-0.000 282 005 914 361 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 361(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
-0.000 282 005 914 361(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:

|-0.000 282 005 914 361| = 0.000 282 005 914 361


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: 0.000 282 005 914 361.

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) 0.000 282 005 914 361 × 2 = 0 + 0.000 564 011 828 722;
  • 2) 0.000 564 011 828 722 × 2 = 0 + 0.001 128 023 657 444;
  • 3) 0.001 128 023 657 444 × 2 = 0 + 0.002 256 047 314 888;
  • 4) 0.002 256 047 314 888 × 2 = 0 + 0.004 512 094 629 776;
  • 5) 0.004 512 094 629 776 × 2 = 0 + 0.009 024 189 259 552;
  • 6) 0.009 024 189 259 552 × 2 = 0 + 0.018 048 378 519 104;
  • 7) 0.018 048 378 519 104 × 2 = 0 + 0.036 096 757 038 208;
  • 8) 0.036 096 757 038 208 × 2 = 0 + 0.072 193 514 076 416;
  • 9) 0.072 193 514 076 416 × 2 = 0 + 0.144 387 028 152 832;
  • 10) 0.144 387 028 152 832 × 2 = 0 + 0.288 774 056 305 664;
  • 11) 0.288 774 056 305 664 × 2 = 0 + 0.577 548 112 611 328;
  • 12) 0.577 548 112 611 328 × 2 = 1 + 0.155 096 225 222 656;
  • 13) 0.155 096 225 222 656 × 2 = 0 + 0.310 192 450 445 312;
  • 14) 0.310 192 450 445 312 × 2 = 0 + 0.620 384 900 890 624;
  • 15) 0.620 384 900 890 624 × 2 = 1 + 0.240 769 801 781 248;
  • 16) 0.240 769 801 781 248 × 2 = 0 + 0.481 539 603 562 496;
  • 17) 0.481 539 603 562 496 × 2 = 0 + 0.963 079 207 124 992;
  • 18) 0.963 079 207 124 992 × 2 = 1 + 0.926 158 414 249 984;
  • 19) 0.926 158 414 249 984 × 2 = 1 + 0.852 316 828 499 968;
  • 20) 0.852 316 828 499 968 × 2 = 1 + 0.704 633 656 999 936;
  • 21) 0.704 633 656 999 936 × 2 = 1 + 0.409 267 313 999 872;
  • 22) 0.409 267 313 999 872 × 2 = 0 + 0.818 534 627 999 744;
  • 23) 0.818 534 627 999 744 × 2 = 1 + 0.637 069 255 999 488;
  • 24) 0.637 069 255 999 488 × 2 = 1 + 0.274 138 511 998 976;
  • 25) 0.274 138 511 998 976 × 2 = 0 + 0.548 277 023 997 952;
  • 26) 0.548 277 023 997 952 × 2 = 1 + 0.096 554 047 995 904;
  • 27) 0.096 554 047 995 904 × 2 = 0 + 0.193 108 095 991 808;
  • 28) 0.193 108 095 991 808 × 2 = 0 + 0.386 216 191 983 616;
  • 29) 0.386 216 191 983 616 × 2 = 0 + 0.772 432 383 967 232;
  • 30) 0.772 432 383 967 232 × 2 = 1 + 0.544 864 767 934 464;
  • 31) 0.544 864 767 934 464 × 2 = 1 + 0.089 729 535 868 928;
  • 32) 0.089 729 535 868 928 × 2 = 0 + 0.179 459 071 737 856;
  • 33) 0.179 459 071 737 856 × 2 = 0 + 0.358 918 143 475 712;
  • 34) 0.358 918 143 475 712 × 2 = 0 + 0.717 836 286 951 424;
  • 35) 0.717 836 286 951 424 × 2 = 1 + 0.435 672 573 902 848;
  • 36) 0.435 672 573 902 848 × 2 = 0 + 0.871 345 147 805 696;
  • 37) 0.871 345 147 805 696 × 2 = 1 + 0.742 690 295 611 392;
  • 38) 0.742 690 295 611 392 × 2 = 1 + 0.485 380 591 222 784;
  • 39) 0.485 380 591 222 784 × 2 = 0 + 0.970 761 182 445 568;
  • 40) 0.970 761 182 445 568 × 2 = 1 + 0.941 522 364 891 136;
  • 41) 0.941 522 364 891 136 × 2 = 1 + 0.883 044 729 782 272;
  • 42) 0.883 044 729 782 272 × 2 = 1 + 0.766 089 459 564 544;
  • 43) 0.766 089 459 564 544 × 2 = 1 + 0.532 178 919 129 088;
  • 44) 0.532 178 919 129 088 × 2 = 1 + 0.064 357 838 258 176;
  • 45) 0.064 357 838 258 176 × 2 = 0 + 0.128 715 676 516 352;
  • 46) 0.128 715 676 516 352 × 2 = 0 + 0.257 431 353 032 704;
  • 47) 0.257 431 353 032 704 × 2 = 0 + 0.514 862 706 065 408;
  • 48) 0.514 862 706 065 408 × 2 = 1 + 0.029 725 412 130 816;
  • 49) 0.029 725 412 130 816 × 2 = 0 + 0.059 450 824 261 632;
  • 50) 0.059 450 824 261 632 × 2 = 0 + 0.118 901 648 523 264;
  • 51) 0.118 901 648 523 264 × 2 = 0 + 0.237 803 297 046 528;
  • 52) 0.237 803 297 046 528 × 2 = 0 + 0.475 606 594 093 056;
  • 53) 0.475 606 594 093 056 × 2 = 0 + 0.951 213 188 186 112;
  • 54) 0.951 213 188 186 112 × 2 = 1 + 0.902 426 376 372 224;
  • 55) 0.902 426 376 372 224 × 2 = 1 + 0.804 852 752 744 448;
  • 56) 0.804 852 752 744 448 × 2 = 1 + 0.609 705 505 488 896;
  • 57) 0.609 705 505 488 896 × 2 = 1 + 0.219 411 010 977 792;
  • 58) 0.219 411 010 977 792 × 2 = 0 + 0.438 822 021 955 584;
  • 59) 0.438 822 021 955 584 × 2 = 0 + 0.877 644 043 911 168;
  • 60) 0.877 644 043 911 168 × 2 = 1 + 0.755 288 087 822 336;
  • 61) 0.755 288 087 822 336 × 2 = 1 + 0.510 576 175 644 672;
  • 62) 0.510 576 175 644 672 × 2 = 1 + 0.021 152 351 289 344;
  • 63) 0.021 152 351 289 344 × 2 = 0 + 0.042 304 702 578 688;
  • 64) 0.042 304 702 578 688 × 2 = 0 + 0.084 609 405 157 376;

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:


0.000 282 005 914 361(10) =

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

6. Positive number before normalization:

0.000 282 005 914 361(10) =

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


0.000 282 005 914 361(10) =

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

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

1.0010 0111 1011 0100 0110 0010 1101 1111 0001 0000 0111 1001 1100(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 0001 0000 0111 1001 1100


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 0001 0000 0111 1001 1100 =

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


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 0001 0000 0111 1001 1100


Decimal number -0.000 282 005 914 361 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1101 1111 0001 0000 0111 1001 1100

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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