64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.707 106 781 186 547 524 400 3 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.707 106 781 186 547 524 400 3(10) converted and written in 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: 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;

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.


0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.707 106 781 186 547 524 400 3.

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) 0.707 106 781 186 547 524 400 3 × 2 = 1 + 0.414 213 562 373 095 048 800 6;
  • 2) 0.414 213 562 373 095 048 800 6 × 2 = 0 + 0.828 427 124 746 190 097 601 2;
  • 3) 0.828 427 124 746 190 097 601 2 × 2 = 1 + 0.656 854 249 492 380 195 202 4;
  • 4) 0.656 854 249 492 380 195 202 4 × 2 = 1 + 0.313 708 498 984 760 390 404 8;
  • 5) 0.313 708 498 984 760 390 404 8 × 2 = 0 + 0.627 416 997 969 520 780 809 6;
  • 6) 0.627 416 997 969 520 780 809 6 × 2 = 1 + 0.254 833 995 939 041 561 619 2;
  • 7) 0.254 833 995 939 041 561 619 2 × 2 = 0 + 0.509 667 991 878 083 123 238 4;
  • 8) 0.509 667 991 878 083 123 238 4 × 2 = 1 + 0.019 335 983 756 166 246 476 8;
  • 9) 0.019 335 983 756 166 246 476 8 × 2 = 0 + 0.038 671 967 512 332 492 953 6;
  • 10) 0.038 671 967 512 332 492 953 6 × 2 = 0 + 0.077 343 935 024 664 985 907 2;
  • 11) 0.077 343 935 024 664 985 907 2 × 2 = 0 + 0.154 687 870 049 329 971 814 4;
  • 12) 0.154 687 870 049 329 971 814 4 × 2 = 0 + 0.309 375 740 098 659 943 628 8;
  • 13) 0.309 375 740 098 659 943 628 8 × 2 = 0 + 0.618 751 480 197 319 887 257 6;
  • 14) 0.618 751 480 197 319 887 257 6 × 2 = 1 + 0.237 502 960 394 639 774 515 2;
  • 15) 0.237 502 960 394 639 774 515 2 × 2 = 0 + 0.475 005 920 789 279 549 030 4;
  • 16) 0.475 005 920 789 279 549 030 4 × 2 = 0 + 0.950 011 841 578 559 098 060 8;
  • 17) 0.950 011 841 578 559 098 060 8 × 2 = 1 + 0.900 023 683 157 118 196 121 6;
  • 18) 0.900 023 683 157 118 196 121 6 × 2 = 1 + 0.800 047 366 314 236 392 243 2;
  • 19) 0.800 047 366 314 236 392 243 2 × 2 = 1 + 0.600 094 732 628 472 784 486 4;
  • 20) 0.600 094 732 628 472 784 486 4 × 2 = 1 + 0.200 189 465 256 945 568 972 8;
  • 21) 0.200 189 465 256 945 568 972 8 × 2 = 0 + 0.400 378 930 513 891 137 945 6;
  • 22) 0.400 378 930 513 891 137 945 6 × 2 = 0 + 0.800 757 861 027 782 275 891 2;
  • 23) 0.800 757 861 027 782 275 891 2 × 2 = 1 + 0.601 515 722 055 564 551 782 4;
  • 24) 0.601 515 722 055 564 551 782 4 × 2 = 1 + 0.203 031 444 111 129 103 564 8;
  • 25) 0.203 031 444 111 129 103 564 8 × 2 = 0 + 0.406 062 888 222 258 207 129 6;
  • 26) 0.406 062 888 222 258 207 129 6 × 2 = 0 + 0.812 125 776 444 516 414 259 2;
  • 27) 0.812 125 776 444 516 414 259 2 × 2 = 1 + 0.624 251 552 889 032 828 518 4;
  • 28) 0.624 251 552 889 032 828 518 4 × 2 = 1 + 0.248 503 105 778 065 657 036 8;
  • 29) 0.248 503 105 778 065 657 036 8 × 2 = 0 + 0.497 006 211 556 131 314 073 6;
  • 30) 0.497 006 211 556 131 314 073 6 × 2 = 0 + 0.994 012 423 112 262 628 147 2;
  • 31) 0.994 012 423 112 262 628 147 2 × 2 = 1 + 0.988 024 846 224 525 256 294 4;
  • 32) 0.988 024 846 224 525 256 294 4 × 2 = 1 + 0.976 049 692 449 050 512 588 8;
  • 33) 0.976 049 692 449 050 512 588 8 × 2 = 1 + 0.952 099 384 898 101 025 177 6;
  • 34) 0.952 099 384 898 101 025 177 6 × 2 = 1 + 0.904 198 769 796 202 050 355 2;
  • 35) 0.904 198 769 796 202 050 355 2 × 2 = 1 + 0.808 397 539 592 404 100 710 4;
  • 36) 0.808 397 539 592 404 100 710 4 × 2 = 1 + 0.616 795 079 184 808 201 420 8;
  • 37) 0.616 795 079 184 808 201 420 8 × 2 = 1 + 0.233 590 158 369 616 402 841 6;
  • 38) 0.233 590 158 369 616 402 841 6 × 2 = 0 + 0.467 180 316 739 232 805 683 2;
  • 39) 0.467 180 316 739 232 805 683 2 × 2 = 0 + 0.934 360 633 478 465 611 366 4;
  • 40) 0.934 360 633 478 465 611 366 4 × 2 = 1 + 0.868 721 266 956 931 222 732 8;
  • 41) 0.868 721 266 956 931 222 732 8 × 2 = 1 + 0.737 442 533 913 862 445 465 6;
  • 42) 0.737 442 533 913 862 445 465 6 × 2 = 1 + 0.474 885 067 827 724 890 931 2;
  • 43) 0.474 885 067 827 724 890 931 2 × 2 = 0 + 0.949 770 135 655 449 781 862 4;
  • 44) 0.949 770 135 655 449 781 862 4 × 2 = 1 + 0.899 540 271 310 899 563 724 8;
  • 45) 0.899 540 271 310 899 563 724 8 × 2 = 1 + 0.799 080 542 621 799 127 449 6;
  • 46) 0.799 080 542 621 799 127 449 6 × 2 = 1 + 0.598 161 085 243 598 254 899 2;
  • 47) 0.598 161 085 243 598 254 899 2 × 2 = 1 + 0.196 322 170 487 196 509 798 4;
  • 48) 0.196 322 170 487 196 509 798 4 × 2 = 0 + 0.392 644 340 974 393 019 596 8;
  • 49) 0.392 644 340 974 393 019 596 8 × 2 = 0 + 0.785 288 681 948 786 039 193 6;
  • 50) 0.785 288 681 948 786 039 193 6 × 2 = 1 + 0.570 577 363 897 572 078 387 2;
  • 51) 0.570 577 363 897 572 078 387 2 × 2 = 1 + 0.141 154 727 795 144 156 774 4;
  • 52) 0.141 154 727 795 144 156 774 4 × 2 = 0 + 0.282 309 455 590 288 313 548 8;
  • 53) 0.282 309 455 590 288 313 548 8 × 2 = 0 + 0.564 618 911 180 576 627 097 6;

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...)


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:


0.707 106 781 186 547 524 400 3(10) =

0.1011 0101 0000 0100 1111 0011 0011 0011 1111 1001 1101 1110 0110 0(2)


5. Positive number before normalization:

0.707 106 781 186 547 524 400 3(10) =

0.1011 0101 0000 0100 1111 0011 0011 0011 1111 1001 1101 1110 0110 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.707 106 781 186 547 524 400 3(10) =

0.1011 0101 0000 0100 1111 0011 0011 0011 1111 1001 1101 1110 0110 0(2) =

0.1011 0101 0000 0100 1111 0011 0011 0011 1111 1001 1101 1110 0110 0(2) × 20 =

1.0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100(2) × 2-1


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): -1

Mantissa (not normalized):
1.0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 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) =

1022(10) =

011 1111 1110(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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100 =

0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100


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 1110

Mantissa (52 bits) =
0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100


The base ten decimal number 0.707 106 781 186 547 524 400 3 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 1110 - 0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

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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