1.414 213 562 373 095 048 801 689 79 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.414 213 562 373 095 048 801 689 79(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 79(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 79.

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 79 × 2 = 0 + 0.828 427 124 746 190 097 603 379 58;
  • 2) 0.828 427 124 746 190 097 603 379 58 × 2 = 1 + 0.656 854 249 492 380 195 206 759 16;
  • 3) 0.656 854 249 492 380 195 206 759 16 × 2 = 1 + 0.313 708 498 984 760 390 413 518 32;
  • 4) 0.313 708 498 984 760 390 413 518 32 × 2 = 0 + 0.627 416 997 969 520 780 827 036 64;
  • 5) 0.627 416 997 969 520 780 827 036 64 × 2 = 1 + 0.254 833 995 939 041 561 654 073 28;
  • 6) 0.254 833 995 939 041 561 654 073 28 × 2 = 0 + 0.509 667 991 878 083 123 308 146 56;
  • 7) 0.509 667 991 878 083 123 308 146 56 × 2 = 1 + 0.019 335 983 756 166 246 616 293 12;
  • 8) 0.019 335 983 756 166 246 616 293 12 × 2 = 0 + 0.038 671 967 512 332 493 232 586 24;
  • 9) 0.038 671 967 512 332 493 232 586 24 × 2 = 0 + 0.077 343 935 024 664 986 465 172 48;
  • 10) 0.077 343 935 024 664 986 465 172 48 × 2 = 0 + 0.154 687 870 049 329 972 930 344 96;
  • 11) 0.154 687 870 049 329 972 930 344 96 × 2 = 0 + 0.309 375 740 098 659 945 860 689 92;
  • 12) 0.309 375 740 098 659 945 860 689 92 × 2 = 0 + 0.618 751 480 197 319 891 721 379 84;
  • 13) 0.618 751 480 197 319 891 721 379 84 × 2 = 1 + 0.237 502 960 394 639 783 442 759 68;
  • 14) 0.237 502 960 394 639 783 442 759 68 × 2 = 0 + 0.475 005 920 789 279 566 885 519 36;
  • 15) 0.475 005 920 789 279 566 885 519 36 × 2 = 0 + 0.950 011 841 578 559 133 771 038 72;
  • 16) 0.950 011 841 578 559 133 771 038 72 × 2 = 1 + 0.900 023 683 157 118 267 542 077 44;
  • 17) 0.900 023 683 157 118 267 542 077 44 × 2 = 1 + 0.800 047 366 314 236 535 084 154 88;
  • 18) 0.800 047 366 314 236 535 084 154 88 × 2 = 1 + 0.600 094 732 628 473 070 168 309 76;
  • 19) 0.600 094 732 628 473 070 168 309 76 × 2 = 1 + 0.200 189 465 256 946 140 336 619 52;
  • 20) 0.200 189 465 256 946 140 336 619 52 × 2 = 0 + 0.400 378 930 513 892 280 673 239 04;
  • 21) 0.400 378 930 513 892 280 673 239 04 × 2 = 0 + 0.800 757 861 027 784 561 346 478 08;
  • 22) 0.800 757 861 027 784 561 346 478 08 × 2 = 1 + 0.601 515 722 055 569 122 692 956 16;
  • 23) 0.601 515 722 055 569 122 692 956 16 × 2 = 1 + 0.203 031 444 111 138 245 385 912 32;
  • 24) 0.203 031 444 111 138 245 385 912 32 × 2 = 0 + 0.406 062 888 222 276 490 771 824 64;
  • 25) 0.406 062 888 222 276 490 771 824 64 × 2 = 0 + 0.812 125 776 444 552 981 543 649 28;
  • 26) 0.812 125 776 444 552 981 543 649 28 × 2 = 1 + 0.624 251 552 889 105 963 087 298 56;
  • 27) 0.624 251 552 889 105 963 087 298 56 × 2 = 1 + 0.248 503 105 778 211 926 174 597 12;
  • 28) 0.248 503 105 778 211 926 174 597 12 × 2 = 0 + 0.497 006 211 556 423 852 349 194 24;
  • 29) 0.497 006 211 556 423 852 349 194 24 × 2 = 0 + 0.994 012 423 112 847 704 698 388 48;
  • 30) 0.994 012 423 112 847 704 698 388 48 × 2 = 1 + 0.988 024 846 225 695 409 396 776 96;
  • 31) 0.988 024 846 225 695 409 396 776 96 × 2 = 1 + 0.976 049 692 451 390 818 793 553 92;
  • 32) 0.976 049 692 451 390 818 793 553 92 × 2 = 1 + 0.952 099 384 902 781 637 587 107 84;
  • 33) 0.952 099 384 902 781 637 587 107 84 × 2 = 1 + 0.904 198 769 805 563 275 174 215 68;
  • 34) 0.904 198 769 805 563 275 174 215 68 × 2 = 1 + 0.808 397 539 611 126 550 348 431 36;
  • 35) 0.808 397 539 611 126 550 348 431 36 × 2 = 1 + 0.616 795 079 222 253 100 696 862 72;
  • 36) 0.616 795 079 222 253 100 696 862 72 × 2 = 1 + 0.233 590 158 444 506 201 393 725 44;
  • 37) 0.233 590 158 444 506 201 393 725 44 × 2 = 0 + 0.467 180 316 889 012 402 787 450 88;
  • 38) 0.467 180 316 889 012 402 787 450 88 × 2 = 0 + 0.934 360 633 778 024 805 574 901 76;
  • 39) 0.934 360 633 778 024 805 574 901 76 × 2 = 1 + 0.868 721 267 556 049 611 149 803 52;
  • 40) 0.868 721 267 556 049 611 149 803 52 × 2 = 1 + 0.737 442 535 112 099 222 299 607 04;
  • 41) 0.737 442 535 112 099 222 299 607 04 × 2 = 1 + 0.474 885 070 224 198 444 599 214 08;
  • 42) 0.474 885 070 224 198 444 599 214 08 × 2 = 0 + 0.949 770 140 448 396 889 198 428 16;
  • 43) 0.949 770 140 448 396 889 198 428 16 × 2 = 1 + 0.899 540 280 896 793 778 396 856 32;
  • 44) 0.899 540 280 896 793 778 396 856 32 × 2 = 1 + 0.799 080 561 793 587 556 793 712 64;
  • 45) 0.799 080 561 793 587 556 793 712 64 × 2 = 1 + 0.598 161 123 587 175 113 587 425 28;
  • 46) 0.598 161 123 587 175 113 587 425 28 × 2 = 1 + 0.196 322 247 174 350 227 174 850 56;
  • 47) 0.196 322 247 174 350 227 174 850 56 × 2 = 0 + 0.392 644 494 348 700 454 349 701 12;
  • 48) 0.392 644 494 348 700 454 349 701 12 × 2 = 0 + 0.785 288 988 697 400 908 699 402 24;
  • 49) 0.785 288 988 697 400 908 699 402 24 × 2 = 1 + 0.570 577 977 394 801 817 398 804 48;
  • 50) 0.570 577 977 394 801 817 398 804 48 × 2 = 1 + 0.141 155 954 789 603 634 797 608 96;
  • 51) 0.141 155 954 789 603 634 797 608 96 × 2 = 0 + 0.282 311 909 579 207 269 595 217 92;
  • 52) 0.282 311 909 579 207 269 595 217 92 × 2 = 0 + 0.564 623 819 158 414 539 190 435 84;
  • 53) 0.564 623 819 158 414 539 190 435 84 × 2 = 1 + 0.129 247 638 316 829 078 380 871 68;

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 79(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 79(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 79(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 79 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