1.414 213 562 373 095 048 801 689 45 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.414 213 562 373 095 048 801 689 45(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
1.414 213 562 373 095 048 801 689 45(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: 0.414 213 562 373 095 048 801 689 45.

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) 0.414 213 562 373 095 048 801 689 45 × 2 = 0 + 0.828 427 124 746 190 097 603 378 9;
  • 2) 0.828 427 124 746 190 097 603 378 9 × 2 = 1 + 0.656 854 249 492 380 195 206 757 8;
  • 3) 0.656 854 249 492 380 195 206 757 8 × 2 = 1 + 0.313 708 498 984 760 390 413 515 6;
  • 4) 0.313 708 498 984 760 390 413 515 6 × 2 = 0 + 0.627 416 997 969 520 780 827 031 2;
  • 5) 0.627 416 997 969 520 780 827 031 2 × 2 = 1 + 0.254 833 995 939 041 561 654 062 4;
  • 6) 0.254 833 995 939 041 561 654 062 4 × 2 = 0 + 0.509 667 991 878 083 123 308 124 8;
  • 7) 0.509 667 991 878 083 123 308 124 8 × 2 = 1 + 0.019 335 983 756 166 246 616 249 6;
  • 8) 0.019 335 983 756 166 246 616 249 6 × 2 = 0 + 0.038 671 967 512 332 493 232 499 2;
  • 9) 0.038 671 967 512 332 493 232 499 2 × 2 = 0 + 0.077 343 935 024 664 986 464 998 4;
  • 10) 0.077 343 935 024 664 986 464 998 4 × 2 = 0 + 0.154 687 870 049 329 972 929 996 8;
  • 11) 0.154 687 870 049 329 972 929 996 8 × 2 = 0 + 0.309 375 740 098 659 945 859 993 6;
  • 12) 0.309 375 740 098 659 945 859 993 6 × 2 = 0 + 0.618 751 480 197 319 891 719 987 2;
  • 13) 0.618 751 480 197 319 891 719 987 2 × 2 = 1 + 0.237 502 960 394 639 783 439 974 4;
  • 14) 0.237 502 960 394 639 783 439 974 4 × 2 = 0 + 0.475 005 920 789 279 566 879 948 8;
  • 15) 0.475 005 920 789 279 566 879 948 8 × 2 = 0 + 0.950 011 841 578 559 133 759 897 6;
  • 16) 0.950 011 841 578 559 133 759 897 6 × 2 = 1 + 0.900 023 683 157 118 267 519 795 2;
  • 17) 0.900 023 683 157 118 267 519 795 2 × 2 = 1 + 0.800 047 366 314 236 535 039 590 4;
  • 18) 0.800 047 366 314 236 535 039 590 4 × 2 = 1 + 0.600 094 732 628 473 070 079 180 8;
  • 19) 0.600 094 732 628 473 070 079 180 8 × 2 = 1 + 0.200 189 465 256 946 140 158 361 6;
  • 20) 0.200 189 465 256 946 140 158 361 6 × 2 = 0 + 0.400 378 930 513 892 280 316 723 2;
  • 21) 0.400 378 930 513 892 280 316 723 2 × 2 = 0 + 0.800 757 861 027 784 560 633 446 4;
  • 22) 0.800 757 861 027 784 560 633 446 4 × 2 = 1 + 0.601 515 722 055 569 121 266 892 8;
  • 23) 0.601 515 722 055 569 121 266 892 8 × 2 = 1 + 0.203 031 444 111 138 242 533 785 6;
  • 24) 0.203 031 444 111 138 242 533 785 6 × 2 = 0 + 0.406 062 888 222 276 485 067 571 2;
  • 25) 0.406 062 888 222 276 485 067 571 2 × 2 = 0 + 0.812 125 776 444 552 970 135 142 4;
  • 26) 0.812 125 776 444 552 970 135 142 4 × 2 = 1 + 0.624 251 552 889 105 940 270 284 8;
  • 27) 0.624 251 552 889 105 940 270 284 8 × 2 = 1 + 0.248 503 105 778 211 880 540 569 6;
  • 28) 0.248 503 105 778 211 880 540 569 6 × 2 = 0 + 0.497 006 211 556 423 761 081 139 2;
  • 29) 0.497 006 211 556 423 761 081 139 2 × 2 = 0 + 0.994 012 423 112 847 522 162 278 4;
  • 30) 0.994 012 423 112 847 522 162 278 4 × 2 = 1 + 0.988 024 846 225 695 044 324 556 8;
  • 31) 0.988 024 846 225 695 044 324 556 8 × 2 = 1 + 0.976 049 692 451 390 088 649 113 6;
  • 32) 0.976 049 692 451 390 088 649 113 6 × 2 = 1 + 0.952 099 384 902 780 177 298 227 2;
  • 33) 0.952 099 384 902 780 177 298 227 2 × 2 = 1 + 0.904 198 769 805 560 354 596 454 4;
  • 34) 0.904 198 769 805 560 354 596 454 4 × 2 = 1 + 0.808 397 539 611 120 709 192 908 8;
  • 35) 0.808 397 539 611 120 709 192 908 8 × 2 = 1 + 0.616 795 079 222 241 418 385 817 6;
  • 36) 0.616 795 079 222 241 418 385 817 6 × 2 = 1 + 0.233 590 158 444 482 836 771 635 2;
  • 37) 0.233 590 158 444 482 836 771 635 2 × 2 = 0 + 0.467 180 316 888 965 673 543 270 4;
  • 38) 0.467 180 316 888 965 673 543 270 4 × 2 = 0 + 0.934 360 633 777 931 347 086 540 8;
  • 39) 0.934 360 633 777 931 347 086 540 8 × 2 = 1 + 0.868 721 267 555 862 694 173 081 6;
  • 40) 0.868 721 267 555 862 694 173 081 6 × 2 = 1 + 0.737 442 535 111 725 388 346 163 2;
  • 41) 0.737 442 535 111 725 388 346 163 2 × 2 = 1 + 0.474 885 070 223 450 776 692 326 4;
  • 42) 0.474 885 070 223 450 776 692 326 4 × 2 = 0 + 0.949 770 140 446 901 553 384 652 8;
  • 43) 0.949 770 140 446 901 553 384 652 8 × 2 = 1 + 0.899 540 280 893 803 106 769 305 6;
  • 44) 0.899 540 280 893 803 106 769 305 6 × 2 = 1 + 0.799 080 561 787 606 213 538 611 2;
  • 45) 0.799 080 561 787 606 213 538 611 2 × 2 = 1 + 0.598 161 123 575 212 427 077 222 4;
  • 46) 0.598 161 123 575 212 427 077 222 4 × 2 = 1 + 0.196 322 247 150 424 854 154 444 8;
  • 47) 0.196 322 247 150 424 854 154 444 8 × 2 = 0 + 0.392 644 494 300 849 708 308 889 6;
  • 48) 0.392 644 494 300 849 708 308 889 6 × 2 = 0 + 0.785 288 988 601 699 416 617 779 2;
  • 49) 0.785 288 988 601 699 416 617 779 2 × 2 = 1 + 0.570 577 977 203 398 833 235 558 4;
  • 50) 0.570 577 977 203 398 833 235 558 4 × 2 = 1 + 0.141 155 954 406 797 666 471 116 8;
  • 51) 0.141 155 954 406 797 666 471 116 8 × 2 = 0 + 0.282 311 908 813 595 332 942 233 6;
  • 52) 0.282 311 908 813 595 332 942 233 6 × 2 = 0 + 0.564 623 817 627 190 665 884 467 2;
  • 53) 0.564 623 817 627 190 665 884 467 2 × 2 = 1 + 0.129 247 635 254 381 331 768 934 4;

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:


0.414 213 562 373 095 048 801 689 45(10) =

0.0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100 1(2)

5. Positive number before normalization:

1.414 213 562 373 095 048 801 689 45(10) =

1.0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100 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:


1.414 213 562 373 095 048 801 689 45(10) =

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

1.0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100 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):
1.0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100 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. 0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 1100 1 =

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 1111

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


Decimal number 1.414 213 562 373 095 048 801 689 45 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 0110 1010 0000 1001 1110 0110 0110 0111 1111 0011 1011 1100 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