-284.011 100 000 001 110 001 110 000 000 010 07 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 110 000 000 010 07(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
-284.011 100 000 001 110 001 110 000 000 010 07(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:

|-284.011 100 000 001 110 001 110 000 000 010 07| = 284.011 100 000 001 110 001 110 000 000 010 07


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: 0.011 100 000 001 110 001 110 000 000 010 07.

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) 0.011 100 000 001 110 001 110 000 000 010 07 × 2 = 0 + 0.022 200 000 002 220 002 220 000 000 020 14;
  • 2) 0.022 200 000 002 220 002 220 000 000 020 14 × 2 = 0 + 0.044 400 000 004 440 004 440 000 000 040 28;
  • 3) 0.044 400 000 004 440 004 440 000 000 040 28 × 2 = 0 + 0.088 800 000 008 880 008 880 000 000 080 56;
  • 4) 0.088 800 000 008 880 008 880 000 000 080 56 × 2 = 0 + 0.177 600 000 017 760 017 760 000 000 161 12;
  • 5) 0.177 600 000 017 760 017 760 000 000 161 12 × 2 = 0 + 0.355 200 000 035 520 035 520 000 000 322 24;
  • 6) 0.355 200 000 035 520 035 520 000 000 322 24 × 2 = 0 + 0.710 400 000 071 040 071 040 000 000 644 48;
  • 7) 0.710 400 000 071 040 071 040 000 000 644 48 × 2 = 1 + 0.420 800 000 142 080 142 080 000 001 288 96;
  • 8) 0.420 800 000 142 080 142 080 000 001 288 96 × 2 = 0 + 0.841 600 000 284 160 284 160 000 002 577 92;
  • 9) 0.841 600 000 284 160 284 160 000 002 577 92 × 2 = 1 + 0.683 200 000 568 320 568 320 000 005 155 84;
  • 10) 0.683 200 000 568 320 568 320 000 005 155 84 × 2 = 1 + 0.366 400 001 136 641 136 640 000 010 311 68;
  • 11) 0.366 400 001 136 641 136 640 000 010 311 68 × 2 = 0 + 0.732 800 002 273 282 273 280 000 020 623 36;
  • 12) 0.732 800 002 273 282 273 280 000 020 623 36 × 2 = 1 + 0.465 600 004 546 564 546 560 000 041 246 72;
  • 13) 0.465 600 004 546 564 546 560 000 041 246 72 × 2 = 0 + 0.931 200 009 093 129 093 120 000 082 493 44;
  • 14) 0.931 200 009 093 129 093 120 000 082 493 44 × 2 = 1 + 0.862 400 018 186 258 186 240 000 164 986 88;
  • 15) 0.862 400 018 186 258 186 240 000 164 986 88 × 2 = 1 + 0.724 800 036 372 516 372 480 000 329 973 76;
  • 16) 0.724 800 036 372 516 372 480 000 329 973 76 × 2 = 1 + 0.449 600 072 745 032 744 960 000 659 947 52;
  • 17) 0.449 600 072 745 032 744 960 000 659 947 52 × 2 = 0 + 0.899 200 145 490 065 489 920 001 319 895 04;
  • 18) 0.899 200 145 490 065 489 920 001 319 895 04 × 2 = 1 + 0.798 400 290 980 130 979 840 002 639 790 08;
  • 19) 0.798 400 290 980 130 979 840 002 639 790 08 × 2 = 1 + 0.596 800 581 960 261 959 680 005 279 580 16;
  • 20) 0.596 800 581 960 261 959 680 005 279 580 16 × 2 = 1 + 0.193 601 163 920 523 919 360 010 559 160 32;
  • 21) 0.193 601 163 920 523 919 360 010 559 160 32 × 2 = 0 + 0.387 202 327 841 047 838 720 021 118 320 64;
  • 22) 0.387 202 327 841 047 838 720 021 118 320 64 × 2 = 0 + 0.774 404 655 682 095 677 440 042 236 641 28;
  • 23) 0.774 404 655 682 095 677 440 042 236 641 28 × 2 = 1 + 0.548 809 311 364 191 354 880 084 473 282 56;
  • 24) 0.548 809 311 364 191 354 880 084 473 282 56 × 2 = 1 + 0.097 618 622 728 382 709 760 168 946 565 12;
  • 25) 0.097 618 622 728 382 709 760 168 946 565 12 × 2 = 0 + 0.195 237 245 456 765 419 520 337 893 130 24;
  • 26) 0.195 237 245 456 765 419 520 337 893 130 24 × 2 = 0 + 0.390 474 490 913 530 839 040 675 786 260 48;
  • 27) 0.390 474 490 913 530 839 040 675 786 260 48 × 2 = 0 + 0.780 948 981 827 061 678 081 351 572 520 96;
  • 28) 0.780 948 981 827 061 678 081 351 572 520 96 × 2 = 1 + 0.561 897 963 654 123 356 162 703 145 041 92;
  • 29) 0.561 897 963 654 123 356 162 703 145 041 92 × 2 = 1 + 0.123 795 927 308 246 712 325 406 290 083 84;
  • 30) 0.123 795 927 308 246 712 325 406 290 083 84 × 2 = 0 + 0.247 591 854 616 493 424 650 812 580 167 68;
  • 31) 0.247 591 854 616 493 424 650 812 580 167 68 × 2 = 0 + 0.495 183 709 232 986 849 301 625 160 335 36;
  • 32) 0.495 183 709 232 986 849 301 625 160 335 36 × 2 = 0 + 0.990 367 418 465 973 698 603 250 320 670 72;
  • 33) 0.990 367 418 465 973 698 603 250 320 670 72 × 2 = 1 + 0.980 734 836 931 947 397 206 500 641 341 44;
  • 34) 0.980 734 836 931 947 397 206 500 641 341 44 × 2 = 1 + 0.961 469 673 863 894 794 413 001 282 682 88;
  • 35) 0.961 469 673 863 894 794 413 001 282 682 88 × 2 = 1 + 0.922 939 347 727 789 588 826 002 565 365 76;
  • 36) 0.922 939 347 727 789 588 826 002 565 365 76 × 2 = 1 + 0.845 878 695 455 579 177 652 005 130 731 52;
  • 37) 0.845 878 695 455 579 177 652 005 130 731 52 × 2 = 1 + 0.691 757 390 911 158 355 304 010 261 463 04;
  • 38) 0.691 757 390 911 158 355 304 010 261 463 04 × 2 = 1 + 0.383 514 781 822 316 710 608 020 522 926 08;
  • 39) 0.383 514 781 822 316 710 608 020 522 926 08 × 2 = 0 + 0.767 029 563 644 633 421 216 041 045 852 16;
  • 40) 0.767 029 563 644 633 421 216 041 045 852 16 × 2 = 1 + 0.534 059 127 289 266 842 432 082 091 704 32;
  • 41) 0.534 059 127 289 266 842 432 082 091 704 32 × 2 = 1 + 0.068 118 254 578 533 684 864 164 183 408 64;
  • 42) 0.068 118 254 578 533 684 864 164 183 408 64 × 2 = 0 + 0.136 236 509 157 067 369 728 328 366 817 28;
  • 43) 0.136 236 509 157 067 369 728 328 366 817 28 × 2 = 0 + 0.272 473 018 314 134 739 456 656 733 634 56;
  • 44) 0.272 473 018 314 134 739 456 656 733 634 56 × 2 = 0 + 0.544 946 036 628 269 478 913 313 467 269 12;
  • 45) 0.544 946 036 628 269 478 913 313 467 269 12 × 2 = 1 + 0.089 892 073 256 538 957 826 626 934 538 24;
  • 46) 0.089 892 073 256 538 957 826 626 934 538 24 × 2 = 0 + 0.179 784 146 513 077 915 653 253 869 076 48;
  • 47) 0.179 784 146 513 077 915 653 253 869 076 48 × 2 = 0 + 0.359 568 293 026 155 831 306 507 738 152 96;
  • 48) 0.359 568 293 026 155 831 306 507 738 152 96 × 2 = 0 + 0.719 136 586 052 311 662 613 015 476 305 92;
  • 49) 0.719 136 586 052 311 662 613 015 476 305 92 × 2 = 1 + 0.438 273 172 104 623 325 226 030 952 611 84;
  • 50) 0.438 273 172 104 623 325 226 030 952 611 84 × 2 = 0 + 0.876 546 344 209 246 650 452 061 905 223 68;
  • 51) 0.876 546 344 209 246 650 452 061 905 223 68 × 2 = 1 + 0.753 092 688 418 493 300 904 123 810 447 36;
  • 52) 0.753 092 688 418 493 300 904 123 810 447 36 × 2 = 1 + 0.506 185 376 836 986 601 808 247 620 894 72;
  • 53) 0.506 185 376 836 986 601 808 247 620 894 72 × 2 = 1 + 0.012 370 753 673 973 203 616 495 241 789 44;

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:


0.011 100 000 001 110 001 110 000 000 010 07(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 110 000 000 010 07(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:


284.011 100 000 001 110 001 110 000 000 010 07(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 -284.011 100 000 001 110 001 110 000 000 010 07 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

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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