3.421 299 999 999 999 563 726 759 796 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 3.421 299 999 999 999 563 726 759 796(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
3.421 299 999 999 999 563 726 759 796(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: 3.
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;
  • 3 ÷ 2 = 1 + 1;
  • 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.

3(10) =

11(2)


3. Convert to binary (base 2) the fractional part: 0.421 299 999 999 999 563 726 759 796.

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) 0.421 299 999 999 999 563 726 759 796 × 2 = 0 + 0.842 599 999 999 999 127 453 519 592;
  • 2) 0.842 599 999 999 999 127 453 519 592 × 2 = 1 + 0.685 199 999 999 998 254 907 039 184;
  • 3) 0.685 199 999 999 998 254 907 039 184 × 2 = 1 + 0.370 399 999 999 996 509 814 078 368;
  • 4) 0.370 399 999 999 996 509 814 078 368 × 2 = 0 + 0.740 799 999 999 993 019 628 156 736;
  • 5) 0.740 799 999 999 993 019 628 156 736 × 2 = 1 + 0.481 599 999 999 986 039 256 313 472;
  • 6) 0.481 599 999 999 986 039 256 313 472 × 2 = 0 + 0.963 199 999 999 972 078 512 626 944;
  • 7) 0.963 199 999 999 972 078 512 626 944 × 2 = 1 + 0.926 399 999 999 944 157 025 253 888;
  • 8) 0.926 399 999 999 944 157 025 253 888 × 2 = 1 + 0.852 799 999 999 888 314 050 507 776;
  • 9) 0.852 799 999 999 888 314 050 507 776 × 2 = 1 + 0.705 599 999 999 776 628 101 015 552;
  • 10) 0.705 599 999 999 776 628 101 015 552 × 2 = 1 + 0.411 199 999 999 553 256 202 031 104;
  • 11) 0.411 199 999 999 553 256 202 031 104 × 2 = 0 + 0.822 399 999 999 106 512 404 062 208;
  • 12) 0.822 399 999 999 106 512 404 062 208 × 2 = 1 + 0.644 799 999 998 213 024 808 124 416;
  • 13) 0.644 799 999 998 213 024 808 124 416 × 2 = 1 + 0.289 599 999 996 426 049 616 248 832;
  • 14) 0.289 599 999 996 426 049 616 248 832 × 2 = 0 + 0.579 199 999 992 852 099 232 497 664;
  • 15) 0.579 199 999 992 852 099 232 497 664 × 2 = 1 + 0.158 399 999 985 704 198 464 995 328;
  • 16) 0.158 399 999 985 704 198 464 995 328 × 2 = 0 + 0.316 799 999 971 408 396 929 990 656;
  • 17) 0.316 799 999 971 408 396 929 990 656 × 2 = 0 + 0.633 599 999 942 816 793 859 981 312;
  • 18) 0.633 599 999 942 816 793 859 981 312 × 2 = 1 + 0.267 199 999 885 633 587 719 962 624;
  • 19) 0.267 199 999 885 633 587 719 962 624 × 2 = 0 + 0.534 399 999 771 267 175 439 925 248;
  • 20) 0.534 399 999 771 267 175 439 925 248 × 2 = 1 + 0.068 799 999 542 534 350 879 850 496;
  • 21) 0.068 799 999 542 534 350 879 850 496 × 2 = 0 + 0.137 599 999 085 068 701 759 700 992;
  • 22) 0.137 599 999 085 068 701 759 700 992 × 2 = 0 + 0.275 199 998 170 137 403 519 401 984;
  • 23) 0.275 199 998 170 137 403 519 401 984 × 2 = 0 + 0.550 399 996 340 274 807 038 803 968;
  • 24) 0.550 399 996 340 274 807 038 803 968 × 2 = 1 + 0.100 799 992 680 549 614 077 607 936;
  • 25) 0.100 799 992 680 549 614 077 607 936 × 2 = 0 + 0.201 599 985 361 099 228 155 215 872;
  • 26) 0.201 599 985 361 099 228 155 215 872 × 2 = 0 + 0.403 199 970 722 198 456 310 431 744;
  • 27) 0.403 199 970 722 198 456 310 431 744 × 2 = 0 + 0.806 399 941 444 396 912 620 863 488;
  • 28) 0.806 399 941 444 396 912 620 863 488 × 2 = 1 + 0.612 799 882 888 793 825 241 726 976;
  • 29) 0.612 799 882 888 793 825 241 726 976 × 2 = 1 + 0.225 599 765 777 587 650 483 453 952;
  • 30) 0.225 599 765 777 587 650 483 453 952 × 2 = 0 + 0.451 199 531 555 175 300 966 907 904;
  • 31) 0.451 199 531 555 175 300 966 907 904 × 2 = 0 + 0.902 399 063 110 350 601 933 815 808;
  • 32) 0.902 399 063 110 350 601 933 815 808 × 2 = 1 + 0.804 798 126 220 701 203 867 631 616;
  • 33) 0.804 798 126 220 701 203 867 631 616 × 2 = 1 + 0.609 596 252 441 402 407 735 263 232;
  • 34) 0.609 596 252 441 402 407 735 263 232 × 2 = 1 + 0.219 192 504 882 804 815 470 526 464;
  • 35) 0.219 192 504 882 804 815 470 526 464 × 2 = 0 + 0.438 385 009 765 609 630 941 052 928;
  • 36) 0.438 385 009 765 609 630 941 052 928 × 2 = 0 + 0.876 770 019 531 219 261 882 105 856;
  • 37) 0.876 770 019 531 219 261 882 105 856 × 2 = 1 + 0.753 540 039 062 438 523 764 211 712;
  • 38) 0.753 540 039 062 438 523 764 211 712 × 2 = 1 + 0.507 080 078 124 877 047 528 423 424;
  • 39) 0.507 080 078 124 877 047 528 423 424 × 2 = 1 + 0.014 160 156 249 754 095 056 846 848;
  • 40) 0.014 160 156 249 754 095 056 846 848 × 2 = 0 + 0.028 320 312 499 508 190 113 693 696;
  • 41) 0.028 320 312 499 508 190 113 693 696 × 2 = 0 + 0.056 640 624 999 016 380 227 387 392;
  • 42) 0.056 640 624 999 016 380 227 387 392 × 2 = 0 + 0.113 281 249 998 032 760 454 774 784;
  • 43) 0.113 281 249 998 032 760 454 774 784 × 2 = 0 + 0.226 562 499 996 065 520 909 549 568;
  • 44) 0.226 562 499 996 065 520 909 549 568 × 2 = 0 + 0.453 124 999 992 131 041 819 099 136;
  • 45) 0.453 124 999 992 131 041 819 099 136 × 2 = 0 + 0.906 249 999 984 262 083 638 198 272;
  • 46) 0.906 249 999 984 262 083 638 198 272 × 2 = 1 + 0.812 499 999 968 524 167 276 396 544;
  • 47) 0.812 499 999 968 524 167 276 396 544 × 2 = 1 + 0.624 999 999 937 048 334 552 793 088;
  • 48) 0.624 999 999 937 048 334 552 793 088 × 2 = 1 + 0.249 999 999 874 096 669 105 586 176;
  • 49) 0.249 999 999 874 096 669 105 586 176 × 2 = 0 + 0.499 999 999 748 193 338 211 172 352;
  • 50) 0.499 999 999 748 193 338 211 172 352 × 2 = 0 + 0.999 999 999 496 386 676 422 344 704;
  • 51) 0.999 999 999 496 386 676 422 344 704 × 2 = 1 + 0.999 999 998 992 773 352 844 689 408;
  • 52) 0.999 999 998 992 773 352 844 689 408 × 2 = 1 + 0.999 999 997 985 546 705 689 378 816;
  • 53) 0.999 999 997 985 546 705 689 378 816 × 2 = 1 + 0.999 999 995 971 093 411 378 757 632;

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:


0.421 299 999 999 999 563 726 759 796(10) =

0.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0011 1(2)

5. Positive number before normalization:

3.421 299 999 999 999 563 726 759 796(10) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0011 1(2)

6. Normalize the binary representation of the number.

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


3.421 299 999 999 999 563 726 759 796(10) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0011 1(2) =

11.0110 1011 1101 1010 0101 0001 0001 1001 1100 1110 0000 0111 0011 1(2) × 20 =

1.1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001 11(2) × 21


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):
1.1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001 11


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) =

1 024(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;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1024(10) =

100 0000 0000(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 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001 11 =

1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001


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) =
100 0000 0000

Mantissa (52 bits) =
1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001


Decimal number 3.421 299 999 999 999 563 726 759 796 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 1011 0101 1110 1101 0010 1000 1000 1100 1110 0111 0000 0011 1001

» Convert decimal 3.421 299 999 999 999 563 726 759 764 to 64 bit double precision IEEE 754 binary floating point representation standard

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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