-0.000 282 005 914 388 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 388(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
-0.000 282 005 914 388(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:

|-0.000 282 005 914 388| = 0.000 282 005 914 388


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: 0.000 282 005 914 388.

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) 0.000 282 005 914 388 × 2 = 0 + 0.000 564 011 828 776;
  • 2) 0.000 564 011 828 776 × 2 = 0 + 0.001 128 023 657 552;
  • 3) 0.001 128 023 657 552 × 2 = 0 + 0.002 256 047 315 104;
  • 4) 0.002 256 047 315 104 × 2 = 0 + 0.004 512 094 630 208;
  • 5) 0.004 512 094 630 208 × 2 = 0 + 0.009 024 189 260 416;
  • 6) 0.009 024 189 260 416 × 2 = 0 + 0.018 048 378 520 832;
  • 7) 0.018 048 378 520 832 × 2 = 0 + 0.036 096 757 041 664;
  • 8) 0.036 096 757 041 664 × 2 = 0 + 0.072 193 514 083 328;
  • 9) 0.072 193 514 083 328 × 2 = 0 + 0.144 387 028 166 656;
  • 10) 0.144 387 028 166 656 × 2 = 0 + 0.288 774 056 333 312;
  • 11) 0.288 774 056 333 312 × 2 = 0 + 0.577 548 112 666 624;
  • 12) 0.577 548 112 666 624 × 2 = 1 + 0.155 096 225 333 248;
  • 13) 0.155 096 225 333 248 × 2 = 0 + 0.310 192 450 666 496;
  • 14) 0.310 192 450 666 496 × 2 = 0 + 0.620 384 901 332 992;
  • 15) 0.620 384 901 332 992 × 2 = 1 + 0.240 769 802 665 984;
  • 16) 0.240 769 802 665 984 × 2 = 0 + 0.481 539 605 331 968;
  • 17) 0.481 539 605 331 968 × 2 = 0 + 0.963 079 210 663 936;
  • 18) 0.963 079 210 663 936 × 2 = 1 + 0.926 158 421 327 872;
  • 19) 0.926 158 421 327 872 × 2 = 1 + 0.852 316 842 655 744;
  • 20) 0.852 316 842 655 744 × 2 = 1 + 0.704 633 685 311 488;
  • 21) 0.704 633 685 311 488 × 2 = 1 + 0.409 267 370 622 976;
  • 22) 0.409 267 370 622 976 × 2 = 0 + 0.818 534 741 245 952;
  • 23) 0.818 534 741 245 952 × 2 = 1 + 0.637 069 482 491 904;
  • 24) 0.637 069 482 491 904 × 2 = 1 + 0.274 138 964 983 808;
  • 25) 0.274 138 964 983 808 × 2 = 0 + 0.548 277 929 967 616;
  • 26) 0.548 277 929 967 616 × 2 = 1 + 0.096 555 859 935 232;
  • 27) 0.096 555 859 935 232 × 2 = 0 + 0.193 111 719 870 464;
  • 28) 0.193 111 719 870 464 × 2 = 0 + 0.386 223 439 740 928;
  • 29) 0.386 223 439 740 928 × 2 = 0 + 0.772 446 879 481 856;
  • 30) 0.772 446 879 481 856 × 2 = 1 + 0.544 893 758 963 712;
  • 31) 0.544 893 758 963 712 × 2 = 1 + 0.089 787 517 927 424;
  • 32) 0.089 787 517 927 424 × 2 = 0 + 0.179 575 035 854 848;
  • 33) 0.179 575 035 854 848 × 2 = 0 + 0.359 150 071 709 696;
  • 34) 0.359 150 071 709 696 × 2 = 0 + 0.718 300 143 419 392;
  • 35) 0.718 300 143 419 392 × 2 = 1 + 0.436 600 286 838 784;
  • 36) 0.436 600 286 838 784 × 2 = 0 + 0.873 200 573 677 568;
  • 37) 0.873 200 573 677 568 × 2 = 1 + 0.746 401 147 355 136;
  • 38) 0.746 401 147 355 136 × 2 = 1 + 0.492 802 294 710 272;
  • 39) 0.492 802 294 710 272 × 2 = 0 + 0.985 604 589 420 544;
  • 40) 0.985 604 589 420 544 × 2 = 1 + 0.971 209 178 841 088;
  • 41) 0.971 209 178 841 088 × 2 = 1 + 0.942 418 357 682 176;
  • 42) 0.942 418 357 682 176 × 2 = 1 + 0.884 836 715 364 352;
  • 43) 0.884 836 715 364 352 × 2 = 1 + 0.769 673 430 728 704;
  • 44) 0.769 673 430 728 704 × 2 = 1 + 0.539 346 861 457 408;
  • 45) 0.539 346 861 457 408 × 2 = 1 + 0.078 693 722 914 816;
  • 46) 0.078 693 722 914 816 × 2 = 0 + 0.157 387 445 829 632;
  • 47) 0.157 387 445 829 632 × 2 = 0 + 0.314 774 891 659 264;
  • 48) 0.314 774 891 659 264 × 2 = 0 + 0.629 549 783 318 528;
  • 49) 0.629 549 783 318 528 × 2 = 1 + 0.259 099 566 637 056;
  • 50) 0.259 099 566 637 056 × 2 = 0 + 0.518 199 133 274 112;
  • 51) 0.518 199 133 274 112 × 2 = 1 + 0.036 398 266 548 224;
  • 52) 0.036 398 266 548 224 × 2 = 0 + 0.072 796 533 096 448;
  • 53) 0.072 796 533 096 448 × 2 = 0 + 0.145 593 066 192 896;
  • 54) 0.145 593 066 192 896 × 2 = 0 + 0.291 186 132 385 792;
  • 55) 0.291 186 132 385 792 × 2 = 0 + 0.582 372 264 771 584;
  • 56) 0.582 372 264 771 584 × 2 = 1 + 0.164 744 529 543 168;
  • 57) 0.164 744 529 543 168 × 2 = 0 + 0.329 489 059 086 336;
  • 58) 0.329 489 059 086 336 × 2 = 0 + 0.658 978 118 172 672;
  • 59) 0.658 978 118 172 672 × 2 = 1 + 0.317 956 236 345 344;
  • 60) 0.317 956 236 345 344 × 2 = 0 + 0.635 912 472 690 688;
  • 61) 0.635 912 472 690 688 × 2 = 1 + 0.271 824 945 381 376;
  • 62) 0.271 824 945 381 376 × 2 = 0 + 0.543 649 890 762 752;
  • 63) 0.543 649 890 762 752 × 2 = 1 + 0.087 299 781 525 504;
  • 64) 0.087 299 781 525 504 × 2 = 0 + 0.174 599 563 051 008;

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:


0.000 282 005 914 388(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 1010 0001 0010 1010(2)

6. Positive number before normalization:

0.000 282 005 914 388(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 1010 0001 0010 1010(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:


0.000 282 005 914 388(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 1010 0001 0010 1010(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1111 1000 1010 0001 0010 1010(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1111 1000 1010 0001 0010 1010(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 1000 1010 0001 0010 1010


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 1000 1010 0001 0010 1010 =

0010 0111 1011 0100 0110 0010 1101 1111 1000 1010 0001 0010 1010


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 1000 1010 0001 0010 1010


Decimal number -0.000 282 005 914 388 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 1000 1010 0001 0010 1010

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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