2.633 913 364 817 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 2.633 913 364 817(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
2.633 913 364 817(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: 2.
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;
  • 2 ÷ 2 = 1 + 0;
  • 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.

2(10) =

10(2)


3. Convert to binary (base 2) the fractional part: 0.633 913 364 817.

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) 0.633 913 364 817 × 2 = 1 + 0.267 826 729 634;
  • 2) 0.267 826 729 634 × 2 = 0 + 0.535 653 459 268;
  • 3) 0.535 653 459 268 × 2 = 1 + 0.071 306 918 536;
  • 4) 0.071 306 918 536 × 2 = 0 + 0.142 613 837 072;
  • 5) 0.142 613 837 072 × 2 = 0 + 0.285 227 674 144;
  • 6) 0.285 227 674 144 × 2 = 0 + 0.570 455 348 288;
  • 7) 0.570 455 348 288 × 2 = 1 + 0.140 910 696 576;
  • 8) 0.140 910 696 576 × 2 = 0 + 0.281 821 393 152;
  • 9) 0.281 821 393 152 × 2 = 0 + 0.563 642 786 304;
  • 10) 0.563 642 786 304 × 2 = 1 + 0.127 285 572 608;
  • 11) 0.127 285 572 608 × 2 = 0 + 0.254 571 145 216;
  • 12) 0.254 571 145 216 × 2 = 0 + 0.509 142 290 432;
  • 13) 0.509 142 290 432 × 2 = 1 + 0.018 284 580 864;
  • 14) 0.018 284 580 864 × 2 = 0 + 0.036 569 161 728;
  • 15) 0.036 569 161 728 × 2 = 0 + 0.073 138 323 456;
  • 16) 0.073 138 323 456 × 2 = 0 + 0.146 276 646 912;
  • 17) 0.146 276 646 912 × 2 = 0 + 0.292 553 293 824;
  • 18) 0.292 553 293 824 × 2 = 0 + 0.585 106 587 648;
  • 19) 0.585 106 587 648 × 2 = 1 + 0.170 213 175 296;
  • 20) 0.170 213 175 296 × 2 = 0 + 0.340 426 350 592;
  • 21) 0.340 426 350 592 × 2 = 0 + 0.680 852 701 184;
  • 22) 0.680 852 701 184 × 2 = 1 + 0.361 705 402 368;
  • 23) 0.361 705 402 368 × 2 = 0 + 0.723 410 804 736;
  • 24) 0.723 410 804 736 × 2 = 1 + 0.446 821 609 472;
  • 25) 0.446 821 609 472 × 2 = 0 + 0.893 643 218 944;
  • 26) 0.893 643 218 944 × 2 = 1 + 0.787 286 437 888;
  • 27) 0.787 286 437 888 × 2 = 1 + 0.574 572 875 776;
  • 28) 0.574 572 875 776 × 2 = 1 + 0.149 145 751 552;
  • 29) 0.149 145 751 552 × 2 = 0 + 0.298 291 503 104;
  • 30) 0.298 291 503 104 × 2 = 0 + 0.596 583 006 208;
  • 31) 0.596 583 006 208 × 2 = 1 + 0.193 166 012 416;
  • 32) 0.193 166 012 416 × 2 = 0 + 0.386 332 024 832;
  • 33) 0.386 332 024 832 × 2 = 0 + 0.772 664 049 664;
  • 34) 0.772 664 049 664 × 2 = 1 + 0.545 328 099 328;
  • 35) 0.545 328 099 328 × 2 = 1 + 0.090 656 198 656;
  • 36) 0.090 656 198 656 × 2 = 0 + 0.181 312 397 312;
  • 37) 0.181 312 397 312 × 2 = 0 + 0.362 624 794 624;
  • 38) 0.362 624 794 624 × 2 = 0 + 0.725 249 589 248;
  • 39) 0.725 249 589 248 × 2 = 1 + 0.450 499 178 496;
  • 40) 0.450 499 178 496 × 2 = 0 + 0.900 998 356 992;
  • 41) 0.900 998 356 992 × 2 = 1 + 0.801 996 713 984;
  • 42) 0.801 996 713 984 × 2 = 1 + 0.603 993 427 968;
  • 43) 0.603 993 427 968 × 2 = 1 + 0.207 986 855 936;
  • 44) 0.207 986 855 936 × 2 = 0 + 0.415 973 711 872;
  • 45) 0.415 973 711 872 × 2 = 0 + 0.831 947 423 744;
  • 46) 0.831 947 423 744 × 2 = 1 + 0.663 894 847 488;
  • 47) 0.663 894 847 488 × 2 = 1 + 0.327 789 694 976;
  • 48) 0.327 789 694 976 × 2 = 0 + 0.655 579 389 952;
  • 49) 0.655 579 389 952 × 2 = 1 + 0.311 158 779 904;
  • 50) 0.311 158 779 904 × 2 = 0 + 0.622 317 559 808;
  • 51) 0.622 317 559 808 × 2 = 1 + 0.244 635 119 616;
  • 52) 0.244 635 119 616 × 2 = 0 + 0.489 270 239 232;
  • 53) 0.489 270 239 232 × 2 = 0 + 0.978 540 478 464;

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:


0.633 913 364 817(10) =

0.1010 0010 0100 1000 0010 0101 0111 0010 0110 0010 1110 0110 1010 0(2)

5. Positive number before normalization:

2.633 913 364 817(10) =

10.1010 0010 0100 1000 0010 0101 0111 0010 0110 0010 1110 0110 1010 0(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:


2.633 913 364 817(10) =

10.1010 0010 0100 1000 0010 0101 0111 0010 0110 0010 1110 0110 1010 0(2) =

10.1010 0010 0100 1000 0010 0101 0111 0010 0110 0010 1110 0110 1010 0(2) × 20 =

1.0101 0001 0010 0100 0001 0010 1011 1001 0011 0001 0111 0011 0101 00(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):
1.0101 0001 0010 0100 0001 0010 1011 1001 0011 0001 0111 0011 0101 00


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. 0101 0001 0010 0100 0001 0010 1011 1001 0011 0001 0111 0011 0101 00 =

0101 0001 0010 0100 0001 0010 1011 1001 0011 0001 0111 0011 0101


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) =
0101 0001 0010 0100 0001 0010 1011 1001 0011 0001 0111 0011 0101


Decimal number 2.633 913 364 817 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0000 - 0101 0001 0010 0100 0001 0010 1011 1001 0011 0001 0111 0011 0101

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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