-284.011 100 000 001 110 001 110 053 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 053(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
-284.011 100 000 001 110 001 110 053(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:

|-284.011 100 000 001 110 001 110 053| = 284.011 100 000 001 110 001 110 053


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: 0.011 100 000 001 110 001 110 053.

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) 0.011 100 000 001 110 001 110 053 × 2 = 0 + 0.022 200 000 002 220 002 220 106;
  • 2) 0.022 200 000 002 220 002 220 106 × 2 = 0 + 0.044 400 000 004 440 004 440 212;
  • 3) 0.044 400 000 004 440 004 440 212 × 2 = 0 + 0.088 800 000 008 880 008 880 424;
  • 4) 0.088 800 000 008 880 008 880 424 × 2 = 0 + 0.177 600 000 017 760 017 760 848;
  • 5) 0.177 600 000 017 760 017 760 848 × 2 = 0 + 0.355 200 000 035 520 035 521 696;
  • 6) 0.355 200 000 035 520 035 521 696 × 2 = 0 + 0.710 400 000 071 040 071 043 392;
  • 7) 0.710 400 000 071 040 071 043 392 × 2 = 1 + 0.420 800 000 142 080 142 086 784;
  • 8) 0.420 800 000 142 080 142 086 784 × 2 = 0 + 0.841 600 000 284 160 284 173 568;
  • 9) 0.841 600 000 284 160 284 173 568 × 2 = 1 + 0.683 200 000 568 320 568 347 136;
  • 10) 0.683 200 000 568 320 568 347 136 × 2 = 1 + 0.366 400 001 136 641 136 694 272;
  • 11) 0.366 400 001 136 641 136 694 272 × 2 = 0 + 0.732 800 002 273 282 273 388 544;
  • 12) 0.732 800 002 273 282 273 388 544 × 2 = 1 + 0.465 600 004 546 564 546 777 088;
  • 13) 0.465 600 004 546 564 546 777 088 × 2 = 0 + 0.931 200 009 093 129 093 554 176;
  • 14) 0.931 200 009 093 129 093 554 176 × 2 = 1 + 0.862 400 018 186 258 187 108 352;
  • 15) 0.862 400 018 186 258 187 108 352 × 2 = 1 + 0.724 800 036 372 516 374 216 704;
  • 16) 0.724 800 036 372 516 374 216 704 × 2 = 1 + 0.449 600 072 745 032 748 433 408;
  • 17) 0.449 600 072 745 032 748 433 408 × 2 = 0 + 0.899 200 145 490 065 496 866 816;
  • 18) 0.899 200 145 490 065 496 866 816 × 2 = 1 + 0.798 400 290 980 130 993 733 632;
  • 19) 0.798 400 290 980 130 993 733 632 × 2 = 1 + 0.596 800 581 960 261 987 467 264;
  • 20) 0.596 800 581 960 261 987 467 264 × 2 = 1 + 0.193 601 163 920 523 974 934 528;
  • 21) 0.193 601 163 920 523 974 934 528 × 2 = 0 + 0.387 202 327 841 047 949 869 056;
  • 22) 0.387 202 327 841 047 949 869 056 × 2 = 0 + 0.774 404 655 682 095 899 738 112;
  • 23) 0.774 404 655 682 095 899 738 112 × 2 = 1 + 0.548 809 311 364 191 799 476 224;
  • 24) 0.548 809 311 364 191 799 476 224 × 2 = 1 + 0.097 618 622 728 383 598 952 448;
  • 25) 0.097 618 622 728 383 598 952 448 × 2 = 0 + 0.195 237 245 456 767 197 904 896;
  • 26) 0.195 237 245 456 767 197 904 896 × 2 = 0 + 0.390 474 490 913 534 395 809 792;
  • 27) 0.390 474 490 913 534 395 809 792 × 2 = 0 + 0.780 948 981 827 068 791 619 584;
  • 28) 0.780 948 981 827 068 791 619 584 × 2 = 1 + 0.561 897 963 654 137 583 239 168;
  • 29) 0.561 897 963 654 137 583 239 168 × 2 = 1 + 0.123 795 927 308 275 166 478 336;
  • 30) 0.123 795 927 308 275 166 478 336 × 2 = 0 + 0.247 591 854 616 550 332 956 672;
  • 31) 0.247 591 854 616 550 332 956 672 × 2 = 0 + 0.495 183 709 233 100 665 913 344;
  • 32) 0.495 183 709 233 100 665 913 344 × 2 = 0 + 0.990 367 418 466 201 331 826 688;
  • 33) 0.990 367 418 466 201 331 826 688 × 2 = 1 + 0.980 734 836 932 402 663 653 376;
  • 34) 0.980 734 836 932 402 663 653 376 × 2 = 1 + 0.961 469 673 864 805 327 306 752;
  • 35) 0.961 469 673 864 805 327 306 752 × 2 = 1 + 0.922 939 347 729 610 654 613 504;
  • 36) 0.922 939 347 729 610 654 613 504 × 2 = 1 + 0.845 878 695 459 221 309 227 008;
  • 37) 0.845 878 695 459 221 309 227 008 × 2 = 1 + 0.691 757 390 918 442 618 454 016;
  • 38) 0.691 757 390 918 442 618 454 016 × 2 = 1 + 0.383 514 781 836 885 236 908 032;
  • 39) 0.383 514 781 836 885 236 908 032 × 2 = 0 + 0.767 029 563 673 770 473 816 064;
  • 40) 0.767 029 563 673 770 473 816 064 × 2 = 1 + 0.534 059 127 347 540 947 632 128;
  • 41) 0.534 059 127 347 540 947 632 128 × 2 = 1 + 0.068 118 254 695 081 895 264 256;
  • 42) 0.068 118 254 695 081 895 264 256 × 2 = 0 + 0.136 236 509 390 163 790 528 512;
  • 43) 0.136 236 509 390 163 790 528 512 × 2 = 0 + 0.272 473 018 780 327 581 057 024;
  • 44) 0.272 473 018 780 327 581 057 024 × 2 = 0 + 0.544 946 037 560 655 162 114 048;
  • 45) 0.544 946 037 560 655 162 114 048 × 2 = 1 + 0.089 892 075 121 310 324 228 096;
  • 46) 0.089 892 075 121 310 324 228 096 × 2 = 0 + 0.179 784 150 242 620 648 456 192;
  • 47) 0.179 784 150 242 620 648 456 192 × 2 = 0 + 0.359 568 300 485 241 296 912 384;
  • 48) 0.359 568 300 485 241 296 912 384 × 2 = 0 + 0.719 136 600 970 482 593 824 768;
  • 49) 0.719 136 600 970 482 593 824 768 × 2 = 1 + 0.438 273 201 940 965 187 649 536;
  • 50) 0.438 273 201 940 965 187 649 536 × 2 = 0 + 0.876 546 403 881 930 375 299 072;
  • 51) 0.876 546 403 881 930 375 299 072 × 2 = 1 + 0.753 092 807 763 860 750 598 144;
  • 52) 0.753 092 807 763 860 750 598 144 × 2 = 1 + 0.506 185 615 527 721 501 196 288;
  • 53) 0.506 185 615 527 721 501 196 288 × 2 = 1 + 0.012 371 231 055 443 002 392 576;

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:


0.011 100 000 001 110 001 110 053(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 053(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:


284.011 100 000 001 110 001 110 053(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 -284.011 100 000 001 110 001 110 053 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