-0.000 282 005 914 217 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 217(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
-0.000 282 005 914 217(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:

|-0.000 282 005 914 217| = 0.000 282 005 914 217


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: 0.000 282 005 914 217.

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) 0.000 282 005 914 217 × 2 = 0 + 0.000 564 011 828 434;
  • 2) 0.000 564 011 828 434 × 2 = 0 + 0.001 128 023 656 868;
  • 3) 0.001 128 023 656 868 × 2 = 0 + 0.002 256 047 313 736;
  • 4) 0.002 256 047 313 736 × 2 = 0 + 0.004 512 094 627 472;
  • 5) 0.004 512 094 627 472 × 2 = 0 + 0.009 024 189 254 944;
  • 6) 0.009 024 189 254 944 × 2 = 0 + 0.018 048 378 509 888;
  • 7) 0.018 048 378 509 888 × 2 = 0 + 0.036 096 757 019 776;
  • 8) 0.036 096 757 019 776 × 2 = 0 + 0.072 193 514 039 552;
  • 9) 0.072 193 514 039 552 × 2 = 0 + 0.144 387 028 079 104;
  • 10) 0.144 387 028 079 104 × 2 = 0 + 0.288 774 056 158 208;
  • 11) 0.288 774 056 158 208 × 2 = 0 + 0.577 548 112 316 416;
  • 12) 0.577 548 112 316 416 × 2 = 1 + 0.155 096 224 632 832;
  • 13) 0.155 096 224 632 832 × 2 = 0 + 0.310 192 449 265 664;
  • 14) 0.310 192 449 265 664 × 2 = 0 + 0.620 384 898 531 328;
  • 15) 0.620 384 898 531 328 × 2 = 1 + 0.240 769 797 062 656;
  • 16) 0.240 769 797 062 656 × 2 = 0 + 0.481 539 594 125 312;
  • 17) 0.481 539 594 125 312 × 2 = 0 + 0.963 079 188 250 624;
  • 18) 0.963 079 188 250 624 × 2 = 1 + 0.926 158 376 501 248;
  • 19) 0.926 158 376 501 248 × 2 = 1 + 0.852 316 753 002 496;
  • 20) 0.852 316 753 002 496 × 2 = 1 + 0.704 633 506 004 992;
  • 21) 0.704 633 506 004 992 × 2 = 1 + 0.409 267 012 009 984;
  • 22) 0.409 267 012 009 984 × 2 = 0 + 0.818 534 024 019 968;
  • 23) 0.818 534 024 019 968 × 2 = 1 + 0.637 068 048 039 936;
  • 24) 0.637 068 048 039 936 × 2 = 1 + 0.274 136 096 079 872;
  • 25) 0.274 136 096 079 872 × 2 = 0 + 0.548 272 192 159 744;
  • 26) 0.548 272 192 159 744 × 2 = 1 + 0.096 544 384 319 488;
  • 27) 0.096 544 384 319 488 × 2 = 0 + 0.193 088 768 638 976;
  • 28) 0.193 088 768 638 976 × 2 = 0 + 0.386 177 537 277 952;
  • 29) 0.386 177 537 277 952 × 2 = 0 + 0.772 355 074 555 904;
  • 30) 0.772 355 074 555 904 × 2 = 1 + 0.544 710 149 111 808;
  • 31) 0.544 710 149 111 808 × 2 = 1 + 0.089 420 298 223 616;
  • 32) 0.089 420 298 223 616 × 2 = 0 + 0.178 840 596 447 232;
  • 33) 0.178 840 596 447 232 × 2 = 0 + 0.357 681 192 894 464;
  • 34) 0.357 681 192 894 464 × 2 = 0 + 0.715 362 385 788 928;
  • 35) 0.715 362 385 788 928 × 2 = 1 + 0.430 724 771 577 856;
  • 36) 0.430 724 771 577 856 × 2 = 0 + 0.861 449 543 155 712;
  • 37) 0.861 449 543 155 712 × 2 = 1 + 0.722 899 086 311 424;
  • 38) 0.722 899 086 311 424 × 2 = 1 + 0.445 798 172 622 848;
  • 39) 0.445 798 172 622 848 × 2 = 0 + 0.891 596 345 245 696;
  • 40) 0.891 596 345 245 696 × 2 = 1 + 0.783 192 690 491 392;
  • 41) 0.783 192 690 491 392 × 2 = 1 + 0.566 385 380 982 784;
  • 42) 0.566 385 380 982 784 × 2 = 1 + 0.132 770 761 965 568;
  • 43) 0.132 770 761 965 568 × 2 = 0 + 0.265 541 523 931 136;
  • 44) 0.265 541 523 931 136 × 2 = 0 + 0.531 083 047 862 272;
  • 45) 0.531 083 047 862 272 × 2 = 1 + 0.062 166 095 724 544;
  • 46) 0.062 166 095 724 544 × 2 = 0 + 0.124 332 191 449 088;
  • 47) 0.124 332 191 449 088 × 2 = 0 + 0.248 664 382 898 176;
  • 48) 0.248 664 382 898 176 × 2 = 0 + 0.497 328 765 796 352;
  • 49) 0.497 328 765 796 352 × 2 = 0 + 0.994 657 531 592 704;
  • 50) 0.994 657 531 592 704 × 2 = 1 + 0.989 315 063 185 408;
  • 51) 0.989 315 063 185 408 × 2 = 1 + 0.978 630 126 370 816;
  • 52) 0.978 630 126 370 816 × 2 = 1 + 0.957 260 252 741 632;
  • 53) 0.957 260 252 741 632 × 2 = 1 + 0.914 520 505 483 264;
  • 54) 0.914 520 505 483 264 × 2 = 1 + 0.829 041 010 966 528;
  • 55) 0.829 041 010 966 528 × 2 = 1 + 0.658 082 021 933 056;
  • 56) 0.658 082 021 933 056 × 2 = 1 + 0.316 164 043 866 112;
  • 57) 0.316 164 043 866 112 × 2 = 0 + 0.632 328 087 732 224;
  • 58) 0.632 328 087 732 224 × 2 = 1 + 0.264 656 175 464 448;
  • 59) 0.264 656 175 464 448 × 2 = 0 + 0.529 312 350 928 896;
  • 60) 0.529 312 350 928 896 × 2 = 1 + 0.058 624 701 857 792;
  • 61) 0.058 624 701 857 792 × 2 = 0 + 0.117 249 403 715 584;
  • 62) 0.117 249 403 715 584 × 2 = 0 + 0.234 498 807 431 168;
  • 63) 0.234 498 807 431 168 × 2 = 0 + 0.468 997 614 862 336;
  • 64) 0.468 997 614 862 336 × 2 = 0 + 0.937 995 229 724 672;

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:


0.000 282 005 914 217(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1000 0111 1111 0101 0000(2)

6. Positive number before normalization:

0.000 282 005 914 217(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1000 0111 1111 0101 0000(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:


0.000 282 005 914 217(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1000 0111 1111 0101 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1100 1000 0111 1111 0101 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1100 1000 0111 1111 0101 0000(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 1100 1000 0111 1111 0101 0000


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 1100 1000 0111 1111 0101 0000 =

0010 0111 1011 0100 0110 0010 1101 1100 1000 0111 1111 0101 0000


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 1100 1000 0111 1111 0101 0000


Decimal number -0.000 282 005 914 217 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1100 1000 0111 1111 0101 0000

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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