6.999 999 999 999 999 555 910 669 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.999 999 999 999 999 555 910 669(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
6.999 999 999 999 999 555 910 669(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: 0.999 999 999 999 999 555 910 669.

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) 0.999 999 999 999 999 555 910 669 × 2 = 1 + 0.999 999 999 999 999 111 821 338;
  • 2) 0.999 999 999 999 999 111 821 338 × 2 = 1 + 0.999 999 999 999 998 223 642 676;
  • 3) 0.999 999 999 999 998 223 642 676 × 2 = 1 + 0.999 999 999 999 996 447 285 352;
  • 4) 0.999 999 999 999 996 447 285 352 × 2 = 1 + 0.999 999 999 999 992 894 570 704;
  • 5) 0.999 999 999 999 992 894 570 704 × 2 = 1 + 0.999 999 999 999 985 789 141 408;
  • 6) 0.999 999 999 999 985 789 141 408 × 2 = 1 + 0.999 999 999 999 971 578 282 816;
  • 7) 0.999 999 999 999 971 578 282 816 × 2 = 1 + 0.999 999 999 999 943 156 565 632;
  • 8) 0.999 999 999 999 943 156 565 632 × 2 = 1 + 0.999 999 999 999 886 313 131 264;
  • 9) 0.999 999 999 999 886 313 131 264 × 2 = 1 + 0.999 999 999 999 772 626 262 528;
  • 10) 0.999 999 999 999 772 626 262 528 × 2 = 1 + 0.999 999 999 999 545 252 525 056;
  • 11) 0.999 999 999 999 545 252 525 056 × 2 = 1 + 0.999 999 999 999 090 505 050 112;
  • 12) 0.999 999 999 999 090 505 050 112 × 2 = 1 + 0.999 999 999 998 181 010 100 224;
  • 13) 0.999 999 999 998 181 010 100 224 × 2 = 1 + 0.999 999 999 996 362 020 200 448;
  • 14) 0.999 999 999 996 362 020 200 448 × 2 = 1 + 0.999 999 999 992 724 040 400 896;
  • 15) 0.999 999 999 992 724 040 400 896 × 2 = 1 + 0.999 999 999 985 448 080 801 792;
  • 16) 0.999 999 999 985 448 080 801 792 × 2 = 1 + 0.999 999 999 970 896 161 603 584;
  • 17) 0.999 999 999 970 896 161 603 584 × 2 = 1 + 0.999 999 999 941 792 323 207 168;
  • 18) 0.999 999 999 941 792 323 207 168 × 2 = 1 + 0.999 999 999 883 584 646 414 336;
  • 19) 0.999 999 999 883 584 646 414 336 × 2 = 1 + 0.999 999 999 767 169 292 828 672;
  • 20) 0.999 999 999 767 169 292 828 672 × 2 = 1 + 0.999 999 999 534 338 585 657 344;
  • 21) 0.999 999 999 534 338 585 657 344 × 2 = 1 + 0.999 999 999 068 677 171 314 688;
  • 22) 0.999 999 999 068 677 171 314 688 × 2 = 1 + 0.999 999 998 137 354 342 629 376;
  • 23) 0.999 999 998 137 354 342 629 376 × 2 = 1 + 0.999 999 996 274 708 685 258 752;
  • 24) 0.999 999 996 274 708 685 258 752 × 2 = 1 + 0.999 999 992 549 417 370 517 504;
  • 25) 0.999 999 992 549 417 370 517 504 × 2 = 1 + 0.999 999 985 098 834 741 035 008;
  • 26) 0.999 999 985 098 834 741 035 008 × 2 = 1 + 0.999 999 970 197 669 482 070 016;
  • 27) 0.999 999 970 197 669 482 070 016 × 2 = 1 + 0.999 999 940 395 338 964 140 032;
  • 28) 0.999 999 940 395 338 964 140 032 × 2 = 1 + 0.999 999 880 790 677 928 280 064;
  • 29) 0.999 999 880 790 677 928 280 064 × 2 = 1 + 0.999 999 761 581 355 856 560 128;
  • 30) 0.999 999 761 581 355 856 560 128 × 2 = 1 + 0.999 999 523 162 711 713 120 256;
  • 31) 0.999 999 523 162 711 713 120 256 × 2 = 1 + 0.999 999 046 325 423 426 240 512;
  • 32) 0.999 999 046 325 423 426 240 512 × 2 = 1 + 0.999 998 092 650 846 852 481 024;
  • 33) 0.999 998 092 650 846 852 481 024 × 2 = 1 + 0.999 996 185 301 693 704 962 048;
  • 34) 0.999 996 185 301 693 704 962 048 × 2 = 1 + 0.999 992 370 603 387 409 924 096;
  • 35) 0.999 992 370 603 387 409 924 096 × 2 = 1 + 0.999 984 741 206 774 819 848 192;
  • 36) 0.999 984 741 206 774 819 848 192 × 2 = 1 + 0.999 969 482 413 549 639 696 384;
  • 37) 0.999 969 482 413 549 639 696 384 × 2 = 1 + 0.999 938 964 827 099 279 392 768;
  • 38) 0.999 938 964 827 099 279 392 768 × 2 = 1 + 0.999 877 929 654 198 558 785 536;
  • 39) 0.999 877 929 654 198 558 785 536 × 2 = 1 + 0.999 755 859 308 397 117 571 072;
  • 40) 0.999 755 859 308 397 117 571 072 × 2 = 1 + 0.999 511 718 616 794 235 142 144;
  • 41) 0.999 511 718 616 794 235 142 144 × 2 = 1 + 0.999 023 437 233 588 470 284 288;
  • 42) 0.999 023 437 233 588 470 284 288 × 2 = 1 + 0.998 046 874 467 176 940 568 576;
  • 43) 0.998 046 874 467 176 940 568 576 × 2 = 1 + 0.996 093 748 934 353 881 137 152;
  • 44) 0.996 093 748 934 353 881 137 152 × 2 = 1 + 0.992 187 497 868 707 762 274 304;
  • 45) 0.992 187 497 868 707 762 274 304 × 2 = 1 + 0.984 374 995 737 415 524 548 608;
  • 46) 0.984 374 995 737 415 524 548 608 × 2 = 1 + 0.968 749 991 474 831 049 097 216;
  • 47) 0.968 749 991 474 831 049 097 216 × 2 = 1 + 0.937 499 982 949 662 098 194 432;
  • 48) 0.937 499 982 949 662 098 194 432 × 2 = 1 + 0.874 999 965 899 324 196 388 864;
  • 49) 0.874 999 965 899 324 196 388 864 × 2 = 1 + 0.749 999 931 798 648 392 777 728;
  • 50) 0.749 999 931 798 648 392 777 728 × 2 = 1 + 0.499 999 863 597 296 785 555 456;
  • 51) 0.499 999 863 597 296 785 555 456 × 2 = 0 + 0.999 999 727 194 593 571 110 912;
  • 52) 0.999 999 727 194 593 571 110 912 × 2 = 1 + 0.999 999 454 389 187 142 221 824;
  • 53) 0.999 999 454 389 187 142 221 824 × 2 = 1 + 0.999 998 908 778 374 284 443 648;

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:


0.999 999 999 999 999 555 910 669(10) =

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

5. Positive number before normalization:

6.999 999 999 999 999 555 910 669(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:


6.999 999 999 999 999 555 910 669(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 6.999 999 999 999 999 555 910 669 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 6.999 999 999 999 999 555 910 628 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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