1.745 459 324 169 999 826 281 696 064 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 064(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
1.745 459 324 169 999 826 281 696 064(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: 0.745 459 324 169 999 826 281 696 064.

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) 0.745 459 324 169 999 826 281 696 064 × 2 = 1 + 0.490 918 648 339 999 652 563 392 128;
  • 2) 0.490 918 648 339 999 652 563 392 128 × 2 = 0 + 0.981 837 296 679 999 305 126 784 256;
  • 3) 0.981 837 296 679 999 305 126 784 256 × 2 = 1 + 0.963 674 593 359 998 610 253 568 512;
  • 4) 0.963 674 593 359 998 610 253 568 512 × 2 = 1 + 0.927 349 186 719 997 220 507 137 024;
  • 5) 0.927 349 186 719 997 220 507 137 024 × 2 = 1 + 0.854 698 373 439 994 441 014 274 048;
  • 6) 0.854 698 373 439 994 441 014 274 048 × 2 = 1 + 0.709 396 746 879 988 882 028 548 096;
  • 7) 0.709 396 746 879 988 882 028 548 096 × 2 = 1 + 0.418 793 493 759 977 764 057 096 192;
  • 8) 0.418 793 493 759 977 764 057 096 192 × 2 = 0 + 0.837 586 987 519 955 528 114 192 384;
  • 9) 0.837 586 987 519 955 528 114 192 384 × 2 = 1 + 0.675 173 975 039 911 056 228 384 768;
  • 10) 0.675 173 975 039 911 056 228 384 768 × 2 = 1 + 0.350 347 950 079 822 112 456 769 536;
  • 11) 0.350 347 950 079 822 112 456 769 536 × 2 = 0 + 0.700 695 900 159 644 224 913 539 072;
  • 12) 0.700 695 900 159 644 224 913 539 072 × 2 = 1 + 0.401 391 800 319 288 449 827 078 144;
  • 13) 0.401 391 800 319 288 449 827 078 144 × 2 = 0 + 0.802 783 600 638 576 899 654 156 288;
  • 14) 0.802 783 600 638 576 899 654 156 288 × 2 = 1 + 0.605 567 201 277 153 799 308 312 576;
  • 15) 0.605 567 201 277 153 799 308 312 576 × 2 = 1 + 0.211 134 402 554 307 598 616 625 152;
  • 16) 0.211 134 402 554 307 598 616 625 152 × 2 = 0 + 0.422 268 805 108 615 197 233 250 304;
  • 17) 0.422 268 805 108 615 197 233 250 304 × 2 = 0 + 0.844 537 610 217 230 394 466 500 608;
  • 18) 0.844 537 610 217 230 394 466 500 608 × 2 = 1 + 0.689 075 220 434 460 788 933 001 216;
  • 19) 0.689 075 220 434 460 788 933 001 216 × 2 = 1 + 0.378 150 440 868 921 577 866 002 432;
  • 20) 0.378 150 440 868 921 577 866 002 432 × 2 = 0 + 0.756 300 881 737 843 155 732 004 864;
  • 21) 0.756 300 881 737 843 155 732 004 864 × 2 = 1 + 0.512 601 763 475 686 311 464 009 728;
  • 22) 0.512 601 763 475 686 311 464 009 728 × 2 = 1 + 0.025 203 526 951 372 622 928 019 456;
  • 23) 0.025 203 526 951 372 622 928 019 456 × 2 = 0 + 0.050 407 053 902 745 245 856 038 912;
  • 24) 0.050 407 053 902 745 245 856 038 912 × 2 = 0 + 0.100 814 107 805 490 491 712 077 824;
  • 25) 0.100 814 107 805 490 491 712 077 824 × 2 = 0 + 0.201 628 215 610 980 983 424 155 648;
  • 26) 0.201 628 215 610 980 983 424 155 648 × 2 = 0 + 0.403 256 431 221 961 966 848 311 296;
  • 27) 0.403 256 431 221 961 966 848 311 296 × 2 = 0 + 0.806 512 862 443 923 933 696 622 592;
  • 28) 0.806 512 862 443 923 933 696 622 592 × 2 = 1 + 0.613 025 724 887 847 867 393 245 184;
  • 29) 0.613 025 724 887 847 867 393 245 184 × 2 = 1 + 0.226 051 449 775 695 734 786 490 368;
  • 30) 0.226 051 449 775 695 734 786 490 368 × 2 = 0 + 0.452 102 899 551 391 469 572 980 736;
  • 31) 0.452 102 899 551 391 469 572 980 736 × 2 = 0 + 0.904 205 799 102 782 939 145 961 472;
  • 32) 0.904 205 799 102 782 939 145 961 472 × 2 = 1 + 0.808 411 598 205 565 878 291 922 944;
  • 33) 0.808 411 598 205 565 878 291 922 944 × 2 = 1 + 0.616 823 196 411 131 756 583 845 888;
  • 34) 0.616 823 196 411 131 756 583 845 888 × 2 = 1 + 0.233 646 392 822 263 513 167 691 776;
  • 35) 0.233 646 392 822 263 513 167 691 776 × 2 = 0 + 0.467 292 785 644 527 026 335 383 552;
  • 36) 0.467 292 785 644 527 026 335 383 552 × 2 = 0 + 0.934 585 571 289 054 052 670 767 104;
  • 37) 0.934 585 571 289 054 052 670 767 104 × 2 = 1 + 0.869 171 142 578 108 105 341 534 208;
  • 38) 0.869 171 142 578 108 105 341 534 208 × 2 = 1 + 0.738 342 285 156 216 210 683 068 416;
  • 39) 0.738 342 285 156 216 210 683 068 416 × 2 = 1 + 0.476 684 570 312 432 421 366 136 832;
  • 40) 0.476 684 570 312 432 421 366 136 832 × 2 = 0 + 0.953 369 140 624 864 842 732 273 664;
  • 41) 0.953 369 140 624 864 842 732 273 664 × 2 = 1 + 0.906 738 281 249 729 685 464 547 328;
  • 42) 0.906 738 281 249 729 685 464 547 328 × 2 = 1 + 0.813 476 562 499 459 370 929 094 656;
  • 43) 0.813 476 562 499 459 370 929 094 656 × 2 = 1 + 0.626 953 124 998 918 741 858 189 312;
  • 44) 0.626 953 124 998 918 741 858 189 312 × 2 = 1 + 0.253 906 249 997 837 483 716 378 624;
  • 45) 0.253 906 249 997 837 483 716 378 624 × 2 = 0 + 0.507 812 499 995 674 967 432 757 248;
  • 46) 0.507 812 499 995 674 967 432 757 248 × 2 = 1 + 0.015 624 999 991 349 934 865 514 496;
  • 47) 0.015 624 999 991 349 934 865 514 496 × 2 = 0 + 0.031 249 999 982 699 869 731 028 992;
  • 48) 0.031 249 999 982 699 869 731 028 992 × 2 = 0 + 0.062 499 999 965 399 739 462 057 984;
  • 49) 0.062 499 999 965 399 739 462 057 984 × 2 = 0 + 0.124 999 999 930 799 478 924 115 968;
  • 50) 0.124 999 999 930 799 478 924 115 968 × 2 = 0 + 0.249 999 999 861 598 957 848 231 936;
  • 51) 0.249 999 999 861 598 957 848 231 936 × 2 = 0 + 0.499 999 999 723 197 915 696 463 872;
  • 52) 0.499 999 999 723 197 915 696 463 872 × 2 = 0 + 0.999 999 999 446 395 831 392 927 744;
  • 53) 0.999 999 999 446 395 831 392 927 744 × 2 = 1 + 0.999 999 998 892 791 662 785 855 488;

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:


0.745 459 324 169 999 826 281 696 064(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 064(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(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:


1.745 459 324 169 999 826 281 696 064(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(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):
1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1


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. 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


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) =
1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


Decimal number 1.745 459 324 169 999 826 281 696 064 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000

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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