-284.011 100 000 001 110 001 110 056 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 056 2(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
-284.011 100 000 001 110 001 110 056 2(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:

|-284.011 100 000 001 110 001 110 056 2| = 284.011 100 000 001 110 001 110 056 2


2. First, convert to binary (in base 2) the integer part: 284.
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;
  • 284 ÷ 2 = 142 + 0;
  • 142 ÷ 2 = 71 + 0;
  • 71 ÷ 2 = 35 + 1;
  • 35 ÷ 2 = 17 + 1;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.

284(10) =

1 0001 1100(2)


4. Convert to binary (base 2) the fractional part: 0.011 100 000 001 110 001 110 056 2.

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) 0.011 100 000 001 110 001 110 056 2 × 2 = 0 + 0.022 200 000 002 220 002 220 112 4;
  • 2) 0.022 200 000 002 220 002 220 112 4 × 2 = 0 + 0.044 400 000 004 440 004 440 224 8;
  • 3) 0.044 400 000 004 440 004 440 224 8 × 2 = 0 + 0.088 800 000 008 880 008 880 449 6;
  • 4) 0.088 800 000 008 880 008 880 449 6 × 2 = 0 + 0.177 600 000 017 760 017 760 899 2;
  • 5) 0.177 600 000 017 760 017 760 899 2 × 2 = 0 + 0.355 200 000 035 520 035 521 798 4;
  • 6) 0.355 200 000 035 520 035 521 798 4 × 2 = 0 + 0.710 400 000 071 040 071 043 596 8;
  • 7) 0.710 400 000 071 040 071 043 596 8 × 2 = 1 + 0.420 800 000 142 080 142 087 193 6;
  • 8) 0.420 800 000 142 080 142 087 193 6 × 2 = 0 + 0.841 600 000 284 160 284 174 387 2;
  • 9) 0.841 600 000 284 160 284 174 387 2 × 2 = 1 + 0.683 200 000 568 320 568 348 774 4;
  • 10) 0.683 200 000 568 320 568 348 774 4 × 2 = 1 + 0.366 400 001 136 641 136 697 548 8;
  • 11) 0.366 400 001 136 641 136 697 548 8 × 2 = 0 + 0.732 800 002 273 282 273 395 097 6;
  • 12) 0.732 800 002 273 282 273 395 097 6 × 2 = 1 + 0.465 600 004 546 564 546 790 195 2;
  • 13) 0.465 600 004 546 564 546 790 195 2 × 2 = 0 + 0.931 200 009 093 129 093 580 390 4;
  • 14) 0.931 200 009 093 129 093 580 390 4 × 2 = 1 + 0.862 400 018 186 258 187 160 780 8;
  • 15) 0.862 400 018 186 258 187 160 780 8 × 2 = 1 + 0.724 800 036 372 516 374 321 561 6;
  • 16) 0.724 800 036 372 516 374 321 561 6 × 2 = 1 + 0.449 600 072 745 032 748 643 123 2;
  • 17) 0.449 600 072 745 032 748 643 123 2 × 2 = 0 + 0.899 200 145 490 065 497 286 246 4;
  • 18) 0.899 200 145 490 065 497 286 246 4 × 2 = 1 + 0.798 400 290 980 130 994 572 492 8;
  • 19) 0.798 400 290 980 130 994 572 492 8 × 2 = 1 + 0.596 800 581 960 261 989 144 985 6;
  • 20) 0.596 800 581 960 261 989 144 985 6 × 2 = 1 + 0.193 601 163 920 523 978 289 971 2;
  • 21) 0.193 601 163 920 523 978 289 971 2 × 2 = 0 + 0.387 202 327 841 047 956 579 942 4;
  • 22) 0.387 202 327 841 047 956 579 942 4 × 2 = 0 + 0.774 404 655 682 095 913 159 884 8;
  • 23) 0.774 404 655 682 095 913 159 884 8 × 2 = 1 + 0.548 809 311 364 191 826 319 769 6;
  • 24) 0.548 809 311 364 191 826 319 769 6 × 2 = 1 + 0.097 618 622 728 383 652 639 539 2;
  • 25) 0.097 618 622 728 383 652 639 539 2 × 2 = 0 + 0.195 237 245 456 767 305 279 078 4;
  • 26) 0.195 237 245 456 767 305 279 078 4 × 2 = 0 + 0.390 474 490 913 534 610 558 156 8;
  • 27) 0.390 474 490 913 534 610 558 156 8 × 2 = 0 + 0.780 948 981 827 069 221 116 313 6;
  • 28) 0.780 948 981 827 069 221 116 313 6 × 2 = 1 + 0.561 897 963 654 138 442 232 627 2;
  • 29) 0.561 897 963 654 138 442 232 627 2 × 2 = 1 + 0.123 795 927 308 276 884 465 254 4;
  • 30) 0.123 795 927 308 276 884 465 254 4 × 2 = 0 + 0.247 591 854 616 553 768 930 508 8;
  • 31) 0.247 591 854 616 553 768 930 508 8 × 2 = 0 + 0.495 183 709 233 107 537 861 017 6;
  • 32) 0.495 183 709 233 107 537 861 017 6 × 2 = 0 + 0.990 367 418 466 215 075 722 035 2;
  • 33) 0.990 367 418 466 215 075 722 035 2 × 2 = 1 + 0.980 734 836 932 430 151 444 070 4;
  • 34) 0.980 734 836 932 430 151 444 070 4 × 2 = 1 + 0.961 469 673 864 860 302 888 140 8;
  • 35) 0.961 469 673 864 860 302 888 140 8 × 2 = 1 + 0.922 939 347 729 720 605 776 281 6;
  • 36) 0.922 939 347 729 720 605 776 281 6 × 2 = 1 + 0.845 878 695 459 441 211 552 563 2;
  • 37) 0.845 878 695 459 441 211 552 563 2 × 2 = 1 + 0.691 757 390 918 882 423 105 126 4;
  • 38) 0.691 757 390 918 882 423 105 126 4 × 2 = 1 + 0.383 514 781 837 764 846 210 252 8;
  • 39) 0.383 514 781 837 764 846 210 252 8 × 2 = 0 + 0.767 029 563 675 529 692 420 505 6;
  • 40) 0.767 029 563 675 529 692 420 505 6 × 2 = 1 + 0.534 059 127 351 059 384 841 011 2;
  • 41) 0.534 059 127 351 059 384 841 011 2 × 2 = 1 + 0.068 118 254 702 118 769 682 022 4;
  • 42) 0.068 118 254 702 118 769 682 022 4 × 2 = 0 + 0.136 236 509 404 237 539 364 044 8;
  • 43) 0.136 236 509 404 237 539 364 044 8 × 2 = 0 + 0.272 473 018 808 475 078 728 089 6;
  • 44) 0.272 473 018 808 475 078 728 089 6 × 2 = 0 + 0.544 946 037 616 950 157 456 179 2;
  • 45) 0.544 946 037 616 950 157 456 179 2 × 2 = 1 + 0.089 892 075 233 900 314 912 358 4;
  • 46) 0.089 892 075 233 900 314 912 358 4 × 2 = 0 + 0.179 784 150 467 800 629 824 716 8;
  • 47) 0.179 784 150 467 800 629 824 716 8 × 2 = 0 + 0.359 568 300 935 601 259 649 433 6;
  • 48) 0.359 568 300 935 601 259 649 433 6 × 2 = 0 + 0.719 136 601 871 202 519 298 867 2;
  • 49) 0.719 136 601 871 202 519 298 867 2 × 2 = 1 + 0.438 273 203 742 405 038 597 734 4;
  • 50) 0.438 273 203 742 405 038 597 734 4 × 2 = 0 + 0.876 546 407 484 810 077 195 468 8;
  • 51) 0.876 546 407 484 810 077 195 468 8 × 2 = 1 + 0.753 092 814 969 620 154 390 937 6;
  • 52) 0.753 092 814 969 620 154 390 937 6 × 2 = 1 + 0.506 185 629 939 240 308 781 875 2;
  • 53) 0.506 185 629 939 240 308 781 875 2 × 2 = 1 + 0.012 371 259 878 480 617 563 750 4;

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:


0.011 100 000 001 110 001 110 056 2(10) =

0.0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2)

6. Positive number before normalization:

284.011 100 000 001 110 001 110 056 2(10) =

1 0001 1100.0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the left, so that only one non zero digit remains to the left of it:


284.011 100 000 001 110 001 110 056 2(10) =

1 0001 1100.0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2) =

1 0001 1100.0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2) × 20 =

1.0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1(2) × 28


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): 8

Mantissa (not normalized):
1.0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1000 1011 1


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

8 + 2(11-1) - 1 =

(8 + 1 023)(10) =

1 031(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;
  • 1 031 ÷ 2 = 515 + 1;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1031(10) =

100 0000 0111(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000 1 0001 0111 =

0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000


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) =
100 0000 0111

Mantissa (52 bits) =
0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000


Decimal number -284.011 100 000 001 110 001 110 056 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 100 0000 0111 - 0001 1100 0000 0010 1101 0111 0111 0011 0001 1000 1111 1101 1000

» Convert decimal -284.011 100 000 001 110 001 110 059 5 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 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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