1.414 213 562 373 095 048 801 691 31 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.414 213 562 373 095 048 801 691 31(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
1.414 213 562 373 095 048 801 691 31(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: 0.414 213 562 373 095 048 801 691 31.

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) 0.414 213 562 373 095 048 801 691 31 × 2 = 0 + 0.828 427 124 746 190 097 603 382 62;
  • 2) 0.828 427 124 746 190 097 603 382 62 × 2 = 1 + 0.656 854 249 492 380 195 206 765 24;
  • 3) 0.656 854 249 492 380 195 206 765 24 × 2 = 1 + 0.313 708 498 984 760 390 413 530 48;
  • 4) 0.313 708 498 984 760 390 413 530 48 × 2 = 0 + 0.627 416 997 969 520 780 827 060 96;
  • 5) 0.627 416 997 969 520 780 827 060 96 × 2 = 1 + 0.254 833 995 939 041 561 654 121 92;
  • 6) 0.254 833 995 939 041 561 654 121 92 × 2 = 0 + 0.509 667 991 878 083 123 308 243 84;
  • 7) 0.509 667 991 878 083 123 308 243 84 × 2 = 1 + 0.019 335 983 756 166 246 616 487 68;
  • 8) 0.019 335 983 756 166 246 616 487 68 × 2 = 0 + 0.038 671 967 512 332 493 232 975 36;
  • 9) 0.038 671 967 512 332 493 232 975 36 × 2 = 0 + 0.077 343 935 024 664 986 465 950 72;
  • 10) 0.077 343 935 024 664 986 465 950 72 × 2 = 0 + 0.154 687 870 049 329 972 931 901 44;
  • 11) 0.154 687 870 049 329 972 931 901 44 × 2 = 0 + 0.309 375 740 098 659 945 863 802 88;
  • 12) 0.309 375 740 098 659 945 863 802 88 × 2 = 0 + 0.618 751 480 197 319 891 727 605 76;
  • 13) 0.618 751 480 197 319 891 727 605 76 × 2 = 1 + 0.237 502 960 394 639 783 455 211 52;
  • 14) 0.237 502 960 394 639 783 455 211 52 × 2 = 0 + 0.475 005 920 789 279 566 910 423 04;
  • 15) 0.475 005 920 789 279 566 910 423 04 × 2 = 0 + 0.950 011 841 578 559 133 820 846 08;
  • 16) 0.950 011 841 578 559 133 820 846 08 × 2 = 1 + 0.900 023 683 157 118 267 641 692 16;
  • 17) 0.900 023 683 157 118 267 641 692 16 × 2 = 1 + 0.800 047 366 314 236 535 283 384 32;
  • 18) 0.800 047 366 314 236 535 283 384 32 × 2 = 1 + 0.600 094 732 628 473 070 566 768 64;
  • 19) 0.600 094 732 628 473 070 566 768 64 × 2 = 1 + 0.200 189 465 256 946 141 133 537 28;
  • 20) 0.200 189 465 256 946 141 133 537 28 × 2 = 0 + 0.400 378 930 513 892 282 267 074 56;
  • 21) 0.400 378 930 513 892 282 267 074 56 × 2 = 0 + 0.800 757 861 027 784 564 534 149 12;
  • 22) 0.800 757 861 027 784 564 534 149 12 × 2 = 1 + 0.601 515 722 055 569 129 068 298 24;
  • 23) 0.601 515 722 055 569 129 068 298 24 × 2 = 1 + 0.203 031 444 111 138 258 136 596 48;
  • 24) 0.203 031 444 111 138 258 136 596 48 × 2 = 0 + 0.406 062 888 222 276 516 273 192 96;
  • 25) 0.406 062 888 222 276 516 273 192 96 × 2 = 0 + 0.812 125 776 444 553 032 546 385 92;
  • 26) 0.812 125 776 444 553 032 546 385 92 × 2 = 1 + 0.624 251 552 889 106 065 092 771 84;
  • 27) 0.624 251 552 889 106 065 092 771 84 × 2 = 1 + 0.248 503 105 778 212 130 185 543 68;
  • 28) 0.248 503 105 778 212 130 185 543 68 × 2 = 0 + 0.497 006 211 556 424 260 371 087 36;
  • 29) 0.497 006 211 556 424 260 371 087 36 × 2 = 0 + 0.994 012 423 112 848 520 742 174 72;
  • 30) 0.994 012 423 112 848 520 742 174 72 × 2 = 1 + 0.988 024 846 225 697 041 484 349 44;
  • 31) 0.988 024 846 225 697 041 484 349 44 × 2 = 1 + 0.976 049 692 451 394 082 968 698 88;
  • 32) 0.976 049 692 451 394 082 968 698 88 × 2 = 1 + 0.952 099 384 902 788 165 937 397 76;
  • 33) 0.952 099 384 902 788 165 937 397 76 × 2 = 1 + 0.904 198 769 805 576 331 874 795 52;
  • 34) 0.904 198 769 805 576 331 874 795 52 × 2 = 1 + 0.808 397 539 611 152 663 749 591 04;
  • 35) 0.808 397 539 611 152 663 749 591 04 × 2 = 1 + 0.616 795 079 222 305 327 499 182 08;
  • 36) 0.616 795 079 222 305 327 499 182 08 × 2 = 1 + 0.233 590 158 444 610 654 998 364 16;
  • 37) 0.233 590 158 444 610 654 998 364 16 × 2 = 0 + 0.467 180 316 889 221 309 996 728 32;
  • 38) 0.467 180 316 889 221 309 996 728 32 × 2 = 0 + 0.934 360 633 778 442 619 993 456 64;
  • 39) 0.934 360 633 778 442 619 993 456 64 × 2 = 1 + 0.868 721 267 556 885 239 986 913 28;
  • 40) 0.868 721 267 556 885 239 986 913 28 × 2 = 1 + 0.737 442 535 113 770 479 973 826 56;
  • 41) 0.737 442 535 113 770 479 973 826 56 × 2 = 1 + 0.474 885 070 227 540 959 947 653 12;
  • 42) 0.474 885 070 227 540 959 947 653 12 × 2 = 0 + 0.949 770 140 455 081 919 895 306 24;
  • 43) 0.949 770 140 455 081 919 895 306 24 × 2 = 1 + 0.899 540 280 910 163 839 790 612 48;
  • 44) 0.899 540 280 910 163 839 790 612 48 × 2 = 1 + 0.799 080 561 820 327 679 581 224 96;
  • 45) 0.799 080 561 820 327 679 581 224 96 × 2 = 1 + 0.598 161 123 640 655 359 162 449 92;
  • 46) 0.598 161 123 640 655 359 162 449 92 × 2 = 1 + 0.196 322 247 281 310 718 324 899 84;
  • 47) 0.196 322 247 281 310 718 324 899 84 × 2 = 0 + 0.392 644 494 562 621 436 649 799 68;
  • 48) 0.392 644 494 562 621 436 649 799 68 × 2 = 0 + 0.785 288 989 125 242 873 299 599 36;
  • 49) 0.785 288 989 125 242 873 299 599 36 × 2 = 1 + 0.570 577 978 250 485 746 599 198 72;
  • 50) 0.570 577 978 250 485 746 599 198 72 × 2 = 1 + 0.141 155 956 500 971 493 198 397 44;
  • 51) 0.141 155 956 500 971 493 198 397 44 × 2 = 0 + 0.282 311 913 001 942 986 396 794 88;
  • 52) 0.282 311 913 001 942 986 396 794 88 × 2 = 0 + 0.564 623 826 003 885 972 793 589 76;
  • 53) 0.564 623 826 003 885 972 793 589 76 × 2 = 1 + 0.129 247 652 007 771 945 587 179 52;

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:


0.414 213 562 373 095 048 801 691 31(10) =

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

5. Positive number before normalization:

1.414 213 562 373 095 048 801 691 31(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:


1.414 213 562 373 095 048 801 691 31(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 1.414 213 562 373 095 048 801 691 31 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