6.833 210 784 070 638 623 841 119 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 6.833 210 784 070 638 623 841 119(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
6.833 210 784 070 638 623 841 119(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: 0.833 210 784 070 638 623 841 119.

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) 0.833 210 784 070 638 623 841 119 × 2 = 1 + 0.666 421 568 141 277 247 682 238;
  • 2) 0.666 421 568 141 277 247 682 238 × 2 = 1 + 0.332 843 136 282 554 495 364 476;
  • 3) 0.332 843 136 282 554 495 364 476 × 2 = 0 + 0.665 686 272 565 108 990 728 952;
  • 4) 0.665 686 272 565 108 990 728 952 × 2 = 1 + 0.331 372 545 130 217 981 457 904;
  • 5) 0.331 372 545 130 217 981 457 904 × 2 = 0 + 0.662 745 090 260 435 962 915 808;
  • 6) 0.662 745 090 260 435 962 915 808 × 2 = 1 + 0.325 490 180 520 871 925 831 616;
  • 7) 0.325 490 180 520 871 925 831 616 × 2 = 0 + 0.650 980 361 041 743 851 663 232;
  • 8) 0.650 980 361 041 743 851 663 232 × 2 = 1 + 0.301 960 722 083 487 703 326 464;
  • 9) 0.301 960 722 083 487 703 326 464 × 2 = 0 + 0.603 921 444 166 975 406 652 928;
  • 10) 0.603 921 444 166 975 406 652 928 × 2 = 1 + 0.207 842 888 333 950 813 305 856;
  • 11) 0.207 842 888 333 950 813 305 856 × 2 = 0 + 0.415 685 776 667 901 626 611 712;
  • 12) 0.415 685 776 667 901 626 611 712 × 2 = 0 + 0.831 371 553 335 803 253 223 424;
  • 13) 0.831 371 553 335 803 253 223 424 × 2 = 1 + 0.662 743 106 671 606 506 446 848;
  • 14) 0.662 743 106 671 606 506 446 848 × 2 = 1 + 0.325 486 213 343 213 012 893 696;
  • 15) 0.325 486 213 343 213 012 893 696 × 2 = 0 + 0.650 972 426 686 426 025 787 392;
  • 16) 0.650 972 426 686 426 025 787 392 × 2 = 1 + 0.301 944 853 372 852 051 574 784;
  • 17) 0.301 944 853 372 852 051 574 784 × 2 = 0 + 0.603 889 706 745 704 103 149 568;
  • 18) 0.603 889 706 745 704 103 149 568 × 2 = 1 + 0.207 779 413 491 408 206 299 136;
  • 19) 0.207 779 413 491 408 206 299 136 × 2 = 0 + 0.415 558 826 982 816 412 598 272;
  • 20) 0.415 558 826 982 816 412 598 272 × 2 = 0 + 0.831 117 653 965 632 825 196 544;
  • 21) 0.831 117 653 965 632 825 196 544 × 2 = 1 + 0.662 235 307 931 265 650 393 088;
  • 22) 0.662 235 307 931 265 650 393 088 × 2 = 1 + 0.324 470 615 862 531 300 786 176;
  • 23) 0.324 470 615 862 531 300 786 176 × 2 = 0 + 0.648 941 231 725 062 601 572 352;
  • 24) 0.648 941 231 725 062 601 572 352 × 2 = 1 + 0.297 882 463 450 125 203 144 704;
  • 25) 0.297 882 463 450 125 203 144 704 × 2 = 0 + 0.595 764 926 900 250 406 289 408;
  • 26) 0.595 764 926 900 250 406 289 408 × 2 = 1 + 0.191 529 853 800 500 812 578 816;
  • 27) 0.191 529 853 800 500 812 578 816 × 2 = 0 + 0.383 059 707 601 001 625 157 632;
  • 28) 0.383 059 707 601 001 625 157 632 × 2 = 0 + 0.766 119 415 202 003 250 315 264;
  • 29) 0.766 119 415 202 003 250 315 264 × 2 = 1 + 0.532 238 830 404 006 500 630 528;
  • 30) 0.532 238 830 404 006 500 630 528 × 2 = 1 + 0.064 477 660 808 013 001 261 056;
  • 31) 0.064 477 660 808 013 001 261 056 × 2 = 0 + 0.128 955 321 616 026 002 522 112;
  • 32) 0.128 955 321 616 026 002 522 112 × 2 = 0 + 0.257 910 643 232 052 005 044 224;
  • 33) 0.257 910 643 232 052 005 044 224 × 2 = 0 + 0.515 821 286 464 104 010 088 448;
  • 34) 0.515 821 286 464 104 010 088 448 × 2 = 1 + 0.031 642 572 928 208 020 176 896;
  • 35) 0.031 642 572 928 208 020 176 896 × 2 = 0 + 0.063 285 145 856 416 040 353 792;
  • 36) 0.063 285 145 856 416 040 353 792 × 2 = 0 + 0.126 570 291 712 832 080 707 584;
  • 37) 0.126 570 291 712 832 080 707 584 × 2 = 0 + 0.253 140 583 425 664 161 415 168;
  • 38) 0.253 140 583 425 664 161 415 168 × 2 = 0 + 0.506 281 166 851 328 322 830 336;
  • 39) 0.506 281 166 851 328 322 830 336 × 2 = 1 + 0.012 562 333 702 656 645 660 672;
  • 40) 0.012 562 333 702 656 645 660 672 × 2 = 0 + 0.025 124 667 405 313 291 321 344;
  • 41) 0.025 124 667 405 313 291 321 344 × 2 = 0 + 0.050 249 334 810 626 582 642 688;
  • 42) 0.050 249 334 810 626 582 642 688 × 2 = 0 + 0.100 498 669 621 253 165 285 376;
  • 43) 0.100 498 669 621 253 165 285 376 × 2 = 0 + 0.200 997 339 242 506 330 570 752;
  • 44) 0.200 997 339 242 506 330 570 752 × 2 = 0 + 0.401 994 678 485 012 661 141 504;
  • 45) 0.401 994 678 485 012 661 141 504 × 2 = 0 + 0.803 989 356 970 025 322 283 008;
  • 46) 0.803 989 356 970 025 322 283 008 × 2 = 1 + 0.607 978 713 940 050 644 566 016;
  • 47) 0.607 978 713 940 050 644 566 016 × 2 = 1 + 0.215 957 427 880 101 289 132 032;
  • 48) 0.215 957 427 880 101 289 132 032 × 2 = 0 + 0.431 914 855 760 202 578 264 064;
  • 49) 0.431 914 855 760 202 578 264 064 × 2 = 0 + 0.863 829 711 520 405 156 528 128;
  • 50) 0.863 829 711 520 405 156 528 128 × 2 = 1 + 0.727 659 423 040 810 313 056 256;
  • 51) 0.727 659 423 040 810 313 056 256 × 2 = 1 + 0.455 318 846 081 620 626 112 512;
  • 52) 0.455 318 846 081 620 626 112 512 × 2 = 0 + 0.910 637 692 163 241 252 225 024;
  • 53) 0.910 637 692 163 241 252 225 024 × 2 = 1 + 0.821 275 384 326 482 504 450 048;

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:


0.833 210 784 070 638 623 841 119(10) =

0.1101 0101 0100 1101 0100 1101 0100 1100 0100 0010 0000 0110 0110 1(2)

5. Positive number before normalization:

6.833 210 784 070 638 623 841 119(10) =

110.1101 0101 0100 1101 0100 1101 0100 1100 0100 0010 0000 0110 0110 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:


6.833 210 784 070 638 623 841 119(10) =

110.1101 0101 0100 1101 0100 1101 0100 1100 0100 0010 0000 0110 0110 1(2) =

110.1101 0101 0100 1101 0100 1101 0100 1100 0100 0010 0000 0110 0110 1(2) × 20 =

1.1011 0101 0101 0011 0101 0011 0101 0011 0001 0000 1000 0001 1001 101(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 0101 0101 0011 0101 0011 0101 0011 0001 0000 1000 0001 1001 101


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 0101 0101 0011 0101 0011 0101 0011 0001 0000 1000 0001 1001 101 =

1011 0101 0101 0011 0101 0011 0101 0011 0001 0000 1000 0001 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 0001

Mantissa (52 bits) =
1011 0101 0101 0011 0101 0011 0101 0011 0001 0000 1000 0001 1001


Decimal number 6.833 210 784 070 638 623 841 119 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0001 - 1011 0101 0101 0011 0101 0011 0101 0011 0001 0000 1000 0001 1001

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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