-284.011 100 000 001 110 001 110 034 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 034(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
-284.011 100 000 001 110 001 110 034(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:

|-284.011 100 000 001 110 001 110 034| = 284.011 100 000 001 110 001 110 034


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: 0.011 100 000 001 110 001 110 034.

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) 0.011 100 000 001 110 001 110 034 × 2 = 0 + 0.022 200 000 002 220 002 220 068;
  • 2) 0.022 200 000 002 220 002 220 068 × 2 = 0 + 0.044 400 000 004 440 004 440 136;
  • 3) 0.044 400 000 004 440 004 440 136 × 2 = 0 + 0.088 800 000 008 880 008 880 272;
  • 4) 0.088 800 000 008 880 008 880 272 × 2 = 0 + 0.177 600 000 017 760 017 760 544;
  • 5) 0.177 600 000 017 760 017 760 544 × 2 = 0 + 0.355 200 000 035 520 035 521 088;
  • 6) 0.355 200 000 035 520 035 521 088 × 2 = 0 + 0.710 400 000 071 040 071 042 176;
  • 7) 0.710 400 000 071 040 071 042 176 × 2 = 1 + 0.420 800 000 142 080 142 084 352;
  • 8) 0.420 800 000 142 080 142 084 352 × 2 = 0 + 0.841 600 000 284 160 284 168 704;
  • 9) 0.841 600 000 284 160 284 168 704 × 2 = 1 + 0.683 200 000 568 320 568 337 408;
  • 10) 0.683 200 000 568 320 568 337 408 × 2 = 1 + 0.366 400 001 136 641 136 674 816;
  • 11) 0.366 400 001 136 641 136 674 816 × 2 = 0 + 0.732 800 002 273 282 273 349 632;
  • 12) 0.732 800 002 273 282 273 349 632 × 2 = 1 + 0.465 600 004 546 564 546 699 264;
  • 13) 0.465 600 004 546 564 546 699 264 × 2 = 0 + 0.931 200 009 093 129 093 398 528;
  • 14) 0.931 200 009 093 129 093 398 528 × 2 = 1 + 0.862 400 018 186 258 186 797 056;
  • 15) 0.862 400 018 186 258 186 797 056 × 2 = 1 + 0.724 800 036 372 516 373 594 112;
  • 16) 0.724 800 036 372 516 373 594 112 × 2 = 1 + 0.449 600 072 745 032 747 188 224;
  • 17) 0.449 600 072 745 032 747 188 224 × 2 = 0 + 0.899 200 145 490 065 494 376 448;
  • 18) 0.899 200 145 490 065 494 376 448 × 2 = 1 + 0.798 400 290 980 130 988 752 896;
  • 19) 0.798 400 290 980 130 988 752 896 × 2 = 1 + 0.596 800 581 960 261 977 505 792;
  • 20) 0.596 800 581 960 261 977 505 792 × 2 = 1 + 0.193 601 163 920 523 955 011 584;
  • 21) 0.193 601 163 920 523 955 011 584 × 2 = 0 + 0.387 202 327 841 047 910 023 168;
  • 22) 0.387 202 327 841 047 910 023 168 × 2 = 0 + 0.774 404 655 682 095 820 046 336;
  • 23) 0.774 404 655 682 095 820 046 336 × 2 = 1 + 0.548 809 311 364 191 640 092 672;
  • 24) 0.548 809 311 364 191 640 092 672 × 2 = 1 + 0.097 618 622 728 383 280 185 344;
  • 25) 0.097 618 622 728 383 280 185 344 × 2 = 0 + 0.195 237 245 456 766 560 370 688;
  • 26) 0.195 237 245 456 766 560 370 688 × 2 = 0 + 0.390 474 490 913 533 120 741 376;
  • 27) 0.390 474 490 913 533 120 741 376 × 2 = 0 + 0.780 948 981 827 066 241 482 752;
  • 28) 0.780 948 981 827 066 241 482 752 × 2 = 1 + 0.561 897 963 654 132 482 965 504;
  • 29) 0.561 897 963 654 132 482 965 504 × 2 = 1 + 0.123 795 927 308 264 965 931 008;
  • 30) 0.123 795 927 308 264 965 931 008 × 2 = 0 + 0.247 591 854 616 529 931 862 016;
  • 31) 0.247 591 854 616 529 931 862 016 × 2 = 0 + 0.495 183 709 233 059 863 724 032;
  • 32) 0.495 183 709 233 059 863 724 032 × 2 = 0 + 0.990 367 418 466 119 727 448 064;
  • 33) 0.990 367 418 466 119 727 448 064 × 2 = 1 + 0.980 734 836 932 239 454 896 128;
  • 34) 0.980 734 836 932 239 454 896 128 × 2 = 1 + 0.961 469 673 864 478 909 792 256;
  • 35) 0.961 469 673 864 478 909 792 256 × 2 = 1 + 0.922 939 347 728 957 819 584 512;
  • 36) 0.922 939 347 728 957 819 584 512 × 2 = 1 + 0.845 878 695 457 915 639 169 024;
  • 37) 0.845 878 695 457 915 639 169 024 × 2 = 1 + 0.691 757 390 915 831 278 338 048;
  • 38) 0.691 757 390 915 831 278 338 048 × 2 = 1 + 0.383 514 781 831 662 556 676 096;
  • 39) 0.383 514 781 831 662 556 676 096 × 2 = 0 + 0.767 029 563 663 325 113 352 192;
  • 40) 0.767 029 563 663 325 113 352 192 × 2 = 1 + 0.534 059 127 326 650 226 704 384;
  • 41) 0.534 059 127 326 650 226 704 384 × 2 = 1 + 0.068 118 254 653 300 453 408 768;
  • 42) 0.068 118 254 653 300 453 408 768 × 2 = 0 + 0.136 236 509 306 600 906 817 536;
  • 43) 0.136 236 509 306 600 906 817 536 × 2 = 0 + 0.272 473 018 613 201 813 635 072;
  • 44) 0.272 473 018 613 201 813 635 072 × 2 = 0 + 0.544 946 037 226 403 627 270 144;
  • 45) 0.544 946 037 226 403 627 270 144 × 2 = 1 + 0.089 892 074 452 807 254 540 288;
  • 46) 0.089 892 074 452 807 254 540 288 × 2 = 0 + 0.179 784 148 905 614 509 080 576;
  • 47) 0.179 784 148 905 614 509 080 576 × 2 = 0 + 0.359 568 297 811 229 018 161 152;
  • 48) 0.359 568 297 811 229 018 161 152 × 2 = 0 + 0.719 136 595 622 458 036 322 304;
  • 49) 0.719 136 595 622 458 036 322 304 × 2 = 1 + 0.438 273 191 244 916 072 644 608;
  • 50) 0.438 273 191 244 916 072 644 608 × 2 = 0 + 0.876 546 382 489 832 145 289 216;
  • 51) 0.876 546 382 489 832 145 289 216 × 2 = 1 + 0.753 092 764 979 664 290 578 432;
  • 52) 0.753 092 764 979 664 290 578 432 × 2 = 1 + 0.506 185 529 959 328 581 156 864;
  • 53) 0.506 185 529 959 328 581 156 864 × 2 = 1 + 0.012 371 059 918 657 162 313 728;

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:


0.011 100 000 001 110 001 110 034(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 034(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:


284.011 100 000 001 110 001 110 034(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 -284.011 100 000 001 110 001 110 034 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 -284.011 100 000 001 110 001 110 095 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