-0.000 282 005 914 266 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 266(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
-0.000 282 005 914 266(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:

|-0.000 282 005 914 266| = 0.000 282 005 914 266


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: 0.000 282 005 914 266.

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) 0.000 282 005 914 266 × 2 = 0 + 0.000 564 011 828 532;
  • 2) 0.000 564 011 828 532 × 2 = 0 + 0.001 128 023 657 064;
  • 3) 0.001 128 023 657 064 × 2 = 0 + 0.002 256 047 314 128;
  • 4) 0.002 256 047 314 128 × 2 = 0 + 0.004 512 094 628 256;
  • 5) 0.004 512 094 628 256 × 2 = 0 + 0.009 024 189 256 512;
  • 6) 0.009 024 189 256 512 × 2 = 0 + 0.018 048 378 513 024;
  • 7) 0.018 048 378 513 024 × 2 = 0 + 0.036 096 757 026 048;
  • 8) 0.036 096 757 026 048 × 2 = 0 + 0.072 193 514 052 096;
  • 9) 0.072 193 514 052 096 × 2 = 0 + 0.144 387 028 104 192;
  • 10) 0.144 387 028 104 192 × 2 = 0 + 0.288 774 056 208 384;
  • 11) 0.288 774 056 208 384 × 2 = 0 + 0.577 548 112 416 768;
  • 12) 0.577 548 112 416 768 × 2 = 1 + 0.155 096 224 833 536;
  • 13) 0.155 096 224 833 536 × 2 = 0 + 0.310 192 449 667 072;
  • 14) 0.310 192 449 667 072 × 2 = 0 + 0.620 384 899 334 144;
  • 15) 0.620 384 899 334 144 × 2 = 1 + 0.240 769 798 668 288;
  • 16) 0.240 769 798 668 288 × 2 = 0 + 0.481 539 597 336 576;
  • 17) 0.481 539 597 336 576 × 2 = 0 + 0.963 079 194 673 152;
  • 18) 0.963 079 194 673 152 × 2 = 1 + 0.926 158 389 346 304;
  • 19) 0.926 158 389 346 304 × 2 = 1 + 0.852 316 778 692 608;
  • 20) 0.852 316 778 692 608 × 2 = 1 + 0.704 633 557 385 216;
  • 21) 0.704 633 557 385 216 × 2 = 1 + 0.409 267 114 770 432;
  • 22) 0.409 267 114 770 432 × 2 = 0 + 0.818 534 229 540 864;
  • 23) 0.818 534 229 540 864 × 2 = 1 + 0.637 068 459 081 728;
  • 24) 0.637 068 459 081 728 × 2 = 1 + 0.274 136 918 163 456;
  • 25) 0.274 136 918 163 456 × 2 = 0 + 0.548 273 836 326 912;
  • 26) 0.548 273 836 326 912 × 2 = 1 + 0.096 547 672 653 824;
  • 27) 0.096 547 672 653 824 × 2 = 0 + 0.193 095 345 307 648;
  • 28) 0.193 095 345 307 648 × 2 = 0 + 0.386 190 690 615 296;
  • 29) 0.386 190 690 615 296 × 2 = 0 + 0.772 381 381 230 592;
  • 30) 0.772 381 381 230 592 × 2 = 1 + 0.544 762 762 461 184;
  • 31) 0.544 762 762 461 184 × 2 = 1 + 0.089 525 524 922 368;
  • 32) 0.089 525 524 922 368 × 2 = 0 + 0.179 051 049 844 736;
  • 33) 0.179 051 049 844 736 × 2 = 0 + 0.358 102 099 689 472;
  • 34) 0.358 102 099 689 472 × 2 = 0 + 0.716 204 199 378 944;
  • 35) 0.716 204 199 378 944 × 2 = 1 + 0.432 408 398 757 888;
  • 36) 0.432 408 398 757 888 × 2 = 0 + 0.864 816 797 515 776;
  • 37) 0.864 816 797 515 776 × 2 = 1 + 0.729 633 595 031 552;
  • 38) 0.729 633 595 031 552 × 2 = 1 + 0.459 267 190 063 104;
  • 39) 0.459 267 190 063 104 × 2 = 0 + 0.918 534 380 126 208;
  • 40) 0.918 534 380 126 208 × 2 = 1 + 0.837 068 760 252 416;
  • 41) 0.837 068 760 252 416 × 2 = 1 + 0.674 137 520 504 832;
  • 42) 0.674 137 520 504 832 × 2 = 1 + 0.348 275 041 009 664;
  • 43) 0.348 275 041 009 664 × 2 = 0 + 0.696 550 082 019 328;
  • 44) 0.696 550 082 019 328 × 2 = 1 + 0.393 100 164 038 656;
  • 45) 0.393 100 164 038 656 × 2 = 0 + 0.786 200 328 077 312;
  • 46) 0.786 200 328 077 312 × 2 = 1 + 0.572 400 656 154 624;
  • 47) 0.572 400 656 154 624 × 2 = 1 + 0.144 801 312 309 248;
  • 48) 0.144 801 312 309 248 × 2 = 0 + 0.289 602 624 618 496;
  • 49) 0.289 602 624 618 496 × 2 = 0 + 0.579 205 249 236 992;
  • 50) 0.579 205 249 236 992 × 2 = 1 + 0.158 410 498 473 984;
  • 51) 0.158 410 498 473 984 × 2 = 0 + 0.316 820 996 947 968;
  • 52) 0.316 820 996 947 968 × 2 = 0 + 0.633 641 993 895 936;
  • 53) 0.633 641 993 895 936 × 2 = 1 + 0.267 283 987 791 872;
  • 54) 0.267 283 987 791 872 × 2 = 0 + 0.534 567 975 583 744;
  • 55) 0.534 567 975 583 744 × 2 = 1 + 0.069 135 951 167 488;
  • 56) 0.069 135 951 167 488 × 2 = 0 + 0.138 271 902 334 976;
  • 57) 0.138 271 902 334 976 × 2 = 0 + 0.276 543 804 669 952;
  • 58) 0.276 543 804 669 952 × 2 = 0 + 0.553 087 609 339 904;
  • 59) 0.553 087 609 339 904 × 2 = 1 + 0.106 175 218 679 808;
  • 60) 0.106 175 218 679 808 × 2 = 0 + 0.212 350 437 359 616;
  • 61) 0.212 350 437 359 616 × 2 = 0 + 0.424 700 874 719 232;
  • 62) 0.424 700 874 719 232 × 2 = 0 + 0.849 401 749 438 464;
  • 63) 0.849 401 749 438 464 × 2 = 1 + 0.698 803 498 876 928;
  • 64) 0.698 803 498 876 928 × 2 = 1 + 0.397 606 997 753 856;

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:


0.000 282 005 914 266(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0110 0100 1010 0010 0011(2)

6. Positive number before normalization:

0.000 282 005 914 266(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0110 0100 1010 0010 0011(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:


0.000 282 005 914 266(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0110 0100 1010 0010 0011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1101 0110 0100 1010 0010 0011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1101 0110 0100 1010 0010 0011(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 1101 0110 0100 1010 0010 0011


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 1101 0110 0100 1010 0010 0011 =

0010 0111 1011 0100 0110 0010 1101 1101 0110 0100 1010 0010 0011


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 1101 0110 0100 1010 0010 0011


Decimal number -0.000 282 005 914 266 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1101 0110 0100 1010 0010 0011

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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