1.595 242 277 440 550 187 010 811 934 418 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.595 242 277 440 550 187 010 811 934 418(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
1.595 242 277 440 550 187 010 811 934 418(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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.595 242 277 440 550 187 010 811 934 418.

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) 0.595 242 277 440 550 187 010 811 934 418 × 2 = 1 + 0.190 484 554 881 100 374 021 623 868 836;
  • 2) 0.190 484 554 881 100 374 021 623 868 836 × 2 = 0 + 0.380 969 109 762 200 748 043 247 737 672;
  • 3) 0.380 969 109 762 200 748 043 247 737 672 × 2 = 0 + 0.761 938 219 524 401 496 086 495 475 344;
  • 4) 0.761 938 219 524 401 496 086 495 475 344 × 2 = 1 + 0.523 876 439 048 802 992 172 990 950 688;
  • 5) 0.523 876 439 048 802 992 172 990 950 688 × 2 = 1 + 0.047 752 878 097 605 984 345 981 901 376;
  • 6) 0.047 752 878 097 605 984 345 981 901 376 × 2 = 0 + 0.095 505 756 195 211 968 691 963 802 752;
  • 7) 0.095 505 756 195 211 968 691 963 802 752 × 2 = 0 + 0.191 011 512 390 423 937 383 927 605 504;
  • 8) 0.191 011 512 390 423 937 383 927 605 504 × 2 = 0 + 0.382 023 024 780 847 874 767 855 211 008;
  • 9) 0.382 023 024 780 847 874 767 855 211 008 × 2 = 0 + 0.764 046 049 561 695 749 535 710 422 016;
  • 10) 0.764 046 049 561 695 749 535 710 422 016 × 2 = 1 + 0.528 092 099 123 391 499 071 420 844 032;
  • 11) 0.528 092 099 123 391 499 071 420 844 032 × 2 = 1 + 0.056 184 198 246 782 998 142 841 688 064;
  • 12) 0.056 184 198 246 782 998 142 841 688 064 × 2 = 0 + 0.112 368 396 493 565 996 285 683 376 128;
  • 13) 0.112 368 396 493 565 996 285 683 376 128 × 2 = 0 + 0.224 736 792 987 131 992 571 366 752 256;
  • 14) 0.224 736 792 987 131 992 571 366 752 256 × 2 = 0 + 0.449 473 585 974 263 985 142 733 504 512;
  • 15) 0.449 473 585 974 263 985 142 733 504 512 × 2 = 0 + 0.898 947 171 948 527 970 285 467 009 024;
  • 16) 0.898 947 171 948 527 970 285 467 009 024 × 2 = 1 + 0.797 894 343 897 055 940 570 934 018 048;
  • 17) 0.797 894 343 897 055 940 570 934 018 048 × 2 = 1 + 0.595 788 687 794 111 881 141 868 036 096;
  • 18) 0.595 788 687 794 111 881 141 868 036 096 × 2 = 1 + 0.191 577 375 588 223 762 283 736 072 192;
  • 19) 0.191 577 375 588 223 762 283 736 072 192 × 2 = 0 + 0.383 154 751 176 447 524 567 472 144 384;
  • 20) 0.383 154 751 176 447 524 567 472 144 384 × 2 = 0 + 0.766 309 502 352 895 049 134 944 288 768;
  • 21) 0.766 309 502 352 895 049 134 944 288 768 × 2 = 1 + 0.532 619 004 705 790 098 269 888 577 536;
  • 22) 0.532 619 004 705 790 098 269 888 577 536 × 2 = 1 + 0.065 238 009 411 580 196 539 777 155 072;
  • 23) 0.065 238 009 411 580 196 539 777 155 072 × 2 = 0 + 0.130 476 018 823 160 393 079 554 310 144;
  • 24) 0.130 476 018 823 160 393 079 554 310 144 × 2 = 0 + 0.260 952 037 646 320 786 159 108 620 288;
  • 25) 0.260 952 037 646 320 786 159 108 620 288 × 2 = 0 + 0.521 904 075 292 641 572 318 217 240 576;
  • 26) 0.521 904 075 292 641 572 318 217 240 576 × 2 = 1 + 0.043 808 150 585 283 144 636 434 481 152;
  • 27) 0.043 808 150 585 283 144 636 434 481 152 × 2 = 0 + 0.087 616 301 170 566 289 272 868 962 304;
  • 28) 0.087 616 301 170 566 289 272 868 962 304 × 2 = 0 + 0.175 232 602 341 132 578 545 737 924 608;
  • 29) 0.175 232 602 341 132 578 545 737 924 608 × 2 = 0 + 0.350 465 204 682 265 157 091 475 849 216;
  • 30) 0.350 465 204 682 265 157 091 475 849 216 × 2 = 0 + 0.700 930 409 364 530 314 182 951 698 432;
  • 31) 0.700 930 409 364 530 314 182 951 698 432 × 2 = 1 + 0.401 860 818 729 060 628 365 903 396 864;
  • 32) 0.401 860 818 729 060 628 365 903 396 864 × 2 = 0 + 0.803 721 637 458 121 256 731 806 793 728;
  • 33) 0.803 721 637 458 121 256 731 806 793 728 × 2 = 1 + 0.607 443 274 916 242 513 463 613 587 456;
  • 34) 0.607 443 274 916 242 513 463 613 587 456 × 2 = 1 + 0.214 886 549 832 485 026 927 227 174 912;
  • 35) 0.214 886 549 832 485 026 927 227 174 912 × 2 = 0 + 0.429 773 099 664 970 053 854 454 349 824;
  • 36) 0.429 773 099 664 970 053 854 454 349 824 × 2 = 0 + 0.859 546 199 329 940 107 708 908 699 648;
  • 37) 0.859 546 199 329 940 107 708 908 699 648 × 2 = 1 + 0.719 092 398 659 880 215 417 817 399 296;
  • 38) 0.719 092 398 659 880 215 417 817 399 296 × 2 = 1 + 0.438 184 797 319 760 430 835 634 798 592;
  • 39) 0.438 184 797 319 760 430 835 634 798 592 × 2 = 0 + 0.876 369 594 639 520 861 671 269 597 184;
  • 40) 0.876 369 594 639 520 861 671 269 597 184 × 2 = 1 + 0.752 739 189 279 041 723 342 539 194 368;
  • 41) 0.752 739 189 279 041 723 342 539 194 368 × 2 = 1 + 0.505 478 378 558 083 446 685 078 388 736;
  • 42) 0.505 478 378 558 083 446 685 078 388 736 × 2 = 1 + 0.010 956 757 116 166 893 370 156 777 472;
  • 43) 0.010 956 757 116 166 893 370 156 777 472 × 2 = 0 + 0.021 913 514 232 333 786 740 313 554 944;
  • 44) 0.021 913 514 232 333 786 740 313 554 944 × 2 = 0 + 0.043 827 028 464 667 573 480 627 109 888;
  • 45) 0.043 827 028 464 667 573 480 627 109 888 × 2 = 0 + 0.087 654 056 929 335 146 961 254 219 776;
  • 46) 0.087 654 056 929 335 146 961 254 219 776 × 2 = 0 + 0.175 308 113 858 670 293 922 508 439 552;
  • 47) 0.175 308 113 858 670 293 922 508 439 552 × 2 = 0 + 0.350 616 227 717 340 587 845 016 879 104;
  • 48) 0.350 616 227 717 340 587 845 016 879 104 × 2 = 0 + 0.701 232 455 434 681 175 690 033 758 208;
  • 49) 0.701 232 455 434 681 175 690 033 758 208 × 2 = 1 + 0.402 464 910 869 362 351 380 067 516 416;
  • 50) 0.402 464 910 869 362 351 380 067 516 416 × 2 = 0 + 0.804 929 821 738 724 702 760 135 032 832;
  • 51) 0.804 929 821 738 724 702 760 135 032 832 × 2 = 1 + 0.609 859 643 477 449 405 520 270 065 664;
  • 52) 0.609 859 643 477 449 405 520 270 065 664 × 2 = 1 + 0.219 719 286 954 898 811 040 540 131 328;
  • 53) 0.219 719 286 954 898 811 040 540 131 328 × 2 = 0 + 0.439 438 573 909 797 622 081 080 262 656;

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).


4. 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:


0.595 242 277 440 550 187 010 811 934 418(10) =

0.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0(2)

5. Positive number before normalization:

1.595 242 277 440 550 187 010 811 934 418(10) =

1.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.595 242 277 440 550 187 010 811 934 418(10) =

1.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0(2) =

1.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(10)


9. 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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1023(10) =

011 1111 1111(2)


11. 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. 1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011 0 =

1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 1111

Mantissa (52 bits) =
1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011


Decimal number 1.595 242 277 440 550 187 010 811 934 418 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1001 1000 0110 0001 1100 1100 0100 0010 1100 1101 1100 0000 1011

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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