-284.011 100 000 001 110 001 110 000 000 009 59 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 59(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 59(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 59| = 284.011 100 000 001 110 001 110 000 000 009 59


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 59.

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 59 × 2 = 0 + 0.022 200 000 002 220 002 220 000 000 019 18;
  • 2) 0.022 200 000 002 220 002 220 000 000 019 18 × 2 = 0 + 0.044 400 000 004 440 004 440 000 000 038 36;
  • 3) 0.044 400 000 004 440 004 440 000 000 038 36 × 2 = 0 + 0.088 800 000 008 880 008 880 000 000 076 72;
  • 4) 0.088 800 000 008 880 008 880 000 000 076 72 × 2 = 0 + 0.177 600 000 017 760 017 760 000 000 153 44;
  • 5) 0.177 600 000 017 760 017 760 000 000 153 44 × 2 = 0 + 0.355 200 000 035 520 035 520 000 000 306 88;
  • 6) 0.355 200 000 035 520 035 520 000 000 306 88 × 2 = 0 + 0.710 400 000 071 040 071 040 000 000 613 76;
  • 7) 0.710 400 000 071 040 071 040 000 000 613 76 × 2 = 1 + 0.420 800 000 142 080 142 080 000 001 227 52;
  • 8) 0.420 800 000 142 080 142 080 000 001 227 52 × 2 = 0 + 0.841 600 000 284 160 284 160 000 002 455 04;
  • 9) 0.841 600 000 284 160 284 160 000 002 455 04 × 2 = 1 + 0.683 200 000 568 320 568 320 000 004 910 08;
  • 10) 0.683 200 000 568 320 568 320 000 004 910 08 × 2 = 1 + 0.366 400 001 136 641 136 640 000 009 820 16;
  • 11) 0.366 400 001 136 641 136 640 000 009 820 16 × 2 = 0 + 0.732 800 002 273 282 273 280 000 019 640 32;
  • 12) 0.732 800 002 273 282 273 280 000 019 640 32 × 2 = 1 + 0.465 600 004 546 564 546 560 000 039 280 64;
  • 13) 0.465 600 004 546 564 546 560 000 039 280 64 × 2 = 0 + 0.931 200 009 093 129 093 120 000 078 561 28;
  • 14) 0.931 200 009 093 129 093 120 000 078 561 28 × 2 = 1 + 0.862 400 018 186 258 186 240 000 157 122 56;
  • 15) 0.862 400 018 186 258 186 240 000 157 122 56 × 2 = 1 + 0.724 800 036 372 516 372 480 000 314 245 12;
  • 16) 0.724 800 036 372 516 372 480 000 314 245 12 × 2 = 1 + 0.449 600 072 745 032 744 960 000 628 490 24;
  • 17) 0.449 600 072 745 032 744 960 000 628 490 24 × 2 = 0 + 0.899 200 145 490 065 489 920 001 256 980 48;
  • 18) 0.899 200 145 490 065 489 920 001 256 980 48 × 2 = 1 + 0.798 400 290 980 130 979 840 002 513 960 96;
  • 19) 0.798 400 290 980 130 979 840 002 513 960 96 × 2 = 1 + 0.596 800 581 960 261 959 680 005 027 921 92;
  • 20) 0.596 800 581 960 261 959 680 005 027 921 92 × 2 = 1 + 0.193 601 163 920 523 919 360 010 055 843 84;
  • 21) 0.193 601 163 920 523 919 360 010 055 843 84 × 2 = 0 + 0.387 202 327 841 047 838 720 020 111 687 68;
  • 22) 0.387 202 327 841 047 838 720 020 111 687 68 × 2 = 0 + 0.774 404 655 682 095 677 440 040 223 375 36;
  • 23) 0.774 404 655 682 095 677 440 040 223 375 36 × 2 = 1 + 0.548 809 311 364 191 354 880 080 446 750 72;
  • 24) 0.548 809 311 364 191 354 880 080 446 750 72 × 2 = 1 + 0.097 618 622 728 382 709 760 160 893 501 44;
  • 25) 0.097 618 622 728 382 709 760 160 893 501 44 × 2 = 0 + 0.195 237 245 456 765 419 520 321 787 002 88;
  • 26) 0.195 237 245 456 765 419 520 321 787 002 88 × 2 = 0 + 0.390 474 490 913 530 839 040 643 574 005 76;
  • 27) 0.390 474 490 913 530 839 040 643 574 005 76 × 2 = 0 + 0.780 948 981 827 061 678 081 287 148 011 52;
  • 28) 0.780 948 981 827 061 678 081 287 148 011 52 × 2 = 1 + 0.561 897 963 654 123 356 162 574 296 023 04;
  • 29) 0.561 897 963 654 123 356 162 574 296 023 04 × 2 = 1 + 0.123 795 927 308 246 712 325 148 592 046 08;
  • 30) 0.123 795 927 308 246 712 325 148 592 046 08 × 2 = 0 + 0.247 591 854 616 493 424 650 297 184 092 16;
  • 31) 0.247 591 854 616 493 424 650 297 184 092 16 × 2 = 0 + 0.495 183 709 232 986 849 300 594 368 184 32;
  • 32) 0.495 183 709 232 986 849 300 594 368 184 32 × 2 = 0 + 0.990 367 418 465 973 698 601 188 736 368 64;
  • 33) 0.990 367 418 465 973 698 601 188 736 368 64 × 2 = 1 + 0.980 734 836 931 947 397 202 377 472 737 28;
  • 34) 0.980 734 836 931 947 397 202 377 472 737 28 × 2 = 1 + 0.961 469 673 863 894 794 404 754 945 474 56;
  • 35) 0.961 469 673 863 894 794 404 754 945 474 56 × 2 = 1 + 0.922 939 347 727 789 588 809 509 890 949 12;
  • 36) 0.922 939 347 727 789 588 809 509 890 949 12 × 2 = 1 + 0.845 878 695 455 579 177 619 019 781 898 24;
  • 37) 0.845 878 695 455 579 177 619 019 781 898 24 × 2 = 1 + 0.691 757 390 911 158 355 238 039 563 796 48;
  • 38) 0.691 757 390 911 158 355 238 039 563 796 48 × 2 = 1 + 0.383 514 781 822 316 710 476 079 127 592 96;
  • 39) 0.383 514 781 822 316 710 476 079 127 592 96 × 2 = 0 + 0.767 029 563 644 633 420 952 158 255 185 92;
  • 40) 0.767 029 563 644 633 420 952 158 255 185 92 × 2 = 1 + 0.534 059 127 289 266 841 904 316 510 371 84;
  • 41) 0.534 059 127 289 266 841 904 316 510 371 84 × 2 = 1 + 0.068 118 254 578 533 683 808 633 020 743 68;
  • 42) 0.068 118 254 578 533 683 808 633 020 743 68 × 2 = 0 + 0.136 236 509 157 067 367 617 266 041 487 36;
  • 43) 0.136 236 509 157 067 367 617 266 041 487 36 × 2 = 0 + 0.272 473 018 314 134 735 234 532 082 974 72;
  • 44) 0.272 473 018 314 134 735 234 532 082 974 72 × 2 = 0 + 0.544 946 036 628 269 470 469 064 165 949 44;
  • 45) 0.544 946 036 628 269 470 469 064 165 949 44 × 2 = 1 + 0.089 892 073 256 538 940 938 128 331 898 88;
  • 46) 0.089 892 073 256 538 940 938 128 331 898 88 × 2 = 0 + 0.179 784 146 513 077 881 876 256 663 797 76;
  • 47) 0.179 784 146 513 077 881 876 256 663 797 76 × 2 = 0 + 0.359 568 293 026 155 763 752 513 327 595 52;
  • 48) 0.359 568 293 026 155 763 752 513 327 595 52 × 2 = 0 + 0.719 136 586 052 311 527 505 026 655 191 04;
  • 49) 0.719 136 586 052 311 527 505 026 655 191 04 × 2 = 1 + 0.438 273 172 104 623 055 010 053 310 382 08;
  • 50) 0.438 273 172 104 623 055 010 053 310 382 08 × 2 = 0 + 0.876 546 344 209 246 110 020 106 620 764 16;
  • 51) 0.876 546 344 209 246 110 020 106 620 764 16 × 2 = 1 + 0.753 092 688 418 492 220 040 213 241 528 32;
  • 52) 0.753 092 688 418 492 220 040 213 241 528 32 × 2 = 1 + 0.506 185 376 836 984 440 080 426 483 056 64;
  • 53) 0.506 185 376 836 984 440 080 426 483 056 64 × 2 = 1 + 0.012 370 753 673 968 880 160 852 966 113 28;

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 59(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 59(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 59(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 59 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