-284.011 100 000 001 110 001 110 047 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 047 3(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
-284.011 100 000 001 110 001 110 047 3(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:

|-284.011 100 000 001 110 001 110 047 3| = 284.011 100 000 001 110 001 110 047 3


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: 0.011 100 000 001 110 001 110 047 3.

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) 0.011 100 000 001 110 001 110 047 3 × 2 = 0 + 0.022 200 000 002 220 002 220 094 6;
  • 2) 0.022 200 000 002 220 002 220 094 6 × 2 = 0 + 0.044 400 000 004 440 004 440 189 2;
  • 3) 0.044 400 000 004 440 004 440 189 2 × 2 = 0 + 0.088 800 000 008 880 008 880 378 4;
  • 4) 0.088 800 000 008 880 008 880 378 4 × 2 = 0 + 0.177 600 000 017 760 017 760 756 8;
  • 5) 0.177 600 000 017 760 017 760 756 8 × 2 = 0 + 0.355 200 000 035 520 035 521 513 6;
  • 6) 0.355 200 000 035 520 035 521 513 6 × 2 = 0 + 0.710 400 000 071 040 071 043 027 2;
  • 7) 0.710 400 000 071 040 071 043 027 2 × 2 = 1 + 0.420 800 000 142 080 142 086 054 4;
  • 8) 0.420 800 000 142 080 142 086 054 4 × 2 = 0 + 0.841 600 000 284 160 284 172 108 8;
  • 9) 0.841 600 000 284 160 284 172 108 8 × 2 = 1 + 0.683 200 000 568 320 568 344 217 6;
  • 10) 0.683 200 000 568 320 568 344 217 6 × 2 = 1 + 0.366 400 001 136 641 136 688 435 2;
  • 11) 0.366 400 001 136 641 136 688 435 2 × 2 = 0 + 0.732 800 002 273 282 273 376 870 4;
  • 12) 0.732 800 002 273 282 273 376 870 4 × 2 = 1 + 0.465 600 004 546 564 546 753 740 8;
  • 13) 0.465 600 004 546 564 546 753 740 8 × 2 = 0 + 0.931 200 009 093 129 093 507 481 6;
  • 14) 0.931 200 009 093 129 093 507 481 6 × 2 = 1 + 0.862 400 018 186 258 187 014 963 2;
  • 15) 0.862 400 018 186 258 187 014 963 2 × 2 = 1 + 0.724 800 036 372 516 374 029 926 4;
  • 16) 0.724 800 036 372 516 374 029 926 4 × 2 = 1 + 0.449 600 072 745 032 748 059 852 8;
  • 17) 0.449 600 072 745 032 748 059 852 8 × 2 = 0 + 0.899 200 145 490 065 496 119 705 6;
  • 18) 0.899 200 145 490 065 496 119 705 6 × 2 = 1 + 0.798 400 290 980 130 992 239 411 2;
  • 19) 0.798 400 290 980 130 992 239 411 2 × 2 = 1 + 0.596 800 581 960 261 984 478 822 4;
  • 20) 0.596 800 581 960 261 984 478 822 4 × 2 = 1 + 0.193 601 163 920 523 968 957 644 8;
  • 21) 0.193 601 163 920 523 968 957 644 8 × 2 = 0 + 0.387 202 327 841 047 937 915 289 6;
  • 22) 0.387 202 327 841 047 937 915 289 6 × 2 = 0 + 0.774 404 655 682 095 875 830 579 2;
  • 23) 0.774 404 655 682 095 875 830 579 2 × 2 = 1 + 0.548 809 311 364 191 751 661 158 4;
  • 24) 0.548 809 311 364 191 751 661 158 4 × 2 = 1 + 0.097 618 622 728 383 503 322 316 8;
  • 25) 0.097 618 622 728 383 503 322 316 8 × 2 = 0 + 0.195 237 245 456 767 006 644 633 6;
  • 26) 0.195 237 245 456 767 006 644 633 6 × 2 = 0 + 0.390 474 490 913 534 013 289 267 2;
  • 27) 0.390 474 490 913 534 013 289 267 2 × 2 = 0 + 0.780 948 981 827 068 026 578 534 4;
  • 28) 0.780 948 981 827 068 026 578 534 4 × 2 = 1 + 0.561 897 963 654 136 053 157 068 8;
  • 29) 0.561 897 963 654 136 053 157 068 8 × 2 = 1 + 0.123 795 927 308 272 106 314 137 6;
  • 30) 0.123 795 927 308 272 106 314 137 6 × 2 = 0 + 0.247 591 854 616 544 212 628 275 2;
  • 31) 0.247 591 854 616 544 212 628 275 2 × 2 = 0 + 0.495 183 709 233 088 425 256 550 4;
  • 32) 0.495 183 709 233 088 425 256 550 4 × 2 = 0 + 0.990 367 418 466 176 850 513 100 8;
  • 33) 0.990 367 418 466 176 850 513 100 8 × 2 = 1 + 0.980 734 836 932 353 701 026 201 6;
  • 34) 0.980 734 836 932 353 701 026 201 6 × 2 = 1 + 0.961 469 673 864 707 402 052 403 2;
  • 35) 0.961 469 673 864 707 402 052 403 2 × 2 = 1 + 0.922 939 347 729 414 804 104 806 4;
  • 36) 0.922 939 347 729 414 804 104 806 4 × 2 = 1 + 0.845 878 695 458 829 608 209 612 8;
  • 37) 0.845 878 695 458 829 608 209 612 8 × 2 = 1 + 0.691 757 390 917 659 216 419 225 6;
  • 38) 0.691 757 390 917 659 216 419 225 6 × 2 = 1 + 0.383 514 781 835 318 432 838 451 2;
  • 39) 0.383 514 781 835 318 432 838 451 2 × 2 = 0 + 0.767 029 563 670 636 865 676 902 4;
  • 40) 0.767 029 563 670 636 865 676 902 4 × 2 = 1 + 0.534 059 127 341 273 731 353 804 8;
  • 41) 0.534 059 127 341 273 731 353 804 8 × 2 = 1 + 0.068 118 254 682 547 462 707 609 6;
  • 42) 0.068 118 254 682 547 462 707 609 6 × 2 = 0 + 0.136 236 509 365 094 925 415 219 2;
  • 43) 0.136 236 509 365 094 925 415 219 2 × 2 = 0 + 0.272 473 018 730 189 850 830 438 4;
  • 44) 0.272 473 018 730 189 850 830 438 4 × 2 = 0 + 0.544 946 037 460 379 701 660 876 8;
  • 45) 0.544 946 037 460 379 701 660 876 8 × 2 = 1 + 0.089 892 074 920 759 403 321 753 6;
  • 46) 0.089 892 074 920 759 403 321 753 6 × 2 = 0 + 0.179 784 149 841 518 806 643 507 2;
  • 47) 0.179 784 149 841 518 806 643 507 2 × 2 = 0 + 0.359 568 299 683 037 613 287 014 4;
  • 48) 0.359 568 299 683 037 613 287 014 4 × 2 = 0 + 0.719 136 599 366 075 226 574 028 8;
  • 49) 0.719 136 599 366 075 226 574 028 8 × 2 = 1 + 0.438 273 198 732 150 453 148 057 6;
  • 50) 0.438 273 198 732 150 453 148 057 6 × 2 = 0 + 0.876 546 397 464 300 906 296 115 2;
  • 51) 0.876 546 397 464 300 906 296 115 2 × 2 = 1 + 0.753 092 794 928 601 812 592 230 4;
  • 52) 0.753 092 794 928 601 812 592 230 4 × 2 = 1 + 0.506 185 589 857 203 625 184 460 8;
  • 53) 0.506 185 589 857 203 625 184 460 8 × 2 = 1 + 0.012 371 179 714 407 250 368 921 6;

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:


0.011 100 000 001 110 001 110 047 3(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 047 3(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:


284.011 100 000 001 110 001 110 047 3(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 -284.011 100 000 001 110 001 110 047 3 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 6 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