-0.000 282 005 914 449 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 449(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
-0.000 282 005 914 449(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:

|-0.000 282 005 914 449| = 0.000 282 005 914 449


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: 0.000 282 005 914 449.

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) 0.000 282 005 914 449 × 2 = 0 + 0.000 564 011 828 898;
  • 2) 0.000 564 011 828 898 × 2 = 0 + 0.001 128 023 657 796;
  • 3) 0.001 128 023 657 796 × 2 = 0 + 0.002 256 047 315 592;
  • 4) 0.002 256 047 315 592 × 2 = 0 + 0.004 512 094 631 184;
  • 5) 0.004 512 094 631 184 × 2 = 0 + 0.009 024 189 262 368;
  • 6) 0.009 024 189 262 368 × 2 = 0 + 0.018 048 378 524 736;
  • 7) 0.018 048 378 524 736 × 2 = 0 + 0.036 096 757 049 472;
  • 8) 0.036 096 757 049 472 × 2 = 0 + 0.072 193 514 098 944;
  • 9) 0.072 193 514 098 944 × 2 = 0 + 0.144 387 028 197 888;
  • 10) 0.144 387 028 197 888 × 2 = 0 + 0.288 774 056 395 776;
  • 11) 0.288 774 056 395 776 × 2 = 0 + 0.577 548 112 791 552;
  • 12) 0.577 548 112 791 552 × 2 = 1 + 0.155 096 225 583 104;
  • 13) 0.155 096 225 583 104 × 2 = 0 + 0.310 192 451 166 208;
  • 14) 0.310 192 451 166 208 × 2 = 0 + 0.620 384 902 332 416;
  • 15) 0.620 384 902 332 416 × 2 = 1 + 0.240 769 804 664 832;
  • 16) 0.240 769 804 664 832 × 2 = 0 + 0.481 539 609 329 664;
  • 17) 0.481 539 609 329 664 × 2 = 0 + 0.963 079 218 659 328;
  • 18) 0.963 079 218 659 328 × 2 = 1 + 0.926 158 437 318 656;
  • 19) 0.926 158 437 318 656 × 2 = 1 + 0.852 316 874 637 312;
  • 20) 0.852 316 874 637 312 × 2 = 1 + 0.704 633 749 274 624;
  • 21) 0.704 633 749 274 624 × 2 = 1 + 0.409 267 498 549 248;
  • 22) 0.409 267 498 549 248 × 2 = 0 + 0.818 534 997 098 496;
  • 23) 0.818 534 997 098 496 × 2 = 1 + 0.637 069 994 196 992;
  • 24) 0.637 069 994 196 992 × 2 = 1 + 0.274 139 988 393 984;
  • 25) 0.274 139 988 393 984 × 2 = 0 + 0.548 279 976 787 968;
  • 26) 0.548 279 976 787 968 × 2 = 1 + 0.096 559 953 575 936;
  • 27) 0.096 559 953 575 936 × 2 = 0 + 0.193 119 907 151 872;
  • 28) 0.193 119 907 151 872 × 2 = 0 + 0.386 239 814 303 744;
  • 29) 0.386 239 814 303 744 × 2 = 0 + 0.772 479 628 607 488;
  • 30) 0.772 479 628 607 488 × 2 = 1 + 0.544 959 257 214 976;
  • 31) 0.544 959 257 214 976 × 2 = 1 + 0.089 918 514 429 952;
  • 32) 0.089 918 514 429 952 × 2 = 0 + 0.179 837 028 859 904;
  • 33) 0.179 837 028 859 904 × 2 = 0 + 0.359 674 057 719 808;
  • 34) 0.359 674 057 719 808 × 2 = 0 + 0.719 348 115 439 616;
  • 35) 0.719 348 115 439 616 × 2 = 1 + 0.438 696 230 879 232;
  • 36) 0.438 696 230 879 232 × 2 = 0 + 0.877 392 461 758 464;
  • 37) 0.877 392 461 758 464 × 2 = 1 + 0.754 784 923 516 928;
  • 38) 0.754 784 923 516 928 × 2 = 1 + 0.509 569 847 033 856;
  • 39) 0.509 569 847 033 856 × 2 = 1 + 0.019 139 694 067 712;
  • 40) 0.019 139 694 067 712 × 2 = 0 + 0.038 279 388 135 424;
  • 41) 0.038 279 388 135 424 × 2 = 0 + 0.076 558 776 270 848;
  • 42) 0.076 558 776 270 848 × 2 = 0 + 0.153 117 552 541 696;
  • 43) 0.153 117 552 541 696 × 2 = 0 + 0.306 235 105 083 392;
  • 44) 0.306 235 105 083 392 × 2 = 0 + 0.612 470 210 166 784;
  • 45) 0.612 470 210 166 784 × 2 = 1 + 0.224 940 420 333 568;
  • 46) 0.224 940 420 333 568 × 2 = 0 + 0.449 880 840 667 136;
  • 47) 0.449 880 840 667 136 × 2 = 0 + 0.899 761 681 334 272;
  • 48) 0.899 761 681 334 272 × 2 = 1 + 0.799 523 362 668 544;
  • 49) 0.799 523 362 668 544 × 2 = 1 + 0.599 046 725 337 088;
  • 50) 0.599 046 725 337 088 × 2 = 1 + 0.198 093 450 674 176;
  • 51) 0.198 093 450 674 176 × 2 = 0 + 0.396 186 901 348 352;
  • 52) 0.396 186 901 348 352 × 2 = 0 + 0.792 373 802 696 704;
  • 53) 0.792 373 802 696 704 × 2 = 1 + 0.584 747 605 393 408;
  • 54) 0.584 747 605 393 408 × 2 = 1 + 0.169 495 210 786 816;
  • 55) 0.169 495 210 786 816 × 2 = 0 + 0.338 990 421 573 632;
  • 56) 0.338 990 421 573 632 × 2 = 0 + 0.677 980 843 147 264;
  • 57) 0.677 980 843 147 264 × 2 = 1 + 0.355 961 686 294 528;
  • 58) 0.355 961 686 294 528 × 2 = 0 + 0.711 923 372 589 056;
  • 59) 0.711 923 372 589 056 × 2 = 1 + 0.423 846 745 178 112;
  • 60) 0.423 846 745 178 112 × 2 = 0 + 0.847 693 490 356 224;
  • 61) 0.847 693 490 356 224 × 2 = 1 + 0.695 386 980 712 448;
  • 62) 0.695 386 980 712 448 × 2 = 1 + 0.390 773 961 424 896;
  • 63) 0.390 773 961 424 896 × 2 = 0 + 0.781 547 922 849 792;
  • 64) 0.781 547 922 849 792 × 2 = 1 + 0.563 095 845 699 584;

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:


0.000 282 005 914 449(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1001 1100 1100 1010 1101(2)

6. Positive number before normalization:

0.000 282 005 914 449(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1001 1100 1100 1010 1101(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:


0.000 282 005 914 449(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1001 1100 1100 1010 1101(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0000 1001 1100 1100 1010 1101(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0000 1001 1100 1100 1010 1101(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 1110 0000 1001 1100 1100 1010 1101


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 1110 0000 1001 1100 1100 1010 1101 =

0010 0111 1011 0100 0110 0010 1110 0000 1001 1100 1100 1010 1101


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 1110 0000 1001 1100 1100 1010 1101


Decimal number -0.000 282 005 914 449 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0111 1011 0100 0110 0010 1110 0000 1001 1100 1100 1010 1101

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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