-0.000 000 803 398 337 183 5 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 803 398 337 183 5(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
-0.000 000 803 398 337 183 5(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:

|-0.000 000 803 398 337 183 5| = 0.000 000 803 398 337 183 5


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.000 000 803 398 337 183 5.

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) 0.000 000 803 398 337 183 5 × 2 = 0 + 0.000 001 606 796 674 367;
  • 2) 0.000 001 606 796 674 367 × 2 = 0 + 0.000 003 213 593 348 734;
  • 3) 0.000 003 213 593 348 734 × 2 = 0 + 0.000 006 427 186 697 468;
  • 4) 0.000 006 427 186 697 468 × 2 = 0 + 0.000 012 854 373 394 936;
  • 5) 0.000 012 854 373 394 936 × 2 = 0 + 0.000 025 708 746 789 872;
  • 6) 0.000 025 708 746 789 872 × 2 = 0 + 0.000 051 417 493 579 744;
  • 7) 0.000 051 417 493 579 744 × 2 = 0 + 0.000 102 834 987 159 488;
  • 8) 0.000 102 834 987 159 488 × 2 = 0 + 0.000 205 669 974 318 976;
  • 9) 0.000 205 669 974 318 976 × 2 = 0 + 0.000 411 339 948 637 952;
  • 10) 0.000 411 339 948 637 952 × 2 = 0 + 0.000 822 679 897 275 904;
  • 11) 0.000 822 679 897 275 904 × 2 = 0 + 0.001 645 359 794 551 808;
  • 12) 0.001 645 359 794 551 808 × 2 = 0 + 0.003 290 719 589 103 616;
  • 13) 0.003 290 719 589 103 616 × 2 = 0 + 0.006 581 439 178 207 232;
  • 14) 0.006 581 439 178 207 232 × 2 = 0 + 0.013 162 878 356 414 464;
  • 15) 0.013 162 878 356 414 464 × 2 = 0 + 0.026 325 756 712 828 928;
  • 16) 0.026 325 756 712 828 928 × 2 = 0 + 0.052 651 513 425 657 856;
  • 17) 0.052 651 513 425 657 856 × 2 = 0 + 0.105 303 026 851 315 712;
  • 18) 0.105 303 026 851 315 712 × 2 = 0 + 0.210 606 053 702 631 424;
  • 19) 0.210 606 053 702 631 424 × 2 = 0 + 0.421 212 107 405 262 848;
  • 20) 0.421 212 107 405 262 848 × 2 = 0 + 0.842 424 214 810 525 696;
  • 21) 0.842 424 214 810 525 696 × 2 = 1 + 0.684 848 429 621 051 392;
  • 22) 0.684 848 429 621 051 392 × 2 = 1 + 0.369 696 859 242 102 784;
  • 23) 0.369 696 859 242 102 784 × 2 = 0 + 0.739 393 718 484 205 568;
  • 24) 0.739 393 718 484 205 568 × 2 = 1 + 0.478 787 436 968 411 136;
  • 25) 0.478 787 436 968 411 136 × 2 = 0 + 0.957 574 873 936 822 272;
  • 26) 0.957 574 873 936 822 272 × 2 = 1 + 0.915 149 747 873 644 544;
  • 27) 0.915 149 747 873 644 544 × 2 = 1 + 0.830 299 495 747 289 088;
  • 28) 0.830 299 495 747 289 088 × 2 = 1 + 0.660 598 991 494 578 176;
  • 29) 0.660 598 991 494 578 176 × 2 = 1 + 0.321 197 982 989 156 352;
  • 30) 0.321 197 982 989 156 352 × 2 = 0 + 0.642 395 965 978 312 704;
  • 31) 0.642 395 965 978 312 704 × 2 = 1 + 0.284 791 931 956 625 408;
  • 32) 0.284 791 931 956 625 408 × 2 = 0 + 0.569 583 863 913 250 816;
  • 33) 0.569 583 863 913 250 816 × 2 = 1 + 0.139 167 727 826 501 632;
  • 34) 0.139 167 727 826 501 632 × 2 = 0 + 0.278 335 455 653 003 264;
  • 35) 0.278 335 455 653 003 264 × 2 = 0 + 0.556 670 911 306 006 528;
  • 36) 0.556 670 911 306 006 528 × 2 = 1 + 0.113 341 822 612 013 056;
  • 37) 0.113 341 822 612 013 056 × 2 = 0 + 0.226 683 645 224 026 112;
  • 38) 0.226 683 645 224 026 112 × 2 = 0 + 0.453 367 290 448 052 224;
  • 39) 0.453 367 290 448 052 224 × 2 = 0 + 0.906 734 580 896 104 448;
  • 40) 0.906 734 580 896 104 448 × 2 = 1 + 0.813 469 161 792 208 896;
  • 41) 0.813 469 161 792 208 896 × 2 = 1 + 0.626 938 323 584 417 792;
  • 42) 0.626 938 323 584 417 792 × 2 = 1 + 0.253 876 647 168 835 584;
  • 43) 0.253 876 647 168 835 584 × 2 = 0 + 0.507 753 294 337 671 168;
  • 44) 0.507 753 294 337 671 168 × 2 = 1 + 0.015 506 588 675 342 336;
  • 45) 0.015 506 588 675 342 336 × 2 = 0 + 0.031 013 177 350 684 672;
  • 46) 0.031 013 177 350 684 672 × 2 = 0 + 0.062 026 354 701 369 344;
  • 47) 0.062 026 354 701 369 344 × 2 = 0 + 0.124 052 709 402 738 688;
  • 48) 0.124 052 709 402 738 688 × 2 = 0 + 0.248 105 418 805 477 376;
  • 49) 0.248 105 418 805 477 376 × 2 = 0 + 0.496 210 837 610 954 752;
  • 50) 0.496 210 837 610 954 752 × 2 = 0 + 0.992 421 675 221 909 504;
  • 51) 0.992 421 675 221 909 504 × 2 = 1 + 0.984 843 350 443 819 008;
  • 52) 0.984 843 350 443 819 008 × 2 = 1 + 0.969 686 700 887 638 016;
  • 53) 0.969 686 700 887 638 016 × 2 = 1 + 0.939 373 401 775 276 032;
  • 54) 0.939 373 401 775 276 032 × 2 = 1 + 0.878 746 803 550 552 064;
  • 55) 0.878 746 803 550 552 064 × 2 = 1 + 0.757 493 607 101 104 128;
  • 56) 0.757 493 607 101 104 128 × 2 = 1 + 0.514 987 214 202 208 256;
  • 57) 0.514 987 214 202 208 256 × 2 = 1 + 0.029 974 428 404 416 512;
  • 58) 0.029 974 428 404 416 512 × 2 = 0 + 0.059 948 856 808 833 024;
  • 59) 0.059 948 856 808 833 024 × 2 = 0 + 0.119 897 713 617 666 048;
  • 60) 0.119 897 713 617 666 048 × 2 = 0 + 0.239 795 427 235 332 096;
  • 61) 0.239 795 427 235 332 096 × 2 = 0 + 0.479 590 854 470 664 192;
  • 62) 0.479 590 854 470 664 192 × 2 = 0 + 0.959 181 708 941 328 384;
  • 63) 0.959 181 708 941 328 384 × 2 = 1 + 0.918 363 417 882 656 768;
  • 64) 0.918 363 417 882 656 768 × 2 = 1 + 0.836 726 835 765 313 536;
  • 65) 0.836 726 835 765 313 536 × 2 = 1 + 0.673 453 671 530 627 072;
  • 66) 0.673 453 671 530 627 072 × 2 = 1 + 0.346 907 343 061 254 144;
  • 67) 0.346 907 343 061 254 144 × 2 = 0 + 0.693 814 686 122 508 288;
  • 68) 0.693 814 686 122 508 288 × 2 = 1 + 0.387 629 372 245 016 576;
  • 69) 0.387 629 372 245 016 576 × 2 = 0 + 0.775 258 744 490 033 152;
  • 70) 0.775 258 744 490 033 152 × 2 = 1 + 0.550 517 488 980 066 304;
  • 71) 0.550 517 488 980 066 304 × 2 = 1 + 0.101 034 977 960 132 608;
  • 72) 0.101 034 977 960 132 608 × 2 = 0 + 0.202 069 955 920 265 216;
  • 73) 0.202 069 955 920 265 216 × 2 = 0 + 0.404 139 911 840 530 432;

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:


0.000 000 803 398 337 183 5(10) =

0.0000 0000 0000 0000 0000 1101 0111 1010 1001 0001 1101 0000 0011 1111 1000 0011 1101 0110 0(2)

6. Positive number before normalization:

0.000 000 803 398 337 183 5(10) =

0.0000 0000 0000 0000 0000 1101 0111 1010 1001 0001 1101 0000 0011 1111 1000 0011 1101 0110 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 21 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 803 398 337 183 5(10) =

0.0000 0000 0000 0000 0000 1101 0111 1010 1001 0001 1101 0000 0011 1111 1000 0011 1101 0110 0(2) =

0.0000 0000 0000 0000 0000 1101 0111 1010 1001 0001 1101 0000 0011 1111 1000 0011 1101 0110 0(2) × 20 =

1.1010 1111 0101 0010 0011 1010 0000 0111 1111 0000 0111 1010 1100(2) × 2-21


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): -21

Mantissa (not normalized):
1.1010 1111 0101 0010 0011 1010 0000 0111 1111 0000 0111 1010 1100


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-21 + 2(11-1) - 1 =

(-21 + 1 023)(10) =

1 002(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;
  • 1 002 ÷ 2 = 501 + 0;
  • 501 ÷ 2 = 250 + 1;
  • 250 ÷ 2 = 125 + 0;
  • 125 ÷ 2 = 62 + 1;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1002(10) =

011 1110 1010(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 1010 1111 0101 0010 0011 1010 0000 0111 1111 0000 0111 1010 1100 =

1010 1111 0101 0010 0011 1010 0000 0111 1111 0000 0111 1010 1100


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) =
011 1110 1010

Mantissa (52 bits) =
1010 1111 0101 0010 0011 1010 0000 0111 1111 0000 0111 1010 1100


Decimal number -0.000 000 803 398 337 183 5 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1110 1010 - 1010 1111 0101 0010 0011 1010 0000 0111 1111 0000 0111 1010 1100

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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