-0.000 282 005 914 336 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 336(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
-0.000 282 005 914 336(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:

|-0.000 282 005 914 336| = 0.000 282 005 914 336


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: 0.000 282 005 914 336.

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) 0.000 282 005 914 336 × 2 = 0 + 0.000 564 011 828 672;
  • 2) 0.000 564 011 828 672 × 2 = 0 + 0.001 128 023 657 344;
  • 3) 0.001 128 023 657 344 × 2 = 0 + 0.002 256 047 314 688;
  • 4) 0.002 256 047 314 688 × 2 = 0 + 0.004 512 094 629 376;
  • 5) 0.004 512 094 629 376 × 2 = 0 + 0.009 024 189 258 752;
  • 6) 0.009 024 189 258 752 × 2 = 0 + 0.018 048 378 517 504;
  • 7) 0.018 048 378 517 504 × 2 = 0 + 0.036 096 757 035 008;
  • 8) 0.036 096 757 035 008 × 2 = 0 + 0.072 193 514 070 016;
  • 9) 0.072 193 514 070 016 × 2 = 0 + 0.144 387 028 140 032;
  • 10) 0.144 387 028 140 032 × 2 = 0 + 0.288 774 056 280 064;
  • 11) 0.288 774 056 280 064 × 2 = 0 + 0.577 548 112 560 128;
  • 12) 0.577 548 112 560 128 × 2 = 1 + 0.155 096 225 120 256;
  • 13) 0.155 096 225 120 256 × 2 = 0 + 0.310 192 450 240 512;
  • 14) 0.310 192 450 240 512 × 2 = 0 + 0.620 384 900 481 024;
  • 15) 0.620 384 900 481 024 × 2 = 1 + 0.240 769 800 962 048;
  • 16) 0.240 769 800 962 048 × 2 = 0 + 0.481 539 601 924 096;
  • 17) 0.481 539 601 924 096 × 2 = 0 + 0.963 079 203 848 192;
  • 18) 0.963 079 203 848 192 × 2 = 1 + 0.926 158 407 696 384;
  • 19) 0.926 158 407 696 384 × 2 = 1 + 0.852 316 815 392 768;
  • 20) 0.852 316 815 392 768 × 2 = 1 + 0.704 633 630 785 536;
  • 21) 0.704 633 630 785 536 × 2 = 1 + 0.409 267 261 571 072;
  • 22) 0.409 267 261 571 072 × 2 = 0 + 0.818 534 523 142 144;
  • 23) 0.818 534 523 142 144 × 2 = 1 + 0.637 069 046 284 288;
  • 24) 0.637 069 046 284 288 × 2 = 1 + 0.274 138 092 568 576;
  • 25) 0.274 138 092 568 576 × 2 = 0 + 0.548 276 185 137 152;
  • 26) 0.548 276 185 137 152 × 2 = 1 + 0.096 552 370 274 304;
  • 27) 0.096 552 370 274 304 × 2 = 0 + 0.193 104 740 548 608;
  • 28) 0.193 104 740 548 608 × 2 = 0 + 0.386 209 481 097 216;
  • 29) 0.386 209 481 097 216 × 2 = 0 + 0.772 418 962 194 432;
  • 30) 0.772 418 962 194 432 × 2 = 1 + 0.544 837 924 388 864;
  • 31) 0.544 837 924 388 864 × 2 = 1 + 0.089 675 848 777 728;
  • 32) 0.089 675 848 777 728 × 2 = 0 + 0.179 351 697 555 456;
  • 33) 0.179 351 697 555 456 × 2 = 0 + 0.358 703 395 110 912;
  • 34) 0.358 703 395 110 912 × 2 = 0 + 0.717 406 790 221 824;
  • 35) 0.717 406 790 221 824 × 2 = 1 + 0.434 813 580 443 648;
  • 36) 0.434 813 580 443 648 × 2 = 0 + 0.869 627 160 887 296;
  • 37) 0.869 627 160 887 296 × 2 = 1 + 0.739 254 321 774 592;
  • 38) 0.739 254 321 774 592 × 2 = 1 + 0.478 508 643 549 184;
  • 39) 0.478 508 643 549 184 × 2 = 0 + 0.957 017 287 098 368;
  • 40) 0.957 017 287 098 368 × 2 = 1 + 0.914 034 574 196 736;
  • 41) 0.914 034 574 196 736 × 2 = 1 + 0.828 069 148 393 472;
  • 42) 0.828 069 148 393 472 × 2 = 1 + 0.656 138 296 786 944;
  • 43) 0.656 138 296 786 944 × 2 = 1 + 0.312 276 593 573 888;
  • 44) 0.312 276 593 573 888 × 2 = 0 + 0.624 553 187 147 776;
  • 45) 0.624 553 187 147 776 × 2 = 1 + 0.249 106 374 295 552;
  • 46) 0.249 106 374 295 552 × 2 = 0 + 0.498 212 748 591 104;
  • 47) 0.498 212 748 591 104 × 2 = 0 + 0.996 425 497 182 208;
  • 48) 0.996 425 497 182 208 × 2 = 1 + 0.992 850 994 364 416;
  • 49) 0.992 850 994 364 416 × 2 = 1 + 0.985 701 988 728 832;
  • 50) 0.985 701 988 728 832 × 2 = 1 + 0.971 403 977 457 664;
  • 51) 0.971 403 977 457 664 × 2 = 1 + 0.942 807 954 915 328;
  • 52) 0.942 807 954 915 328 × 2 = 1 + 0.885 615 909 830 656;
  • 53) 0.885 615 909 830 656 × 2 = 1 + 0.771 231 819 661 312;
  • 54) 0.771 231 819 661 312 × 2 = 1 + 0.542 463 639 322 624;
  • 55) 0.542 463 639 322 624 × 2 = 1 + 0.084 927 278 645 248;
  • 56) 0.084 927 278 645 248 × 2 = 0 + 0.169 854 557 290 496;
  • 57) 0.169 854 557 290 496 × 2 = 0 + 0.339 709 114 580 992;
  • 58) 0.339 709 114 580 992 × 2 = 0 + 0.679 418 229 161 984;
  • 59) 0.679 418 229 161 984 × 2 = 1 + 0.358 836 458 323 968;
  • 60) 0.358 836 458 323 968 × 2 = 0 + 0.717 672 916 647 936;
  • 61) 0.717 672 916 647 936 × 2 = 1 + 0.435 345 833 295 872;
  • 62) 0.435 345 833 295 872 × 2 = 0 + 0.870 691 666 591 744;
  • 63) 0.870 691 666 591 744 × 2 = 1 + 0.741 383 333 183 488;
  • 64) 0.741 383 333 183 488 × 2 = 1 + 0.482 766 666 366 976;

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:


0.000 282 005 914 336(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1110 1001 1111 1110 0010 1011(2)

6. Positive number before normalization:

0.000 282 005 914 336(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1110 1001 1111 1110 0010 1011(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:


0.000 282 005 914 336(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1110 1001 1111 1110 0010 1011(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1101 1110 1001 1111 1110 0010 1011(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1101 1110 1001 1111 1110 0010 1011(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 1110 1001 1111 1110 0010 1011


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 1110 1001 1111 1110 0010 1011 =

0010 0111 1011 0100 0110 0010 1101 1110 1001 1111 1110 0010 1011


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 1110 1001 1111 1110 0010 1011


Decimal number -0.000 282 005 914 336 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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