-0.000 282 005 914 574 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 005 914 574(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
-0.000 282 005 914 574(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:

|-0.000 282 005 914 574| = 0.000 282 005 914 574


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: 0.000 282 005 914 574.

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) 0.000 282 005 914 574 × 2 = 0 + 0.000 564 011 829 148;
  • 2) 0.000 564 011 829 148 × 2 = 0 + 0.001 128 023 658 296;
  • 3) 0.001 128 023 658 296 × 2 = 0 + 0.002 256 047 316 592;
  • 4) 0.002 256 047 316 592 × 2 = 0 + 0.004 512 094 633 184;
  • 5) 0.004 512 094 633 184 × 2 = 0 + 0.009 024 189 266 368;
  • 6) 0.009 024 189 266 368 × 2 = 0 + 0.018 048 378 532 736;
  • 7) 0.018 048 378 532 736 × 2 = 0 + 0.036 096 757 065 472;
  • 8) 0.036 096 757 065 472 × 2 = 0 + 0.072 193 514 130 944;
  • 9) 0.072 193 514 130 944 × 2 = 0 + 0.144 387 028 261 888;
  • 10) 0.144 387 028 261 888 × 2 = 0 + 0.288 774 056 523 776;
  • 11) 0.288 774 056 523 776 × 2 = 0 + 0.577 548 113 047 552;
  • 12) 0.577 548 113 047 552 × 2 = 1 + 0.155 096 226 095 104;
  • 13) 0.155 096 226 095 104 × 2 = 0 + 0.310 192 452 190 208;
  • 14) 0.310 192 452 190 208 × 2 = 0 + 0.620 384 904 380 416;
  • 15) 0.620 384 904 380 416 × 2 = 1 + 0.240 769 808 760 832;
  • 16) 0.240 769 808 760 832 × 2 = 0 + 0.481 539 617 521 664;
  • 17) 0.481 539 617 521 664 × 2 = 0 + 0.963 079 235 043 328;
  • 18) 0.963 079 235 043 328 × 2 = 1 + 0.926 158 470 086 656;
  • 19) 0.926 158 470 086 656 × 2 = 1 + 0.852 316 940 173 312;
  • 20) 0.852 316 940 173 312 × 2 = 1 + 0.704 633 880 346 624;
  • 21) 0.704 633 880 346 624 × 2 = 1 + 0.409 267 760 693 248;
  • 22) 0.409 267 760 693 248 × 2 = 0 + 0.818 535 521 386 496;
  • 23) 0.818 535 521 386 496 × 2 = 1 + 0.637 071 042 772 992;
  • 24) 0.637 071 042 772 992 × 2 = 1 + 0.274 142 085 545 984;
  • 25) 0.274 142 085 545 984 × 2 = 0 + 0.548 284 171 091 968;
  • 26) 0.548 284 171 091 968 × 2 = 1 + 0.096 568 342 183 936;
  • 27) 0.096 568 342 183 936 × 2 = 0 + 0.193 136 684 367 872;
  • 28) 0.193 136 684 367 872 × 2 = 0 + 0.386 273 368 735 744;
  • 29) 0.386 273 368 735 744 × 2 = 0 + 0.772 546 737 471 488;
  • 30) 0.772 546 737 471 488 × 2 = 1 + 0.545 093 474 942 976;
  • 31) 0.545 093 474 942 976 × 2 = 1 + 0.090 186 949 885 952;
  • 32) 0.090 186 949 885 952 × 2 = 0 + 0.180 373 899 771 904;
  • 33) 0.180 373 899 771 904 × 2 = 0 + 0.360 747 799 543 808;
  • 34) 0.360 747 799 543 808 × 2 = 0 + 0.721 495 599 087 616;
  • 35) 0.721 495 599 087 616 × 2 = 1 + 0.442 991 198 175 232;
  • 36) 0.442 991 198 175 232 × 2 = 0 + 0.885 982 396 350 464;
  • 37) 0.885 982 396 350 464 × 2 = 1 + 0.771 964 792 700 928;
  • 38) 0.771 964 792 700 928 × 2 = 1 + 0.543 929 585 401 856;
  • 39) 0.543 929 585 401 856 × 2 = 1 + 0.087 859 170 803 712;
  • 40) 0.087 859 170 803 712 × 2 = 0 + 0.175 718 341 607 424;
  • 41) 0.175 718 341 607 424 × 2 = 0 + 0.351 436 683 214 848;
  • 42) 0.351 436 683 214 848 × 2 = 0 + 0.702 873 366 429 696;
  • 43) 0.702 873 366 429 696 × 2 = 1 + 0.405 746 732 859 392;
  • 44) 0.405 746 732 859 392 × 2 = 0 + 0.811 493 465 718 784;
  • 45) 0.811 493 465 718 784 × 2 = 1 + 0.622 986 931 437 568;
  • 46) 0.622 986 931 437 568 × 2 = 1 + 0.245 973 862 875 136;
  • 47) 0.245 973 862 875 136 × 2 = 0 + 0.491 947 725 750 272;
  • 48) 0.491 947 725 750 272 × 2 = 0 + 0.983 895 451 500 544;
  • 49) 0.983 895 451 500 544 × 2 = 1 + 0.967 790 903 001 088;
  • 50) 0.967 790 903 001 088 × 2 = 1 + 0.935 581 806 002 176;
  • 51) 0.935 581 806 002 176 × 2 = 1 + 0.871 163 612 004 352;
  • 52) 0.871 163 612 004 352 × 2 = 1 + 0.742 327 224 008 704;
  • 53) 0.742 327 224 008 704 × 2 = 1 + 0.484 654 448 017 408;
  • 54) 0.484 654 448 017 408 × 2 = 0 + 0.969 308 896 034 816;
  • 55) 0.969 308 896 034 816 × 2 = 1 + 0.938 617 792 069 632;
  • 56) 0.938 617 792 069 632 × 2 = 1 + 0.877 235 584 139 264;
  • 57) 0.877 235 584 139 264 × 2 = 1 + 0.754 471 168 278 528;
  • 58) 0.754 471 168 278 528 × 2 = 1 + 0.508 942 336 557 056;
  • 59) 0.508 942 336 557 056 × 2 = 1 + 0.017 884 673 114 112;
  • 60) 0.017 884 673 114 112 × 2 = 0 + 0.035 769 346 228 224;
  • 61) 0.035 769 346 228 224 × 2 = 0 + 0.071 538 692 456 448;
  • 62) 0.071 538 692 456 448 × 2 = 0 + 0.143 077 384 912 896;
  • 63) 0.143 077 384 912 896 × 2 = 0 + 0.286 154 769 825 792;
  • 64) 0.286 154 769 825 792 × 2 = 0 + 0.572 309 539 651 584;

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:


0.000 282 005 914 574(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 1100 1111 1011 1110 0000(2)

6. Positive number before normalization:

0.000 282 005 914 574(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 1100 1111 1011 1110 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:


0.000 282 005 914 574(10) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 1100 1111 1011 1110 0000(2) =

0.0000 0000 0001 0010 0111 1011 0100 0110 0010 1110 0010 1100 1111 1011 1110 0000(2) × 20 =

1.0010 0111 1011 0100 0110 0010 1110 0010 1100 1111 1011 1110 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 1110 0010 1100 1111 1011 1110 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 1110 0010 1100 1111 1011 1110 0000 =

0010 0111 1011 0100 0110 0010 1110 0010 1100 1111 1011 1110 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 1110 0010 1100 1111 1011 1110 0000


Decimal number -0.000 282 005 914 574 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

Share

» Convert decimal -0.000 282 005 914 516 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Attention Economy - The Battlefield For Your Focus. NotebookLM YouTube Video of 7.50 minutes

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