-284.011 100 000 001 110 001 110 056 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 056 6(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 6(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 6| = 284.011 100 000 001 110 001 110 056 6


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 6.

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 6 × 2 = 0 + 0.022 200 000 002 220 002 220 113 2;
  • 2) 0.022 200 000 002 220 002 220 113 2 × 2 = 0 + 0.044 400 000 004 440 004 440 226 4;
  • 3) 0.044 400 000 004 440 004 440 226 4 × 2 = 0 + 0.088 800 000 008 880 008 880 452 8;
  • 4) 0.088 800 000 008 880 008 880 452 8 × 2 = 0 + 0.177 600 000 017 760 017 760 905 6;
  • 5) 0.177 600 000 017 760 017 760 905 6 × 2 = 0 + 0.355 200 000 035 520 035 521 811 2;
  • 6) 0.355 200 000 035 520 035 521 811 2 × 2 = 0 + 0.710 400 000 071 040 071 043 622 4;
  • 7) 0.710 400 000 071 040 071 043 622 4 × 2 = 1 + 0.420 800 000 142 080 142 087 244 8;
  • 8) 0.420 800 000 142 080 142 087 244 8 × 2 = 0 + 0.841 600 000 284 160 284 174 489 6;
  • 9) 0.841 600 000 284 160 284 174 489 6 × 2 = 1 + 0.683 200 000 568 320 568 348 979 2;
  • 10) 0.683 200 000 568 320 568 348 979 2 × 2 = 1 + 0.366 400 001 136 641 136 697 958 4;
  • 11) 0.366 400 001 136 641 136 697 958 4 × 2 = 0 + 0.732 800 002 273 282 273 395 916 8;
  • 12) 0.732 800 002 273 282 273 395 916 8 × 2 = 1 + 0.465 600 004 546 564 546 791 833 6;
  • 13) 0.465 600 004 546 564 546 791 833 6 × 2 = 0 + 0.931 200 009 093 129 093 583 667 2;
  • 14) 0.931 200 009 093 129 093 583 667 2 × 2 = 1 + 0.862 400 018 186 258 187 167 334 4;
  • 15) 0.862 400 018 186 258 187 167 334 4 × 2 = 1 + 0.724 800 036 372 516 374 334 668 8;
  • 16) 0.724 800 036 372 516 374 334 668 8 × 2 = 1 + 0.449 600 072 745 032 748 669 337 6;
  • 17) 0.449 600 072 745 032 748 669 337 6 × 2 = 0 + 0.899 200 145 490 065 497 338 675 2;
  • 18) 0.899 200 145 490 065 497 338 675 2 × 2 = 1 + 0.798 400 290 980 130 994 677 350 4;
  • 19) 0.798 400 290 980 130 994 677 350 4 × 2 = 1 + 0.596 800 581 960 261 989 354 700 8;
  • 20) 0.596 800 581 960 261 989 354 700 8 × 2 = 1 + 0.193 601 163 920 523 978 709 401 6;
  • 21) 0.193 601 163 920 523 978 709 401 6 × 2 = 0 + 0.387 202 327 841 047 957 418 803 2;
  • 22) 0.387 202 327 841 047 957 418 803 2 × 2 = 0 + 0.774 404 655 682 095 914 837 606 4;
  • 23) 0.774 404 655 682 095 914 837 606 4 × 2 = 1 + 0.548 809 311 364 191 829 675 212 8;
  • 24) 0.548 809 311 364 191 829 675 212 8 × 2 = 1 + 0.097 618 622 728 383 659 350 425 6;
  • 25) 0.097 618 622 728 383 659 350 425 6 × 2 = 0 + 0.195 237 245 456 767 318 700 851 2;
  • 26) 0.195 237 245 456 767 318 700 851 2 × 2 = 0 + 0.390 474 490 913 534 637 401 702 4;
  • 27) 0.390 474 490 913 534 637 401 702 4 × 2 = 0 + 0.780 948 981 827 069 274 803 404 8;
  • 28) 0.780 948 981 827 069 274 803 404 8 × 2 = 1 + 0.561 897 963 654 138 549 606 809 6;
  • 29) 0.561 897 963 654 138 549 606 809 6 × 2 = 1 + 0.123 795 927 308 277 099 213 619 2;
  • 30) 0.123 795 927 308 277 099 213 619 2 × 2 = 0 + 0.247 591 854 616 554 198 427 238 4;
  • 31) 0.247 591 854 616 554 198 427 238 4 × 2 = 0 + 0.495 183 709 233 108 396 854 476 8;
  • 32) 0.495 183 709 233 108 396 854 476 8 × 2 = 0 + 0.990 367 418 466 216 793 708 953 6;
  • 33) 0.990 367 418 466 216 793 708 953 6 × 2 = 1 + 0.980 734 836 932 433 587 417 907 2;
  • 34) 0.980 734 836 932 433 587 417 907 2 × 2 = 1 + 0.961 469 673 864 867 174 835 814 4;
  • 35) 0.961 469 673 864 867 174 835 814 4 × 2 = 1 + 0.922 939 347 729 734 349 671 628 8;
  • 36) 0.922 939 347 729 734 349 671 628 8 × 2 = 1 + 0.845 878 695 459 468 699 343 257 6;
  • 37) 0.845 878 695 459 468 699 343 257 6 × 2 = 1 + 0.691 757 390 918 937 398 686 515 2;
  • 38) 0.691 757 390 918 937 398 686 515 2 × 2 = 1 + 0.383 514 781 837 874 797 373 030 4;
  • 39) 0.383 514 781 837 874 797 373 030 4 × 2 = 0 + 0.767 029 563 675 749 594 746 060 8;
  • 40) 0.767 029 563 675 749 594 746 060 8 × 2 = 1 + 0.534 059 127 351 499 189 492 121 6;
  • 41) 0.534 059 127 351 499 189 492 121 6 × 2 = 1 + 0.068 118 254 702 998 378 984 243 2;
  • 42) 0.068 118 254 702 998 378 984 243 2 × 2 = 0 + 0.136 236 509 405 996 757 968 486 4;
  • 43) 0.136 236 509 405 996 757 968 486 4 × 2 = 0 + 0.272 473 018 811 993 515 936 972 8;
  • 44) 0.272 473 018 811 993 515 936 972 8 × 2 = 0 + 0.544 946 037 623 987 031 873 945 6;
  • 45) 0.544 946 037 623 987 031 873 945 6 × 2 = 1 + 0.089 892 075 247 974 063 747 891 2;
  • 46) 0.089 892 075 247 974 063 747 891 2 × 2 = 0 + 0.179 784 150 495 948 127 495 782 4;
  • 47) 0.179 784 150 495 948 127 495 782 4 × 2 = 0 + 0.359 568 300 991 896 254 991 564 8;
  • 48) 0.359 568 300 991 896 254 991 564 8 × 2 = 0 + 0.719 136 601 983 792 509 983 129 6;
  • 49) 0.719 136 601 983 792 509 983 129 6 × 2 = 1 + 0.438 273 203 967 585 019 966 259 2;
  • 50) 0.438 273 203 967 585 019 966 259 2 × 2 = 0 + 0.876 546 407 935 170 039 932 518 4;
  • 51) 0.876 546 407 935 170 039 932 518 4 × 2 = 1 + 0.753 092 815 870 340 079 865 036 8;
  • 52) 0.753 092 815 870 340 079 865 036 8 × 2 = 1 + 0.506 185 631 740 680 159 730 073 6;
  • 53) 0.506 185 631 740 680 159 730 073 6 × 2 = 1 + 0.012 371 263 481 360 319 460 147 2;

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 6(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 6(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 6(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 6 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 -284.011 100 000 001 110 001 110 050 5 to 64 bit double precision IEEE 754 binary floating point representation standard

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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