874 162 646 351 872;
  • 37) 0.874 162 646 351 872 × 2 = 1 + 0.748 325 292 703 744;
  • 38) 0.748 325 292 703 744 × 2 = 1 + 0.496 650 585 407 488;
  • 39) 0.496 650 585 407 488 × 2 = 0 + 0.993 301 170 814 976;
  • 40) 0.993 301 170 814 976 × 2 = 1 + 0.986 602 341 629 952;
  • 41) 0.986 602 341 629 952 × 2 = 1 + 0.973 204 683 259 904;
  • 42) 0.973 204 683 259 904 × 2 = 1 + 0.946 409 366 519 808;
  • 43) 0.946 409 366 519 808 × 2 = 1 + 0.892 818 733 039 616;
  • 44) 0.892 818 733 039 616 × 2 = 1 + 0.785 637 466 079 232;
  • 45) 0.785 637 466 079 232 × 2 = 1 + 0.571 274 932 158 464;
  • 46) 0.571 274 932 158 464 × 2 = 1 + 0.142 549 864 316 928;
  • 47) 0.142 549 864 316 928 × 2 = 0 + 0.285 099 728 633 856;
  • 48) 0.285 099 728 633 856 × 2 = 0 + 0.570 199 457 267 712;
  • 49) 0.570 199 457 267 712 × 2 = 1 + 0.140 398 914 535 424;
  • 50) 0.140 398 914 535 424 × 2 = 0 + 0.280 797 829 070 848;
  • 51) 0.280 797 829 070 848 × 2 = 0 + 0.561 595 658 141 696;
  • 52) 0.561 595 658 141 696 × 2 = 1 + 0.123 191 316 283 392;
  • 53) 0.123 191 316 283 392 × 2 = 0 + 0.246 382 632 566 784;
  • 54) 0.246 382 632 566 784 × 2 = 0 + 0.492 765 265 133 568;
  • 55) 0.492 765 265 133 568 × 2 = 0 + 0.985 530 530 267 136;
  • 56) 0.985 530 530 267 136 × 2 = 1 + 0.971 061 060 534 272;
  • 57) 0.971 061 060 534 272 × 2 = 1 + 0.942 122 121 068 544;
  • 58) 0.942 122 121 068 544 × 2 = 1 + 0.884 244 242 137 088;
  • 59) 0.884 244 242 137 088 × 2 = 1 + 0.768 488 484 274 176;
  • 60) 0.768 488 484 274 176 × 2 = 1 + 0.536 976 968 548 352;
  • 61) 0.536 976 968 548 352 × 2 = 1 + 0.073 953 937 096 704;
  • 62) 0.073 953 937 096 704 × 2 = 0 + 0.147 907 874 193 408;
  • 63) 0.147 907 874 193 408 × 2 = 0 + 0.295 815 748 386 816;
  • 64) 0.295 815 748 386 816 × 2 = 0 + 0.591 631 496 773 632;

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:


0.000 282 005 914 402(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1100 1001 0001 1111 1000(2)

6. Positive number before normalization:

0.000 282 005 914 402(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1100 1001 0001 1111 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:


0.000 282 005 914 402(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1100 1001 0001 1111 1000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1100 1001 0001 1111 1000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1100 1001 0001 1111 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 0010 1101 1111 1100 1001 0001 1111 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 0010 1101 1111 1100 1001 0001 1111 1000 =

0010 0111 1011 0100 0110 0010 1101 1111 1100 1001 0001 1111 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 0010 1101 1111 1100 1001 0001 1111 1000


Decimal number -0.000 282 005 914 402 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 1100 1001 0001 1111 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