33.780 086 699 999 998 245 402 821 339 632 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 33.780 086 699 999 998 245 402 821 339 632 8(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
33.780 086 699 999 998 245 402 821 339 632 8(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: 33.
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;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 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.

33(10) =

10 0001(2)


3. Convert to binary (base 2) the fractional part: 0.780 086 699 999 998 245 402 821 339 632 8.

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) 0.780 086 699 999 998 245 402 821 339 632 8 × 2 = 1 + 0.560 173 399 999 996 490 805 642 679 265 6;
  • 2) 0.560 173 399 999 996 490 805 642 679 265 6 × 2 = 1 + 0.120 346 799 999 992 981 611 285 358 531 2;
  • 3) 0.120 346 799 999 992 981 611 285 358 531 2 × 2 = 0 + 0.240 693 599 999 985 963 222 570 717 062 4;
  • 4) 0.240 693 599 999 985 963 222 570 717 062 4 × 2 = 0 + 0.481 387 199 999 971 926 445 141 434 124 8;
  • 5) 0.481 387 199 999 971 926 445 141 434 124 8 × 2 = 0 + 0.962 774 399 999 943 852 890 282 868 249 6;
  • 6) 0.962 774 399 999 943 852 890 282 868 249 6 × 2 = 1 + 0.925 548 799 999 887 705 780 565 736 499 2;
  • 7) 0.925 548 799 999 887 705 780 565 736 499 2 × 2 = 1 + 0.851 097 599 999 775 411 561 131 472 998 4;
  • 8) 0.851 097 599 999 775 411 561 131 472 998 4 × 2 = 1 + 0.702 195 199 999 550 823 122 262 945 996 8;
  • 9) 0.702 195 199 999 550 823 122 262 945 996 8 × 2 = 1 + 0.404 390 399 999 101 646 244 525 891 993 6;
  • 10) 0.404 390 399 999 101 646 244 525 891 993 6 × 2 = 0 + 0.808 780 799 998 203 292 489 051 783 987 2;
  • 11) 0.808 780 799 998 203 292 489 051 783 987 2 × 2 = 1 + 0.617 561 599 996 406 584 978 103 567 974 4;
  • 12) 0.617 561 599 996 406 584 978 103 567 974 4 × 2 = 1 + 0.235 123 199 992 813 169 956 207 135 948 8;
  • 13) 0.235 123 199 992 813 169 956 207 135 948 8 × 2 = 0 + 0.470 246 399 985 626 339 912 414 271 897 6;
  • 14) 0.470 246 399 985 626 339 912 414 271 897 6 × 2 = 0 + 0.940 492 799 971 252 679 824 828 543 795 2;
  • 15) 0.940 492 799 971 252 679 824 828 543 795 2 × 2 = 1 + 0.880 985 599 942 505 359 649 657 087 590 4;
  • 16) 0.880 985 599 942 505 359 649 657 087 590 4 × 2 = 1 + 0.761 971 199 885 010 719 299 314 175 180 8;
  • 17) 0.761 971 199 885 010 719 299 314 175 180 8 × 2 = 1 + 0.523 942 399 770 021 438 598 628 350 361 6;
  • 18) 0.523 942 399 770 021 438 598 628 350 361 6 × 2 = 1 + 0.047 884 799 540 042 877 197 256 700 723 2;
  • 19) 0.047 884 799 540 042 877 197 256 700 723 2 × 2 = 0 + 0.095 769 599 080 085 754 394 513 401 446 4;
  • 20) 0.095 769 599 080 085 754 394 513 401 446 4 × 2 = 0 + 0.191 539 198 160 171 508 789 026 802 892 8;
  • 21) 0.191 539 198 160 171 508 789 026 802 892 8 × 2 = 0 + 0.383 078 396 320 343 017 578 053 605 785 6;
  • 22) 0.383 078 396 320 343 017 578 053 605 785 6 × 2 = 0 + 0.766 156 792 640 686 035 156 107 211 571 2;
  • 23) 0.766 156 792 640 686 035 156 107 211 571 2 × 2 = 1 + 0.532 313 585 281 372 070 312 214 423 142 4;
  • 24) 0.532 313 585 281 372 070 312 214 423 142 4 × 2 = 1 + 0.064 627 170 562 744 140 624 428 846 284 8;
  • 25) 0.064 627 170 562 744 140 624 428 846 284 8 × 2 = 0 + 0.129 254 341 125 488 281 248 857 692 569 6;
  • 26) 0.129 254 341 125 488 281 248 857 692 569 6 × 2 = 0 + 0.258 508 682 250 976 562 497 715 385 139 2;
  • 27) 0.258 508 682 250 976 562 497 715 385 139 2 × 2 = 0 + 0.517 017 364 501 953 124 995 430 770 278 4;
  • 28) 0.517 017 364 501 953 124 995 430 770 278 4 × 2 = 1 + 0.034 034 729 003 906 249 990 861 540 556 8;
  • 29) 0.034 034 729 003 906 249 990 861 540 556 8 × 2 = 0 + 0.068 069 458 007 812 499 981 723 081 113 6;
  • 30) 0.068 069 458 007 812 499 981 723 081 113 6 × 2 = 0 + 0.136 138 916 015 624 999 963 446 162 227 2;
  • 31) 0.136 138 916 015 624 999 963 446 162 227 2 × 2 = 0 + 0.272 277 832 031 249 999 926 892 324 454 4;
  • 32) 0.272 277 832 031 249 999 926 892 324 454 4 × 2 = 0 + 0.544 555 664 062 499 999 853 784 648 908 8;
  • 33) 0.544 555 664 062 499 999 853 784 648 908 8 × 2 = 1 + 0.089 111 328 124 999 999 707 569 297 817 6;
  • 34) 0.089 111 328 124 999 999 707 569 297 817 6 × 2 = 0 + 0.178 222 656 249 999 999 415 138 595 635 2;
  • 35) 0.178 222 656 249 999 999 415 138 595 635 2 × 2 = 0 + 0.356 445 312 499 999 998 830 277 191 270 4;
  • 36) 0.356 445 312 499 999 998 830 277 191 270 4 × 2 = 0 + 0.712 890 624 999 999 997 660 554 382 540 8;
  • 37) 0.712 890 624 999 999 997 660 554 382 540 8 × 2 = 1 + 0.425 781 249 999 999 995 321 108 765 081 6;
  • 38) 0.425 781 249 999 999 995 321 108 765 081 6 × 2 = 0 + 0.851 562 499 999 999 990 642 217 530 163 2;
  • 39) 0.851 562 499 999 999 990 642 217 530 163 2 × 2 = 1 + 0.703 124 999 999 999 981 284 435 060 326 4;
  • 40) 0.703 124 999 999 999 981 284 435 060 326 4 × 2 = 1 + 0.406 249 999 999 999 962 568 870 120 652 8;
  • 41) 0.406 249 999 999 999 962 568 870 120 652 8 × 2 = 0 + 0.812 499 999 999 999 925 137 740 241 305 6;
  • 42) 0.812 499 999 999 999 925 137 740 241 305 6 × 2 = 1 + 0.624 999 999 999 999 850 275 480 482 611 2;
  • 43) 0.624 999 999 999 999 850 275 480 482 611 2 × 2 = 1 + 0.249 999 999 999 999 700 550 960 965 222 4;
  • 44) 0.249 999 999 999 999 700 550 960 965 222 4 × 2 = 0 + 0.499 999 999 999 999 401 101 921 930 444 8;
  • 45) 0.499 999 999 999 999 401 101 921 930 444 8 × 2 = 0 + 0.999 999 999 999 998 802 203 843 860 889 6;
  • 46) 0.999 999 999 999 998 802 203 843 860 889 6 × 2 = 1 + 0.999 999 999 999 997 604 407 687 721 779 2;
  • 47) 0.999 999 999 999 997 604 407 687 721 779 2 × 2 = 1 + 0.999 999 999 999 995 208 815 375 443 558 4;
  • 48) 0.999 999 999 999 995 208 815 375 443 558 4 × 2 = 1 + 0.999 999 999 999 990 417 630 750 887 116 8;
  • 49) 0.999 999 999 999 990 417 630 750 887 116 8 × 2 = 1 + 0.999 999 999 999 980 835 261 501 774 233 6;
  • 50) 0.999 999 999 999 980 835 261 501 774 233 6 × 2 = 1 + 0.999 999 999 999 961 670 523 003 548 467 2;
  • 51) 0.999 999 999 999 961 670 523 003 548 467 2 × 2 = 1 + 0.999 999 999 999 923 341 046 007 096 934 4;
  • 52) 0.999 999 999 999 923 341 046 007 096 934 4 × 2 = 1 + 0.999 999 999 999 846 682 092 014 193 868 8;
  • 53) 0.999 999 999 999 846 682 092 014 193 868 8 × 2 = 1 + 0.999 999 999 999 693 364 184 028 387 737 6;

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:


0.780 086 699 999 998 245 402 821 339 632 8(10) =

0.1100 0111 1011 0011 1100 0011 0001 0000 1000 1011 0110 0111 1111 1(2)

5. Positive number before normalization:

33.780 086 699 999 998 245 402 821 339 632 8(10) =

10 0001.1100 0111 1011 0011 1100 0011 0001 0000 1000 1011 0110 0111 1111 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


33.780 086 699 999 998 245 402 821 339 632 8(10) =

10 0001.1100 0111 1011 0011 1100 0011 0001 0000 1000 1011 0110 0111 1111 1(2) =

10 0001.1100 0111 1011 0011 1100 0011 0001 0000 1000 1011 0110 0111 1111 1(2) × 20 =

1.0000 1110 0011 1101 1001 1110 0001 1000 1000 0100 0101 1011 0011 1111 11(2) × 25


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): 5

Mantissa (not normalized):
1.0000 1110 0011 1101 1001 1110 0001 1000 1000 0100 0101 1011 0011 1111 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

5 + 2(11-1) - 1 =

(5 + 1 023)(10) =

1 028(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;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 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) =

1028(10) =

100 0000 0100(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. 0000 1110 0011 1101 1001 1110 0001 1000 1000 0100 0101 1011 0011 11 1111 =

0000 1110 0011 1101 1001 1110 0001 1000 1000 0100 0101 1011 0011


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 0100

Mantissa (52 bits) =
0000 1110 0011 1101 1001 1110 0001 1000 1000 0100 0101 1011 0011


Decimal number 33.780 086 699 999 998 245 402 821 339 632 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0100 - 0000 1110 0011 1101 1001 1110 0001 1000 1000 0100 0101 1011 0011

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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