-284.011 100 000 001 110 001 110 174 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 174(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
-284.011 100 000 001 110 001 110 174(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:

|-284.011 100 000 001 110 001 110 174| = 284.011 100 000 001 110 001 110 174


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: 0.011 100 000 001 110 001 110 174.

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) 0.011 100 000 001 110 001 110 174 × 2 = 0 + 0.022 200 000 002 220 002 220 348;
  • 2) 0.022 200 000 002 220 002 220 348 × 2 = 0 + 0.044 400 000 004 440 004 440 696;
  • 3) 0.044 400 000 004 440 004 440 696 × 2 = 0 + 0.088 800 000 008 880 008 881 392;
  • 4) 0.088 800 000 008 880 008 881 392 × 2 = 0 + 0.177 600 000 017 760 017 762 784;
  • 5) 0.177 600 000 017 760 017 762 784 × 2 = 0 + 0.355 200 000 035 520 035 525 568;
  • 6) 0.355 200 000 035 520 035 525 568 × 2 = 0 + 0.710 400 000 071 040 071 051 136;
  • 7) 0.710 400 000 071 040 071 051 136 × 2 = 1 + 0.420 800 000 142 080 142 102 272;
  • 8) 0.420 800 000 142 080 142 102 272 × 2 = 0 + 0.841 600 000 284 160 284 204 544;
  • 9) 0.841 600 000 284 160 284 204 544 × 2 = 1 + 0.683 200 000 568 320 568 409 088;
  • 10) 0.683 200 000 568 320 568 409 088 × 2 = 1 + 0.366 400 001 136 641 136 818 176;
  • 11) 0.366 400 001 136 641 136 818 176 × 2 = 0 + 0.732 800 002 273 282 273 636 352;
  • 12) 0.732 800 002 273 282 273 636 352 × 2 = 1 + 0.465 600 004 546 564 547 272 704;
  • 13) 0.465 600 004 546 564 547 272 704 × 2 = 0 + 0.931 200 009 093 129 094 545 408;
  • 14) 0.931 200 009 093 129 094 545 408 × 2 = 1 + 0.862 400 018 186 258 189 090 816;
  • 15) 0.862 400 018 186 258 189 090 816 × 2 = 1 + 0.724 800 036 372 516 378 181 632;
  • 16) 0.724 800 036 372 516 378 181 632 × 2 = 1 + 0.449 600 072 745 032 756 363 264;
  • 17) 0.449 600 072 745 032 756 363 264 × 2 = 0 + 0.899 200 145 490 065 512 726 528;
  • 18) 0.899 200 145 490 065 512 726 528 × 2 = 1 + 0.798 400 290 980 131 025 453 056;
  • 19) 0.798 400 290 980 131 025 453 056 × 2 = 1 + 0.596 800 581 960 262 050 906 112;
  • 20) 0.596 800 581 960 262 050 906 112 × 2 = 1 + 0.193 601 163 920 524 101 812 224;
  • 21) 0.193 601 163 920 524 101 812 224 × 2 = 0 + 0.387 202 327 841 048 203 624 448;
  • 22) 0.387 202 327 841 048 203 624 448 × 2 = 0 + 0.774 404 655 682 096 407 248 896;
  • 23) 0.774 404 655 682 096 407 248 896 × 2 = 1 + 0.548 809 311 364 192 814 497 792;
  • 24) 0.548 809 311 364 192 814 497 792 × 2 = 1 + 0.097 618 622 728 385 628 995 584;
  • 25) 0.097 618 622 728 385 628 995 584 × 2 = 0 + 0.195 237 245 456 771 257 991 168;
  • 26) 0.195 237 245 456 771 257 991 168 × 2 = 0 + 0.390 474 490 913 542 515 982 336;
  • 27) 0.390 474 490 913 542 515 982 336 × 2 = 0 + 0.780 948 981 827 085 031 964 672;
  • 28) 0.780 948 981 827 085 031 964 672 × 2 = 1 + 0.561 897 963 654 170 063 929 344;
  • 29) 0.561 897 963 654 170 063 929 344 × 2 = 1 + 0.123 795 927 308 340 127 858 688;
  • 30) 0.123 795 927 308 340 127 858 688 × 2 = 0 + 0.247 591 854 616 680 255 717 376;
  • 31) 0.247 591 854 616 680 255 717 376 × 2 = 0 + 0.495 183 709 233 360 511 434 752;
  • 32) 0.495 183 709 233 360 511 434 752 × 2 = 0 + 0.990 367 418 466 721 022 869 504;
  • 33) 0.990 367 418 466 721 022 869 504 × 2 = 1 + 0.980 734 836 933 442 045 739 008;
  • 34) 0.980 734 836 933 442 045 739 008 × 2 = 1 + 0.961 469 673 866 884 091 478 016;
  • 35) 0.961 469 673 866 884 091 478 016 × 2 = 1 + 0.922 939 347 733 768 182 956 032;
  • 36) 0.922 939 347 733 768 182 956 032 × 2 = 1 + 0.845 878 695 467 536 365 912 064;
  • 37) 0.845 878 695 467 536 365 912 064 × 2 = 1 + 0.691 757 390 935 072 731 824 128;
  • 38) 0.691 757 390 935 072 731 824 128 × 2 = 1 + 0.383 514 781 870 145 463 648 256;
  • 39) 0.383 514 781 870 145 463 648 256 × 2 = 0 + 0.767 029 563 740 290 927 296 512;
  • 40) 0.767 029 563 740 290 927 296 512 × 2 = 1 + 0.534 059 127 480 581 854 593 024;
  • 41) 0.534 059 127 480 581 854 593 024 × 2 = 1 + 0.068 118 254 961 163 709 186 048;
  • 42) 0.068 118 254 961 163 709 186 048 × 2 = 0 + 0.136 236 509 922 327 418 372 096;
  • 43) 0.136 236 509 922 327 418 372 096 × 2 = 0 + 0.272 473 019 844 654 836 744 192;
  • 44) 0.272 473 019 844 654 836 744 192 × 2 = 0 + 0.544 946 039 689 309 673 488 384;
  • 45) 0.544 946 039 689 309 673 488 384 × 2 = 1 + 0.089 892 079 378 619 346 976 768;
  • 46) 0.089 892 079 378 619 346 976 768 × 2 = 0 + 0.179 784 158 757 238 693 953 536;
  • 47) 0.179 784 158 757 238 693 953 536 × 2 = 0 + 0.359 568 317 514 477 387 907 072;
  • 48) 0.359 568 317 514 477 387 907 072 × 2 = 0 + 0.719 136 635 028 954 775 814 144;
  • 49) 0.719 136 635 028 954 775 814 144 × 2 = 1 + 0.438 273 270 057 909 551 628 288;
  • 50) 0.438 273 270 057 909 551 628 288 × 2 = 0 + 0.876 546 540 115 819 103 256 576;
  • 51) 0.876 546 540 115 819 103 256 576 × 2 = 1 + 0.753 093 080 231 638 206 513 152;
  • 52) 0.753 093 080 231 638 206 513 152 × 2 = 1 + 0.506 186 160 463 276 413 026 304;
  • 53) 0.506 186 160 463 276 413 026 304 × 2 = 1 + 0.012 372 320 926 552 826 052 608;

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:


0.011 100 000 001 110 001 110 174(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 174(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:


284.011 100 000 001 110 001 110 174(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 -284.011 100 000 001 110 001 110 174 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 08 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