-284.011 100 000 001 110 001 110 000 000 009 82 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 000 000 009 82(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
-284.011 100 000 001 110 001 110 000 000 009 82(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:

|-284.011 100 000 001 110 001 110 000 000 009 82| = 284.011 100 000 001 110 001 110 000 000 009 82


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: 0.011 100 000 001 110 001 110 000 000 009 82.

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) 0.011 100 000 001 110 001 110 000 000 009 82 × 2 = 0 + 0.022 200 000 002 220 002 220 000 000 019 64;
  • 2) 0.022 200 000 002 220 002 220 000 000 019 64 × 2 = 0 + 0.044 400 000 004 440 004 440 000 000 039 28;
  • 3) 0.044 400 000 004 440 004 440 000 000 039 28 × 2 = 0 + 0.088 800 000 008 880 008 880 000 000 078 56;
  • 4) 0.088 800 000 008 880 008 880 000 000 078 56 × 2 = 0 + 0.177 600 000 017 760 017 760 000 000 157 12;
  • 5) 0.177 600 000 017 760 017 760 000 000 157 12 × 2 = 0 + 0.355 200 000 035 520 035 520 000 000 314 24;
  • 6) 0.355 200 000 035 520 035 520 000 000 314 24 × 2 = 0 + 0.710 400 000 071 040 071 040 000 000 628 48;
  • 7) 0.710 400 000 071 040 071 040 000 000 628 48 × 2 = 1 + 0.420 800 000 142 080 142 080 000 001 256 96;
  • 8) 0.420 800 000 142 080 142 080 000 001 256 96 × 2 = 0 + 0.841 600 000 284 160 284 160 000 002 513 92;
  • 9) 0.841 600 000 284 160 284 160 000 002 513 92 × 2 = 1 + 0.683 200 000 568 320 568 320 000 005 027 84;
  • 10) 0.683 200 000 568 320 568 320 000 005 027 84 × 2 = 1 + 0.366 400 001 136 641 136 640 000 010 055 68;
  • 11) 0.366 400 001 136 641 136 640 000 010 055 68 × 2 = 0 + 0.732 800 002 273 282 273 280 000 020 111 36;
  • 12) 0.732 800 002 273 282 273 280 000 020 111 36 × 2 = 1 + 0.465 600 004 546 564 546 560 000 040 222 72;
  • 13) 0.465 600 004 546 564 546 560 000 040 222 72 × 2 = 0 + 0.931 200 009 093 129 093 120 000 080 445 44;
  • 14) 0.931 200 009 093 129 093 120 000 080 445 44 × 2 = 1 + 0.862 400 018 186 258 186 240 000 160 890 88;
  • 15) 0.862 400 018 186 258 186 240 000 160 890 88 × 2 = 1 + 0.724 800 036 372 516 372 480 000 321 781 76;
  • 16) 0.724 800 036 372 516 372 480 000 321 781 76 × 2 = 1 + 0.449 600 072 745 032 744 960 000 643 563 52;
  • 17) 0.449 600 072 745 032 744 960 000 643 563 52 × 2 = 0 + 0.899 200 145 490 065 489 920 001 287 127 04;
  • 18) 0.899 200 145 490 065 489 920 001 287 127 04 × 2 = 1 + 0.798 400 290 980 130 979 840 002 574 254 08;
  • 19) 0.798 400 290 980 130 979 840 002 574 254 08 × 2 = 1 + 0.596 800 581 960 261 959 680 005 148 508 16;
  • 20) 0.596 800 581 960 261 959 680 005 148 508 16 × 2 = 1 + 0.193 601 163 920 523 919 360 010 297 016 32;
  • 21) 0.193 601 163 920 523 919 360 010 297 016 32 × 2 = 0 + 0.387 202 327 841 047 838 720 020 594 032 64;
  • 22) 0.387 202 327 841 047 838 720 020 594 032 64 × 2 = 0 + 0.774 404 655 682 095 677 440 041 188 065 28;
  • 23) 0.774 404 655 682 095 677 440 041 188 065 28 × 2 = 1 + 0.548 809 311 364 191 354 880 082 376 130 56;
  • 24) 0.548 809 311 364 191 354 880 082 376 130 56 × 2 = 1 + 0.097 618 622 728 382 709 760 164 752 261 12;
  • 25) 0.097 618 622 728 382 709 760 164 752 261 12 × 2 = 0 + 0.195 237 245 456 765 419 520 329 504 522 24;
  • 26) 0.195 237 245 456 765 419 520 329 504 522 24 × 2 = 0 + 0.390 474 490 913 530 839 040 659 009 044 48;
  • 27) 0.390 474 490 913 530 839 040 659 009 044 48 × 2 = 0 + 0.780 948 981 827 061 678 081 318 018 088 96;
  • 28) 0.780 948 981 827 061 678 081 318 018 088 96 × 2 = 1 + 0.561 897 963 654 123 356 162 636 036 177 92;
  • 29) 0.561 897 963 654 123 356 162 636 036 177 92 × 2 = 1 + 0.123 795 927 308 246 712 325 272 072 355 84;
  • 30) 0.123 795 927 308 246 712 325 272 072 355 84 × 2 = 0 + 0.247 591 854 616 493 424 650 544 144 711 68;
  • 31) 0.247 591 854 616 493 424 650 544 144 711 68 × 2 = 0 + 0.495 183 709 232 986 849 301 088 289 423 36;
  • 32) 0.495 183 709 232 986 849 301 088 289 423 36 × 2 = 0 + 0.990 367 418 465 973 698 602 176 578 846 72;
  • 33) 0.990 367 418 465 973 698 602 176 578 846 72 × 2 = 1 + 0.980 734 836 931 947 397 204 353 157 693 44;
  • 34) 0.980 734 836 931 947 397 204 353 157 693 44 × 2 = 1 + 0.961 469 673 863 894 794 408 706 315 386 88;
  • 35) 0.961 469 673 863 894 794 408 706 315 386 88 × 2 = 1 + 0.922 939 347 727 789 588 817 412 630 773 76;
  • 36) 0.922 939 347 727 789 588 817 412 630 773 76 × 2 = 1 + 0.845 878 695 455 579 177 634 825 261 547 52;
  • 37) 0.845 878 695 455 579 177 634 825 261 547 52 × 2 = 1 + 0.691 757 390 911 158 355 269 650 523 095 04;
  • 38) 0.691 757 390 911 158 355 269 650 523 095 04 × 2 = 1 + 0.383 514 781 822 316 710 539 301 046 190 08;
  • 39) 0.383 514 781 822 316 710 539 301 046 190 08 × 2 = 0 + 0.767 029 563 644 633 421 078 602 092 380 16;
  • 40) 0.767 029 563 644 633 421 078 602 092 380 16 × 2 = 1 + 0.534 059 127 289 266 842 157 204 184 760 32;
  • 41) 0.534 059 127 289 266 842 157 204 184 760 32 × 2 = 1 + 0.068 118 254 578 533 684 314 408 369 520 64;
  • 42) 0.068 118 254 578 533 684 314 408 369 520 64 × 2 = 0 + 0.136 236 509 157 067 368 628 816 739 041 28;
  • 43) 0.136 236 509 157 067 368 628 816 739 041 28 × 2 = 0 + 0.272 473 018 314 134 737 257 633 478 082 56;
  • 44) 0.272 473 018 314 134 737 257 633 478 082 56 × 2 = 0 + 0.544 946 036 628 269 474 515 266 956 165 12;
  • 45) 0.544 946 036 628 269 474 515 266 956 165 12 × 2 = 1 + 0.089 892 073 256 538 949 030 533 912 330 24;
  • 46) 0.089 892 073 256 538 949 030 533 912 330 24 × 2 = 0 + 0.179 784 146 513 077 898 061 067 824 660 48;
  • 47) 0.179 784 146 513 077 898 061 067 824 660 48 × 2 = 0 + 0.359 568 293 026 155 796 122 135 649 320 96;
  • 48) 0.359 568 293 026 155 796 122 135 649 320 96 × 2 = 0 + 0.719 136 586 052 311 592 244 271 298 641 92;
  • 49) 0.719 136 586 052 311 592 244 271 298 641 92 × 2 = 1 + 0.438 273 172 104 623 184 488 542 597 283 84;
  • 50) 0.438 273 172 104 623 184 488 542 597 283 84 × 2 = 0 + 0.876 546 344 209 246 368 977 085 194 567 68;
  • 51) 0.876 546 344 209 246 368 977 085 194 567 68 × 2 = 1 + 0.753 092 688 418 492 737 954 170 389 135 36;
  • 52) 0.753 092 688 418 492 737 954 170 389 135 36 × 2 = 1 + 0.506 185 376 836 985 475 908 340 778 270 72;
  • 53) 0.506 185 376 836 985 475 908 340 778 270 72 × 2 = 1 + 0.012 370 753 673 970 951 816 681 556 541 44;

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:


0.011 100 000 001 110 001 110 000 000 009 82(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 000 000 009 82(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:


284.011 100 000 001 110 001 110 000 000 009 82(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 -284.011 100 000 001 110 001 110 000 000 009 82 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

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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