-284.011 100 000 001 110 001 109 999 37 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -284.011 100 000 001 110 001 109 999 37(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
-284.011 100 000 001 110 001 109 999 37(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:

|-284.011 100 000 001 110 001 109 999 37| = 284.011 100 000 001 110 001 109 999 37


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: 0.011 100 000 001 110 001 109 999 37.

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) 0.011 100 000 001 110 001 109 999 37 × 2 = 0 + 0.022 200 000 002 220 002 219 998 74;
  • 2) 0.022 200 000 002 220 002 219 998 74 × 2 = 0 + 0.044 400 000 004 440 004 439 997 48;
  • 3) 0.044 400 000 004 440 004 439 997 48 × 2 = 0 + 0.088 800 000 008 880 008 879 994 96;
  • 4) 0.088 800 000 008 880 008 879 994 96 × 2 = 0 + 0.177 600 000 017 760 017 759 989 92;
  • 5) 0.177 600 000 017 760 017 759 989 92 × 2 = 0 + 0.355 200 000 035 520 035 519 979 84;
  • 6) 0.355 200 000 035 520 035 519 979 84 × 2 = 0 + 0.710 400 000 071 040 071 039 959 68;
  • 7) 0.710 400 000 071 040 071 039 959 68 × 2 = 1 + 0.420 800 000 142 080 142 079 919 36;
  • 8) 0.420 800 000 142 080 142 079 919 36 × 2 = 0 + 0.841 600 000 284 160 284 159 838 72;
  • 9) 0.841 600 000 284 160 284 159 838 72 × 2 = 1 + 0.683 200 000 568 320 568 319 677 44;
  • 10) 0.683 200 000 568 320 568 319 677 44 × 2 = 1 + 0.366 400 001 136 641 136 639 354 88;
  • 11) 0.366 400 001 136 641 136 639 354 88 × 2 = 0 + 0.732 800 002 273 282 273 278 709 76;
  • 12) 0.732 800 002 273 282 273 278 709 76 × 2 = 1 + 0.465 600 004 546 564 546 557 419 52;
  • 13) 0.465 600 004 546 564 546 557 419 52 × 2 = 0 + 0.931 200 009 093 129 093 114 839 04;
  • 14) 0.931 200 009 093 129 093 114 839 04 × 2 = 1 + 0.862 400 018 186 258 186 229 678 08;
  • 15) 0.862 400 018 186 258 186 229 678 08 × 2 = 1 + 0.724 800 036 372 516 372 459 356 16;
  • 16) 0.724 800 036 372 516 372 459 356 16 × 2 = 1 + 0.449 600 072 745 032 744 918 712 32;
  • 17) 0.449 600 072 745 032 744 918 712 32 × 2 = 0 + 0.899 200 145 490 065 489 837 424 64;
  • 18) 0.899 200 145 490 065 489 837 424 64 × 2 = 1 + 0.798 400 290 980 130 979 674 849 28;
  • 19) 0.798 400 290 980 130 979 674 849 28 × 2 = 1 + 0.596 800 581 960 261 959 349 698 56;
  • 20) 0.596 800 581 960 261 959 349 698 56 × 2 = 1 + 0.193 601 163 920 523 918 699 397 12;
  • 21) 0.193 601 163 920 523 918 699 397 12 × 2 = 0 + 0.387 202 327 841 047 837 398 794 24;
  • 22) 0.387 202 327 841 047 837 398 794 24 × 2 = 0 + 0.774 404 655 682 095 674 797 588 48;
  • 23) 0.774 404 655 682 095 674 797 588 48 × 2 = 1 + 0.548 809 311 364 191 349 595 176 96;
  • 24) 0.548 809 311 364 191 349 595 176 96 × 2 = 1 + 0.097 618 622 728 382 699 190 353 92;
  • 25) 0.097 618 622 728 382 699 190 353 92 × 2 = 0 + 0.195 237 245 456 765 398 380 707 84;
  • 26) 0.195 237 245 456 765 398 380 707 84 × 2 = 0 + 0.390 474 490 913 530 796 761 415 68;
  • 27) 0.390 474 490 913 530 796 761 415 68 × 2 = 0 + 0.780 948 981 827 061 593 522 831 36;
  • 28) 0.780 948 981 827 061 593 522 831 36 × 2 = 1 + 0.561 897 963 654 123 187 045 662 72;
  • 29) 0.561 897 963 654 123 187 045 662 72 × 2 = 1 + 0.123 795 927 308 246 374 091 325 44;
  • 30) 0.123 795 927 308 246 374 091 325 44 × 2 = 0 + 0.247 591 854 616 492 748 182 650 88;
  • 31) 0.247 591 854 616 492 748 182 650 88 × 2 = 0 + 0.495 183 709 232 985 496 365 301 76;
  • 32) 0.495 183 709 232 985 496 365 301 76 × 2 = 0 + 0.990 367 418 465 970 992 730 603 52;
  • 33) 0.990 367 418 465 970 992 730 603 52 × 2 = 1 + 0.980 734 836 931 941 985 461 207 04;
  • 34) 0.980 734 836 931 941 985 461 207 04 × 2 = 1 + 0.961 469 673 863 883 970 922 414 08;
  • 35) 0.961 469 673 863 883 970 922 414 08 × 2 = 1 + 0.922 939 347 727 767 941 844 828 16;
  • 36) 0.922 939 347 727 767 941 844 828 16 × 2 = 1 + 0.845 878 695 455 535 883 689 656 32;
  • 37) 0.845 878 695 455 535 883 689 656 32 × 2 = 1 + 0.691 757 390 911 071 767 379 312 64;
  • 38) 0.691 757 390 911 071 767 379 312 64 × 2 = 1 + 0.383 514 781 822 143 534 758 625 28;
  • 39) 0.383 514 781 822 143 534 758 625 28 × 2 = 0 + 0.767 029 563 644 287 069 517 250 56;
  • 40) 0.767 029 563 644 287 069 517 250 56 × 2 = 1 + 0.534 059 127 288 574 139 034 501 12;
  • 41) 0.534 059 127 288 574 139 034 501 12 × 2 = 1 + 0.068 118 254 577 148 278 069 002 24;
  • 42) 0.068 118 254 577 148 278 069 002 24 × 2 = 0 + 0.136 236 509 154 296 556 138 004 48;
  • 43) 0.136 236 509 154 296 556 138 004 48 × 2 = 0 + 0.272 473 018 308 593 112 276 008 96;
  • 44) 0.272 473 018 308 593 112 276 008 96 × 2 = 0 + 0.544 946 036 617 186 224 552 017 92;
  • 45) 0.544 946 036 617 186 224 552 017 92 × 2 = 1 + 0.089 892 073 234 372 449 104 035 84;
  • 46) 0.089 892 073 234 372 449 104 035 84 × 2 = 0 + 0.179 784 146 468 744 898 208 071 68;
  • 47) 0.179 784 146 468 744 898 208 071 68 × 2 = 0 + 0.359 568 292 937 489 796 416 143 36;
  • 48) 0.359 568 292 937 489 796 416 143 36 × 2 = 0 + 0.719 136 585 874 979 592 832 286 72;
  • 49) 0.719 136 585 874 979 592 832 286 72 × 2 = 1 + 0.438 273 171 749 959 185 664 573 44;
  • 50) 0.438 273 171 749 959 185 664 573 44 × 2 = 0 + 0.876 546 343 499 918 371 329 146 88;
  • 51) 0.876 546 343 499 918 371 329 146 88 × 2 = 1 + 0.753 092 686 999 836 742 658 293 76;
  • 52) 0.753 092 686 999 836 742 658 293 76 × 2 = 1 + 0.506 185 373 999 673 485 316 587 52;
  • 53) 0.506 185 373 999 673 485 316 587 52 × 2 = 1 + 0.012 370 747 999 346 970 633 175 04;

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:


0.011 100 000 001 110 001 109 999 37(10) =

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

6. Positive number before normalization:

284.011 100 000 001 110 001 109 999 37(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:


284.011 100 000 001 110 001 109 999 37(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 -284.011 100 000 001 110 001 109 999 37 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