6.999 999 999 999 999 555 910 616 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.999 999 999 999 999 555 910 616(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
6.999 999 999 999 999 555 910 616(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: 6.
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;
  • 6 ÷ 2 = 3 + 0;
  • 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.

6(10) =

110(2)


3. Convert to binary (base 2) the fractional part: 0.999 999 999 999 999 555 910 616.

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) 0.999 999 999 999 999 555 910 616 × 2 = 1 + 0.999 999 999 999 999 111 821 232;
  • 2) 0.999 999 999 999 999 111 821 232 × 2 = 1 + 0.999 999 999 999 998 223 642 464;
  • 3) 0.999 999 999 999 998 223 642 464 × 2 = 1 + 0.999 999 999 999 996 447 284 928;
  • 4) 0.999 999 999 999 996 447 284 928 × 2 = 1 + 0.999 999 999 999 992 894 569 856;
  • 5) 0.999 999 999 999 992 894 569 856 × 2 = 1 + 0.999 999 999 999 985 789 139 712;
  • 6) 0.999 999 999 999 985 789 139 712 × 2 = 1 + 0.999 999 999 999 971 578 279 424;
  • 7) 0.999 999 999 999 971 578 279 424 × 2 = 1 + 0.999 999 999 999 943 156 558 848;
  • 8) 0.999 999 999 999 943 156 558 848 × 2 = 1 + 0.999 999 999 999 886 313 117 696;
  • 9) 0.999 999 999 999 886 313 117 696 × 2 = 1 + 0.999 999 999 999 772 626 235 392;
  • 10) 0.999 999 999 999 772 626 235 392 × 2 = 1 + 0.999 999 999 999 545 252 470 784;
  • 11) 0.999 999 999 999 545 252 470 784 × 2 = 1 + 0.999 999 999 999 090 504 941 568;
  • 12) 0.999 999 999 999 090 504 941 568 × 2 = 1 + 0.999 999 999 998 181 009 883 136;
  • 13) 0.999 999 999 998 181 009 883 136 × 2 = 1 + 0.999 999 999 996 362 019 766 272;
  • 14) 0.999 999 999 996 362 019 766 272 × 2 = 1 + 0.999 999 999 992 724 039 532 544;
  • 15) 0.999 999 999 992 724 039 532 544 × 2 = 1 + 0.999 999 999 985 448 079 065 088;
  • 16) 0.999 999 999 985 448 079 065 088 × 2 = 1 + 0.999 999 999 970 896 158 130 176;
  • 17) 0.999 999 999 970 896 158 130 176 × 2 = 1 + 0.999 999 999 941 792 316 260 352;
  • 18) 0.999 999 999 941 792 316 260 352 × 2 = 1 + 0.999 999 999 883 584 632 520 704;
  • 19) 0.999 999 999 883 584 632 520 704 × 2 = 1 + 0.999 999 999 767 169 265 041 408;
  • 20) 0.999 999 999 767 169 265 041 408 × 2 = 1 + 0.999 999 999 534 338 530 082 816;
  • 21) 0.999 999 999 534 338 530 082 816 × 2 = 1 + 0.999 999 999 068 677 060 165 632;
  • 22) 0.999 999 999 068 677 060 165 632 × 2 = 1 + 0.999 999 998 137 354 120 331 264;
  • 23) 0.999 999 998 137 354 120 331 264 × 2 = 1 + 0.999 999 996 274 708 240 662 528;
  • 24) 0.999 999 996 274 708 240 662 528 × 2 = 1 + 0.999 999 992 549 416 481 325 056;
  • 25) 0.999 999 992 549 416 481 325 056 × 2 = 1 + 0.999 999 985 098 832 962 650 112;
  • 26) 0.999 999 985 098 832 962 650 112 × 2 = 1 + 0.999 999 970 197 665 925 300 224;
  • 27) 0.999 999 970 197 665 925 300 224 × 2 = 1 + 0.999 999 940 395 331 850 600 448;
  • 28) 0.999 999 940 395 331 850 600 448 × 2 = 1 + 0.999 999 880 790 663 701 200 896;
  • 29) 0.999 999 880 790 663 701 200 896 × 2 = 1 + 0.999 999 761 581 327 402 401 792;
  • 30) 0.999 999 761 581 327 402 401 792 × 2 = 1 + 0.999 999 523 162 654 804 803 584;
  • 31) 0.999 999 523 162 654 804 803 584 × 2 = 1 + 0.999 999 046 325 309 609 607 168;
  • 32) 0.999 999 046 325 309 609 607 168 × 2 = 1 + 0.999 998 092 650 619 219 214 336;
  • 33) 0.999 998 092 650 619 219 214 336 × 2 = 1 + 0.999 996 185 301 238 438 428 672;
  • 34) 0.999 996 185 301 238 438 428 672 × 2 = 1 + 0.999 992 370 602 476 876 857 344;
  • 35) 0.999 992 370 602 476 876 857 344 × 2 = 1 + 0.999 984 741 204 953 753 714 688;
  • 36) 0.999 984 741 204 953 753 714 688 × 2 = 1 + 0.999 969 482 409 907 507 429 376;
  • 37) 0.999 969 482 409 907 507 429 376 × 2 = 1 + 0.999 938 964 819 815 014 858 752;
  • 38) 0.999 938 964 819 815 014 858 752 × 2 = 1 + 0.999 877 929 639 630 029 717 504;
  • 39) 0.999 877 929 639 630 029 717 504 × 2 = 1 + 0.999 755 859 279 260 059 435 008;
  • 40) 0.999 755 859 279 260 059 435 008 × 2 = 1 + 0.999 511 718 558 520 118 870 016;
  • 41) 0.999 511 718 558 520 118 870 016 × 2 = 1 + 0.999 023 437 117 040 237 740 032;
  • 42) 0.999 023 437 117 040 237 740 032 × 2 = 1 + 0.998 046 874 234 080 475 480 064;
  • 43) 0.998 046 874 234 080 475 480 064 × 2 = 1 + 0.996 093 748 468 160 950 960 128;
  • 44) 0.996 093 748 468 160 950 960 128 × 2 = 1 + 0.992 187 496 936 321 901 920 256;
  • 45) 0.992 187 496 936 321 901 920 256 × 2 = 1 + 0.984 374 993 872 643 803 840 512;
  • 46) 0.984 374 993 872 643 803 840 512 × 2 = 1 + 0.968 749 987 745 287 607 681 024;
  • 47) 0.968 749 987 745 287 607 681 024 × 2 = 1 + 0.937 499 975 490 575 215 362 048;
  • 48) 0.937 499 975 490 575 215 362 048 × 2 = 1 + 0.874 999 950 981 150 430 724 096;
  • 49) 0.874 999 950 981 150 430 724 096 × 2 = 1 + 0.749 999 901 962 300 861 448 192;
  • 50) 0.749 999 901 962 300 861 448 192 × 2 = 1 + 0.499 999 803 924 601 722 896 384;
  • 51) 0.499 999 803 924 601 722 896 384 × 2 = 0 + 0.999 999 607 849 203 445 792 768;
  • 52) 0.999 999 607 849 203 445 792 768 × 2 = 1 + 0.999 999 215 698 406 891 585 536;
  • 53) 0.999 999 215 698 406 891 585 536 × 2 = 1 + 0.999 998 431 396 813 783 171 072;

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:


0.999 999 999 999 999 555 910 616(10) =

0.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1(2)

5. Positive number before normalization:

6.999 999 999 999 999 555 910 616(10) =

110.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the left, so that only one non zero digit remains to the left of it:


6.999 999 999 999 999 555 910 616(10) =

110.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1(2) =

110.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1(2) × 20 =

1.1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 011(2) × 22


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): 2

Mantissa (not normalized):
1.1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

2 + 2(11-1) - 1 =

(2 + 1 023)(10) =

1 025(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;
  • 1 025 ÷ 2 = 512 + 1;
  • 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) =

1025(10) =

100 0000 0001(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 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 011 =

1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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 0001

Mantissa (52 bits) =
1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


Decimal number 6.999 999 999 999 999 555 910 616 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1011 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

» Convert decimal 6.999 999 999 999 999 555 910 66 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