-284.011 100 000 001 110 001 110 050 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 050 5(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
-284.011 100 000 001 110 001 110 050 5(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:

|-284.011 100 000 001 110 001 110 050 5| = 284.011 100 000 001 110 001 110 050 5


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: 0.011 100 000 001 110 001 110 050 5.

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) 0.011 100 000 001 110 001 110 050 5 × 2 = 0 + 0.022 200 000 002 220 002 220 101;
  • 2) 0.022 200 000 002 220 002 220 101 × 2 = 0 + 0.044 400 000 004 440 004 440 202;
  • 3) 0.044 400 000 004 440 004 440 202 × 2 = 0 + 0.088 800 000 008 880 008 880 404;
  • 4) 0.088 800 000 008 880 008 880 404 × 2 = 0 + 0.177 600 000 017 760 017 760 808;
  • 5) 0.177 600 000 017 760 017 760 808 × 2 = 0 + 0.355 200 000 035 520 035 521 616;
  • 6) 0.355 200 000 035 520 035 521 616 × 2 = 0 + 0.710 400 000 071 040 071 043 232;
  • 7) 0.710 400 000 071 040 071 043 232 × 2 = 1 + 0.420 800 000 142 080 142 086 464;
  • 8) 0.420 800 000 142 080 142 086 464 × 2 = 0 + 0.841 600 000 284 160 284 172 928;
  • 9) 0.841 600 000 284 160 284 172 928 × 2 = 1 + 0.683 200 000 568 320 568 345 856;
  • 10) 0.683 200 000 568 320 568 345 856 × 2 = 1 + 0.366 400 001 136 641 136 691 712;
  • 11) 0.366 400 001 136 641 136 691 712 × 2 = 0 + 0.732 800 002 273 282 273 383 424;
  • 12) 0.732 800 002 273 282 273 383 424 × 2 = 1 + 0.465 600 004 546 564 546 766 848;
  • 13) 0.465 600 004 546 564 546 766 848 × 2 = 0 + 0.931 200 009 093 129 093 533 696;
  • 14) 0.931 200 009 093 129 093 533 696 × 2 = 1 + 0.862 400 018 186 258 187 067 392;
  • 15) 0.862 400 018 186 258 187 067 392 × 2 = 1 + 0.724 800 036 372 516 374 134 784;
  • 16) 0.724 800 036 372 516 374 134 784 × 2 = 1 + 0.449 600 072 745 032 748 269 568;
  • 17) 0.449 600 072 745 032 748 269 568 × 2 = 0 + 0.899 200 145 490 065 496 539 136;
  • 18) 0.899 200 145 490 065 496 539 136 × 2 = 1 + 0.798 400 290 980 130 993 078 272;
  • 19) 0.798 400 290 980 130 993 078 272 × 2 = 1 + 0.596 800 581 960 261 986 156 544;
  • 20) 0.596 800 581 960 261 986 156 544 × 2 = 1 + 0.193 601 163 920 523 972 313 088;
  • 21) 0.193 601 163 920 523 972 313 088 × 2 = 0 + 0.387 202 327 841 047 944 626 176;
  • 22) 0.387 202 327 841 047 944 626 176 × 2 = 0 + 0.774 404 655 682 095 889 252 352;
  • 23) 0.774 404 655 682 095 889 252 352 × 2 = 1 + 0.548 809 311 364 191 778 504 704;
  • 24) 0.548 809 311 364 191 778 504 704 × 2 = 1 + 0.097 618 622 728 383 557 009 408;
  • 25) 0.097 618 622 728 383 557 009 408 × 2 = 0 + 0.195 237 245 456 767 114 018 816;
  • 26) 0.195 237 245 456 767 114 018 816 × 2 = 0 + 0.390 474 490 913 534 228 037 632;
  • 27) 0.390 474 490 913 534 228 037 632 × 2 = 0 + 0.780 948 981 827 068 456 075 264;
  • 28) 0.780 948 981 827 068 456 075 264 × 2 = 1 + 0.561 897 963 654 136 912 150 528;
  • 29) 0.561 897 963 654 136 912 150 528 × 2 = 1 + 0.123 795 927 308 273 824 301 056;
  • 30) 0.123 795 927 308 273 824 301 056 × 2 = 0 + 0.247 591 854 616 547 648 602 112;
  • 31) 0.247 591 854 616 547 648 602 112 × 2 = 0 + 0.495 183 709 233 095 297 204 224;
  • 32) 0.495 183 709 233 095 297 204 224 × 2 = 0 + 0.990 367 418 466 190 594 408 448;
  • 33) 0.990 367 418 466 190 594 408 448 × 2 = 1 + 0.980 734 836 932 381 188 816 896;
  • 34) 0.980 734 836 932 381 188 816 896 × 2 = 1 + 0.961 469 673 864 762 377 633 792;
  • 35) 0.961 469 673 864 762 377 633 792 × 2 = 1 + 0.922 939 347 729 524 755 267 584;
  • 36) 0.922 939 347 729 524 755 267 584 × 2 = 1 + 0.845 878 695 459 049 510 535 168;
  • 37) 0.845 878 695 459 049 510 535 168 × 2 = 1 + 0.691 757 390 918 099 021 070 336;
  • 38) 0.691 757 390 918 099 021 070 336 × 2 = 1 + 0.383 514 781 836 198 042 140 672;
  • 39) 0.383 514 781 836 198 042 140 672 × 2 = 0 + 0.767 029 563 672 396 084 281 344;
  • 40) 0.767 029 563 672 396 084 281 344 × 2 = 1 + 0.534 059 127 344 792 168 562 688;
  • 41) 0.534 059 127 344 792 168 562 688 × 2 = 1 + 0.068 118 254 689 584 337 125 376;
  • 42) 0.068 118 254 689 584 337 125 376 × 2 = 0 + 0.136 236 509 379 168 674 250 752;
  • 43) 0.136 236 509 379 168 674 250 752 × 2 = 0 + 0.272 473 018 758 337 348 501 504;
  • 44) 0.272 473 018 758 337 348 501 504 × 2 = 0 + 0.544 946 037 516 674 697 003 008;
  • 45) 0.544 946 037 516 674 697 003 008 × 2 = 1 + 0.089 892 075 033 349 394 006 016;
  • 46) 0.089 892 075 033 349 394 006 016 × 2 = 0 + 0.179 784 150 066 698 788 012 032;
  • 47) 0.179 784 150 066 698 788 012 032 × 2 = 0 + 0.359 568 300 133 397 576 024 064;
  • 48) 0.359 568 300 133 397 576 024 064 × 2 = 0 + 0.719 136 600 266 795 152 048 128;
  • 49) 0.719 136 600 266 795 152 048 128 × 2 = 1 + 0.438 273 200 533 590 304 096 256;
  • 50) 0.438 273 200 533 590 304 096 256 × 2 = 0 + 0.876 546 401 067 180 608 192 512;
  • 51) 0.876 546 401 067 180 608 192 512 × 2 = 1 + 0.753 092 802 134 361 216 385 024;
  • 52) 0.753 092 802 134 361 216 385 024 × 2 = 1 + 0.506 185 604 268 722 432 770 048;
  • 53) 0.506 185 604 268 722 432 770 048 × 2 = 1 + 0.012 371 208 537 444 865 540 096;

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:


0.011 100 000 001 110 001 110 050 5(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 050 5(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:


284.011 100 000 001 110 001 110 050 5(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 -284.011 100 000 001 110 001 110 050 5 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 -284.011 100 000 001 110 001 110 056 9 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