6.999 999 999 999 999 555 910 533 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.999 999 999 999 999 555 910 533(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
6.999 999 999 999 999 555 910 533(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: 0.999 999 999 999 999 555 910 533.

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) 0.999 999 999 999 999 555 910 533 × 2 = 1 + 0.999 999 999 999 999 111 821 066;
  • 2) 0.999 999 999 999 999 111 821 066 × 2 = 1 + 0.999 999 999 999 998 223 642 132;
  • 3) 0.999 999 999 999 998 223 642 132 × 2 = 1 + 0.999 999 999 999 996 447 284 264;
  • 4) 0.999 999 999 999 996 447 284 264 × 2 = 1 + 0.999 999 999 999 992 894 568 528;
  • 5) 0.999 999 999 999 992 894 568 528 × 2 = 1 + 0.999 999 999 999 985 789 137 056;
  • 6) 0.999 999 999 999 985 789 137 056 × 2 = 1 + 0.999 999 999 999 971 578 274 112;
  • 7) 0.999 999 999 999 971 578 274 112 × 2 = 1 + 0.999 999 999 999 943 156 548 224;
  • 8) 0.999 999 999 999 943 156 548 224 × 2 = 1 + 0.999 999 999 999 886 313 096 448;
  • 9) 0.999 999 999 999 886 313 096 448 × 2 = 1 + 0.999 999 999 999 772 626 192 896;
  • 10) 0.999 999 999 999 772 626 192 896 × 2 = 1 + 0.999 999 999 999 545 252 385 792;
  • 11) 0.999 999 999 999 545 252 385 792 × 2 = 1 + 0.999 999 999 999 090 504 771 584;
  • 12) 0.999 999 999 999 090 504 771 584 × 2 = 1 + 0.999 999 999 998 181 009 543 168;
  • 13) 0.999 999 999 998 181 009 543 168 × 2 = 1 + 0.999 999 999 996 362 019 086 336;
  • 14) 0.999 999 999 996 362 019 086 336 × 2 = 1 + 0.999 999 999 992 724 038 172 672;
  • 15) 0.999 999 999 992 724 038 172 672 × 2 = 1 + 0.999 999 999 985 448 076 345 344;
  • 16) 0.999 999 999 985 448 076 345 344 × 2 = 1 + 0.999 999 999 970 896 152 690 688;
  • 17) 0.999 999 999 970 896 152 690 688 × 2 = 1 + 0.999 999 999 941 792 305 381 376;
  • 18) 0.999 999 999 941 792 305 381 376 × 2 = 1 + 0.999 999 999 883 584 610 762 752;
  • 19) 0.999 999 999 883 584 610 762 752 × 2 = 1 + 0.999 999 999 767 169 221 525 504;
  • 20) 0.999 999 999 767 169 221 525 504 × 2 = 1 + 0.999 999 999 534 338 443 051 008;
  • 21) 0.999 999 999 534 338 443 051 008 × 2 = 1 + 0.999 999 999 068 676 886 102 016;
  • 22) 0.999 999 999 068 676 886 102 016 × 2 = 1 + 0.999 999 998 137 353 772 204 032;
  • 23) 0.999 999 998 137 353 772 204 032 × 2 = 1 + 0.999 999 996 274 707 544 408 064;
  • 24) 0.999 999 996 274 707 544 408 064 × 2 = 1 + 0.999 999 992 549 415 088 816 128;
  • 25) 0.999 999 992 549 415 088 816 128 × 2 = 1 + 0.999 999 985 098 830 177 632 256;
  • 26) 0.999 999 985 098 830 177 632 256 × 2 = 1 + 0.999 999 970 197 660 355 264 512;
  • 27) 0.999 999 970 197 660 355 264 512 × 2 = 1 + 0.999 999 940 395 320 710 529 024;
  • 28) 0.999 999 940 395 320 710 529 024 × 2 = 1 + 0.999 999 880 790 641 421 058 048;
  • 29) 0.999 999 880 790 641 421 058 048 × 2 = 1 + 0.999 999 761 581 282 842 116 096;
  • 30) 0.999 999 761 581 282 842 116 096 × 2 = 1 + 0.999 999 523 162 565 684 232 192;
  • 31) 0.999 999 523 162 565 684 232 192 × 2 = 1 + 0.999 999 046 325 131 368 464 384;
  • 32) 0.999 999 046 325 131 368 464 384 × 2 = 1 + 0.999 998 092 650 262 736 928 768;
  • 33) 0.999 998 092 650 262 736 928 768 × 2 = 1 + 0.999 996 185 300 525 473 857 536;
  • 34) 0.999 996 185 300 525 473 857 536 × 2 = 1 + 0.999 992 370 601 050 947 715 072;
  • 35) 0.999 992 370 601 050 947 715 072 × 2 = 1 + 0.999 984 741 202 101 895 430 144;
  • 36) 0.999 984 741 202 101 895 430 144 × 2 = 1 + 0.999 969 482 404 203 790 860 288;
  • 37) 0.999 969 482 404 203 790 860 288 × 2 = 1 + 0.999 938 964 808 407 581 720 576;
  • 38) 0.999 938 964 808 407 581 720 576 × 2 = 1 + 0.999 877 929 616 815 163 441 152;
  • 39) 0.999 877 929 616 815 163 441 152 × 2 = 1 + 0.999 755 859 233 630 326 882 304;
  • 40) 0.999 755 859 233 630 326 882 304 × 2 = 1 + 0.999 511 718 467 260 653 764 608;
  • 41) 0.999 511 718 467 260 653 764 608 × 2 = 1 + 0.999 023 436 934 521 307 529 216;
  • 42) 0.999 023 436 934 521 307 529 216 × 2 = 1 + 0.998 046 873 869 042 615 058 432;
  • 43) 0.998 046 873 869 042 615 058 432 × 2 = 1 + 0.996 093 747 738 085 230 116 864;
  • 44) 0.996 093 747 738 085 230 116 864 × 2 = 1 + 0.992 187 495 476 170 460 233 728;
  • 45) 0.992 187 495 476 170 460 233 728 × 2 = 1 + 0.984 374 990 952 340 920 467 456;
  • 46) 0.984 374 990 952 340 920 467 456 × 2 = 1 + 0.968 749 981 904 681 840 934 912;
  • 47) 0.968 749 981 904 681 840 934 912 × 2 = 1 + 0.937 499 963 809 363 681 869 824;
  • 48) 0.937 499 963 809 363 681 869 824 × 2 = 1 + 0.874 999 927 618 727 363 739 648;
  • 49) 0.874 999 927 618 727 363 739 648 × 2 = 1 + 0.749 999 855 237 454 727 479 296;
  • 50) 0.749 999 855 237 454 727 479 296 × 2 = 1 + 0.499 999 710 474 909 454 958 592;
  • 51) 0.499 999 710 474 909 454 958 592 × 2 = 0 + 0.999 999 420 949 818 909 917 184;
  • 52) 0.999 999 420 949 818 909 917 184 × 2 = 1 + 0.999 998 841 899 637 819 834 368;
  • 53) 0.999 998 841 899 637 819 834 368 × 2 = 1 + 0.999 997 683 799 275 639 668 736;

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:


0.999 999 999 999 999 555 910 533(10) =

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

5. Positive number before normalization:

6.999 999 999 999 999 555 910 533(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:


6.999 999 999 999 999 555 910 533(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 6.999 999 999 999 999 555 910 533 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 6.999 999 999 999 999 555 910 535 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