-284.011 100 000 001 110 001 110 055 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 055 1(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
-284.011 100 000 001 110 001 110 055 1(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:

|-284.011 100 000 001 110 001 110 055 1| = 284.011 100 000 001 110 001 110 055 1


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: 0.011 100 000 001 110 001 110 055 1.

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) 0.011 100 000 001 110 001 110 055 1 × 2 = 0 + 0.022 200 000 002 220 002 220 110 2;
  • 2) 0.022 200 000 002 220 002 220 110 2 × 2 = 0 + 0.044 400 000 004 440 004 440 220 4;
  • 3) 0.044 400 000 004 440 004 440 220 4 × 2 = 0 + 0.088 800 000 008 880 008 880 440 8;
  • 4) 0.088 800 000 008 880 008 880 440 8 × 2 = 0 + 0.177 600 000 017 760 017 760 881 6;
  • 5) 0.177 600 000 017 760 017 760 881 6 × 2 = 0 + 0.355 200 000 035 520 035 521 763 2;
  • 6) 0.355 200 000 035 520 035 521 763 2 × 2 = 0 + 0.710 400 000 071 040 071 043 526 4;
  • 7) 0.710 400 000 071 040 071 043 526 4 × 2 = 1 + 0.420 800 000 142 080 142 087 052 8;
  • 8) 0.420 800 000 142 080 142 087 052 8 × 2 = 0 + 0.841 600 000 284 160 284 174 105 6;
  • 9) 0.841 600 000 284 160 284 174 105 6 × 2 = 1 + 0.683 200 000 568 320 568 348 211 2;
  • 10) 0.683 200 000 568 320 568 348 211 2 × 2 = 1 + 0.366 400 001 136 641 136 696 422 4;
  • 11) 0.366 400 001 136 641 136 696 422 4 × 2 = 0 + 0.732 800 002 273 282 273 392 844 8;
  • 12) 0.732 800 002 273 282 273 392 844 8 × 2 = 1 + 0.465 600 004 546 564 546 785 689 6;
  • 13) 0.465 600 004 546 564 546 785 689 6 × 2 = 0 + 0.931 200 009 093 129 093 571 379 2;
  • 14) 0.931 200 009 093 129 093 571 379 2 × 2 = 1 + 0.862 400 018 186 258 187 142 758 4;
  • 15) 0.862 400 018 186 258 187 142 758 4 × 2 = 1 + 0.724 800 036 372 516 374 285 516 8;
  • 16) 0.724 800 036 372 516 374 285 516 8 × 2 = 1 + 0.449 600 072 745 032 748 571 033 6;
  • 17) 0.449 600 072 745 032 748 571 033 6 × 2 = 0 + 0.899 200 145 490 065 497 142 067 2;
  • 18) 0.899 200 145 490 065 497 142 067 2 × 2 = 1 + 0.798 400 290 980 130 994 284 134 4;
  • 19) 0.798 400 290 980 130 994 284 134 4 × 2 = 1 + 0.596 800 581 960 261 988 568 268 8;
  • 20) 0.596 800 581 960 261 988 568 268 8 × 2 = 1 + 0.193 601 163 920 523 977 136 537 6;
  • 21) 0.193 601 163 920 523 977 136 537 6 × 2 = 0 + 0.387 202 327 841 047 954 273 075 2;
  • 22) 0.387 202 327 841 047 954 273 075 2 × 2 = 0 + 0.774 404 655 682 095 908 546 150 4;
  • 23) 0.774 404 655 682 095 908 546 150 4 × 2 = 1 + 0.548 809 311 364 191 817 092 300 8;
  • 24) 0.548 809 311 364 191 817 092 300 8 × 2 = 1 + 0.097 618 622 728 383 634 184 601 6;
  • 25) 0.097 618 622 728 383 634 184 601 6 × 2 = 0 + 0.195 237 245 456 767 268 369 203 2;
  • 26) 0.195 237 245 456 767 268 369 203 2 × 2 = 0 + 0.390 474 490 913 534 536 738 406 4;
  • 27) 0.390 474 490 913 534 536 738 406 4 × 2 = 0 + 0.780 948 981 827 069 073 476 812 8;
  • 28) 0.780 948 981 827 069 073 476 812 8 × 2 = 1 + 0.561 897 963 654 138 146 953 625 6;
  • 29) 0.561 897 963 654 138 146 953 625 6 × 2 = 1 + 0.123 795 927 308 276 293 907 251 2;
  • 30) 0.123 795 927 308 276 293 907 251 2 × 2 = 0 + 0.247 591 854 616 552 587 814 502 4;
  • 31) 0.247 591 854 616 552 587 814 502 4 × 2 = 0 + 0.495 183 709 233 105 175 629 004 8;
  • 32) 0.495 183 709 233 105 175 629 004 8 × 2 = 0 + 0.990 367 418 466 210 351 258 009 6;
  • 33) 0.990 367 418 466 210 351 258 009 6 × 2 = 1 + 0.980 734 836 932 420 702 516 019 2;
  • 34) 0.980 734 836 932 420 702 516 019 2 × 2 = 1 + 0.961 469 673 864 841 405 032 038 4;
  • 35) 0.961 469 673 864 841 405 032 038 4 × 2 = 1 + 0.922 939 347 729 682 810 064 076 8;
  • 36) 0.922 939 347 729 682 810 064 076 8 × 2 = 1 + 0.845 878 695 459 365 620 128 153 6;
  • 37) 0.845 878 695 459 365 620 128 153 6 × 2 = 1 + 0.691 757 390 918 731 240 256 307 2;
  • 38) 0.691 757 390 918 731 240 256 307 2 × 2 = 1 + 0.383 514 781 837 462 480 512 614 4;
  • 39) 0.383 514 781 837 462 480 512 614 4 × 2 = 0 + 0.767 029 563 674 924 961 025 228 8;
  • 40) 0.767 029 563 674 924 961 025 228 8 × 2 = 1 + 0.534 059 127 349 849 922 050 457 6;
  • 41) 0.534 059 127 349 849 922 050 457 6 × 2 = 1 + 0.068 118 254 699 699 844 100 915 2;
  • 42) 0.068 118 254 699 699 844 100 915 2 × 2 = 0 + 0.136 236 509 399 399 688 201 830 4;
  • 43) 0.136 236 509 399 399 688 201 830 4 × 2 = 0 + 0.272 473 018 798 799 376 403 660 8;
  • 44) 0.272 473 018 798 799 376 403 660 8 × 2 = 0 + 0.544 946 037 597 598 752 807 321 6;
  • 45) 0.544 946 037 597 598 752 807 321 6 × 2 = 1 + 0.089 892 075 195 197 505 614 643 2;
  • 46) 0.089 892 075 195 197 505 614 643 2 × 2 = 0 + 0.179 784 150 390 395 011 229 286 4;
  • 47) 0.179 784 150 390 395 011 229 286 4 × 2 = 0 + 0.359 568 300 780 790 022 458 572 8;
  • 48) 0.359 568 300 780 790 022 458 572 8 × 2 = 0 + 0.719 136 601 561 580 044 917 145 6;
  • 49) 0.719 136 601 561 580 044 917 145 6 × 2 = 1 + 0.438 273 203 123 160 089 834 291 2;
  • 50) 0.438 273 203 123 160 089 834 291 2 × 2 = 0 + 0.876 546 406 246 320 179 668 582 4;
  • 51) 0.876 546 406 246 320 179 668 582 4 × 2 = 1 + 0.753 092 812 492 640 359 337 164 8;
  • 52) 0.753 092 812 492 640 359 337 164 8 × 2 = 1 + 0.506 185 624 985 280 718 674 329 6;
  • 53) 0.506 185 624 985 280 718 674 329 6 × 2 = 1 + 0.012 371 249 970 561 437 348 659 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:


0.011 100 000 001 110 001 110 055 1(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 055 1(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:


284.011 100 000 001 110 001 110 055 1(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 -284.011 100 000 001 110 001 110 055 1 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 -284.011 100 000 001 110 001 110 064 2 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