-284.011 100 000 001 110 001 110 087 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 087(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
-284.011 100 000 001 110 001 110 087(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:

|-284.011 100 000 001 110 001 110 087| = 284.011 100 000 001 110 001 110 087


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: 0.011 100 000 001 110 001 110 087.

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) 0.011 100 000 001 110 001 110 087 × 2 = 0 + 0.022 200 000 002 220 002 220 174;
  • 2) 0.022 200 000 002 220 002 220 174 × 2 = 0 + 0.044 400 000 004 440 004 440 348;
  • 3) 0.044 400 000 004 440 004 440 348 × 2 = 0 + 0.088 800 000 008 880 008 880 696;
  • 4) 0.088 800 000 008 880 008 880 696 × 2 = 0 + 0.177 600 000 017 760 017 761 392;
  • 5) 0.177 600 000 017 760 017 761 392 × 2 = 0 + 0.355 200 000 035 520 035 522 784;
  • 6) 0.355 200 000 035 520 035 522 784 × 2 = 0 + 0.710 400 000 071 040 071 045 568;
  • 7) 0.710 400 000 071 040 071 045 568 × 2 = 1 + 0.420 800 000 142 080 142 091 136;
  • 8) 0.420 800 000 142 080 142 091 136 × 2 = 0 + 0.841 600 000 284 160 284 182 272;
  • 9) 0.841 600 000 284 160 284 182 272 × 2 = 1 + 0.683 200 000 568 320 568 364 544;
  • 10) 0.683 200 000 568 320 568 364 544 × 2 = 1 + 0.366 400 001 136 641 136 729 088;
  • 11) 0.366 400 001 136 641 136 729 088 × 2 = 0 + 0.732 800 002 273 282 273 458 176;
  • 12) 0.732 800 002 273 282 273 458 176 × 2 = 1 + 0.465 600 004 546 564 546 916 352;
  • 13) 0.465 600 004 546 564 546 916 352 × 2 = 0 + 0.931 200 009 093 129 093 832 704;
  • 14) 0.931 200 009 093 129 093 832 704 × 2 = 1 + 0.862 400 018 186 258 187 665 408;
  • 15) 0.862 400 018 186 258 187 665 408 × 2 = 1 + 0.724 800 036 372 516 375 330 816;
  • 16) 0.724 800 036 372 516 375 330 816 × 2 = 1 + 0.449 600 072 745 032 750 661 632;
  • 17) 0.449 600 072 745 032 750 661 632 × 2 = 0 + 0.899 200 145 490 065 501 323 264;
  • 18) 0.899 200 145 490 065 501 323 264 × 2 = 1 + 0.798 400 290 980 131 002 646 528;
  • 19) 0.798 400 290 980 131 002 646 528 × 2 = 1 + 0.596 800 581 960 262 005 293 056;
  • 20) 0.596 800 581 960 262 005 293 056 × 2 = 1 + 0.193 601 163 920 524 010 586 112;
  • 21) 0.193 601 163 920 524 010 586 112 × 2 = 0 + 0.387 202 327 841 048 021 172 224;
  • 22) 0.387 202 327 841 048 021 172 224 × 2 = 0 + 0.774 404 655 682 096 042 344 448;
  • 23) 0.774 404 655 682 096 042 344 448 × 2 = 1 + 0.548 809 311 364 192 084 688 896;
  • 24) 0.548 809 311 364 192 084 688 896 × 2 = 1 + 0.097 618 622 728 384 169 377 792;
  • 25) 0.097 618 622 728 384 169 377 792 × 2 = 0 + 0.195 237 245 456 768 338 755 584;
  • 26) 0.195 237 245 456 768 338 755 584 × 2 = 0 + 0.390 474 490 913 536 677 511 168;
  • 27) 0.390 474 490 913 536 677 511 168 × 2 = 0 + 0.780 948 981 827 073 355 022 336;
  • 28) 0.780 948 981 827 073 355 022 336 × 2 = 1 + 0.561 897 963 654 146 710 044 672;
  • 29) 0.561 897 963 654 146 710 044 672 × 2 = 1 + 0.123 795 927 308 293 420 089 344;
  • 30) 0.123 795 927 308 293 420 089 344 × 2 = 0 + 0.247 591 854 616 586 840 178 688;
  • 31) 0.247 591 854 616 586 840 178 688 × 2 = 0 + 0.495 183 709 233 173 680 357 376;
  • 32) 0.495 183 709 233 173 680 357 376 × 2 = 0 + 0.990 367 418 466 347 360 714 752;
  • 33) 0.990 367 418 466 347 360 714 752 × 2 = 1 + 0.980 734 836 932 694 721 429 504;
  • 34) 0.980 734 836 932 694 721 429 504 × 2 = 1 + 0.961 469 673 865 389 442 859 008;
  • 35) 0.961 469 673 865 389 442 859 008 × 2 = 1 + 0.922 939 347 730 778 885 718 016;
  • 36) 0.922 939 347 730 778 885 718 016 × 2 = 1 + 0.845 878 695 461 557 771 436 032;
  • 37) 0.845 878 695 461 557 771 436 032 × 2 = 1 + 0.691 757 390 923 115 542 872 064;
  • 38) 0.691 757 390 923 115 542 872 064 × 2 = 1 + 0.383 514 781 846 231 085 744 128;
  • 39) 0.383 514 781 846 231 085 744 128 × 2 = 0 + 0.767 029 563 692 462 171 488 256;
  • 40) 0.767 029 563 692 462 171 488 256 × 2 = 1 + 0.534 059 127 384 924 342 976 512;
  • 41) 0.534 059 127 384 924 342 976 512 × 2 = 1 + 0.068 118 254 769 848 685 953 024;
  • 42) 0.068 118 254 769 848 685 953 024 × 2 = 0 + 0.136 236 509 539 697 371 906 048;
  • 43) 0.136 236 509 539 697 371 906 048 × 2 = 0 + 0.272 473 019 079 394 743 812 096;
  • 44) 0.272 473 019 079 394 743 812 096 × 2 = 0 + 0.544 946 038 158 789 487 624 192;
  • 45) 0.544 946 038 158 789 487 624 192 × 2 = 1 + 0.089 892 076 317 578 975 248 384;
  • 46) 0.089 892 076 317 578 975 248 384 × 2 = 0 + 0.179 784 152 635 157 950 496 768;
  • 47) 0.179 784 152 635 157 950 496 768 × 2 = 0 + 0.359 568 305 270 315 900 993 536;
  • 48) 0.359 568 305 270 315 900 993 536 × 2 = 0 + 0.719 136 610 540 631 801 987 072;
  • 49) 0.719 136 610 540 631 801 987 072 × 2 = 1 + 0.438 273 221 081 263 603 974 144;
  • 50) 0.438 273 221 081 263 603 974 144 × 2 = 0 + 0.876 546 442 162 527 207 948 288;
  • 51) 0.876 546 442 162 527 207 948 288 × 2 = 1 + 0.753 092 884 325 054 415 896 576;
  • 52) 0.753 092 884 325 054 415 896 576 × 2 = 1 + 0.506 185 768 650 108 831 793 152;
  • 53) 0.506 185 768 650 108 831 793 152 × 2 = 1 + 0.012 371 537 300 217 663 586 304;

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:


0.011 100 000 001 110 001 110 087(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 087(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:


284.011 100 000 001 110 001 110 087(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 -284.011 100 000 001 110 001 110 087 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 -284.011 100 000 001 110 001 110 081 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