1.745 459 324 169 999 826 281 696 186 923 84 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 923 84(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
1.745 459 324 169 999 826 281 696 186 923 84(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: 1.
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;
  • 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.745 459 324 169 999 826 281 696 186 923 84.

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) 0.745 459 324 169 999 826 281 696 186 923 84 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 847 68;
  • 2) 0.490 918 648 339 999 652 563 392 373 847 68 × 2 = 0 + 0.981 837 296 679 999 305 126 784 747 695 36;
  • 3) 0.981 837 296 679 999 305 126 784 747 695 36 × 2 = 1 + 0.963 674 593 359 998 610 253 569 495 390 72;
  • 4) 0.963 674 593 359 998 610 253 569 495 390 72 × 2 = 1 + 0.927 349 186 719 997 220 507 138 990 781 44;
  • 5) 0.927 349 186 719 997 220 507 138 990 781 44 × 2 = 1 + 0.854 698 373 439 994 441 014 277 981 562 88;
  • 6) 0.854 698 373 439 994 441 014 277 981 562 88 × 2 = 1 + 0.709 396 746 879 988 882 028 555 963 125 76;
  • 7) 0.709 396 746 879 988 882 028 555 963 125 76 × 2 = 1 + 0.418 793 493 759 977 764 057 111 926 251 52;
  • 8) 0.418 793 493 759 977 764 057 111 926 251 52 × 2 = 0 + 0.837 586 987 519 955 528 114 223 852 503 04;
  • 9) 0.837 586 987 519 955 528 114 223 852 503 04 × 2 = 1 + 0.675 173 975 039 911 056 228 447 705 006 08;
  • 10) 0.675 173 975 039 911 056 228 447 705 006 08 × 2 = 1 + 0.350 347 950 079 822 112 456 895 410 012 16;
  • 11) 0.350 347 950 079 822 112 456 895 410 012 16 × 2 = 0 + 0.700 695 900 159 644 224 913 790 820 024 32;
  • 12) 0.700 695 900 159 644 224 913 790 820 024 32 × 2 = 1 + 0.401 391 800 319 288 449 827 581 640 048 64;
  • 13) 0.401 391 800 319 288 449 827 581 640 048 64 × 2 = 0 + 0.802 783 600 638 576 899 655 163 280 097 28;
  • 14) 0.802 783 600 638 576 899 655 163 280 097 28 × 2 = 1 + 0.605 567 201 277 153 799 310 326 560 194 56;
  • 15) 0.605 567 201 277 153 799 310 326 560 194 56 × 2 = 1 + 0.211 134 402 554 307 598 620 653 120 389 12;
  • 16) 0.211 134 402 554 307 598 620 653 120 389 12 × 2 = 0 + 0.422 268 805 108 615 197 241 306 240 778 24;
  • 17) 0.422 268 805 108 615 197 241 306 240 778 24 × 2 = 0 + 0.844 537 610 217 230 394 482 612 481 556 48;
  • 18) 0.844 537 610 217 230 394 482 612 481 556 48 × 2 = 1 + 0.689 075 220 434 460 788 965 224 963 112 96;
  • 19) 0.689 075 220 434 460 788 965 224 963 112 96 × 2 = 1 + 0.378 150 440 868 921 577 930 449 926 225 92;
  • 20) 0.378 150 440 868 921 577 930 449 926 225 92 × 2 = 0 + 0.756 300 881 737 843 155 860 899 852 451 84;
  • 21) 0.756 300 881 737 843 155 860 899 852 451 84 × 2 = 1 + 0.512 601 763 475 686 311 721 799 704 903 68;
  • 22) 0.512 601 763 475 686 311 721 799 704 903 68 × 2 = 1 + 0.025 203 526 951 372 623 443 599 409 807 36;
  • 23) 0.025 203 526 951 372 623 443 599 409 807 36 × 2 = 0 + 0.050 407 053 902 745 246 887 198 819 614 72;
  • 24) 0.050 407 053 902 745 246 887 198 819 614 72 × 2 = 0 + 0.100 814 107 805 490 493 774 397 639 229 44;
  • 25) 0.100 814 107 805 490 493 774 397 639 229 44 × 2 = 0 + 0.201 628 215 610 980 987 548 795 278 458 88;
  • 26) 0.201 628 215 610 980 987 548 795 278 458 88 × 2 = 0 + 0.403 256 431 221 961 975 097 590 556 917 76;
  • 27) 0.403 256 431 221 961 975 097 590 556 917 76 × 2 = 0 + 0.806 512 862 443 923 950 195 181 113 835 52;
  • 28) 0.806 512 862 443 923 950 195 181 113 835 52 × 2 = 1 + 0.613 025 724 887 847 900 390 362 227 671 04;
  • 29) 0.613 025 724 887 847 900 390 362 227 671 04 × 2 = 1 + 0.226 051 449 775 695 800 780 724 455 342 08;
  • 30) 0.226 051 449 775 695 800 780 724 455 342 08 × 2 = 0 + 0.452 102 899 551 391 601 561 448 910 684 16;
  • 31) 0.452 102 899 551 391 601 561 448 910 684 16 × 2 = 0 + 0.904 205 799 102 783 203 122 897 821 368 32;
  • 32) 0.904 205 799 102 783 203 122 897 821 368 32 × 2 = 1 + 0.808 411 598 205 566 406 245 795 642 736 64;
  • 33) 0.808 411 598 205 566 406 245 795 642 736 64 × 2 = 1 + 0.616 823 196 411 132 812 491 591 285 473 28;
  • 34) 0.616 823 196 411 132 812 491 591 285 473 28 × 2 = 1 + 0.233 646 392 822 265 624 983 182 570 946 56;
  • 35) 0.233 646 392 822 265 624 983 182 570 946 56 × 2 = 0 + 0.467 292 785 644 531 249 966 365 141 893 12;
  • 36) 0.467 292 785 644 531 249 966 365 141 893 12 × 2 = 0 + 0.934 585 571 289 062 499 932 730 283 786 24;
  • 37) 0.934 585 571 289 062 499 932 730 283 786 24 × 2 = 1 + 0.869 171 142 578 124 999 865 460 567 572 48;
  • 38) 0.869 171 142 578 124 999 865 460 567 572 48 × 2 = 1 + 0.738 342 285 156 249 999 730 921 135 144 96;
  • 39) 0.738 342 285 156 249 999 730 921 135 144 96 × 2 = 1 + 0.476 684 570 312 499 999 461 842 270 289 92;
  • 40) 0.476 684 570 312 499 999 461 842 270 289 92 × 2 = 0 + 0.953 369 140 624 999 998 923 684 540 579 84;
  • 41) 0.953 369 140 624 999 998 923 684 540 579 84 × 2 = 1 + 0.906 738 281 249 999 997 847 369 081 159 68;
  • 42) 0.906 738 281 249 999 997 847 369 081 159 68 × 2 = 1 + 0.813 476 562 499 999 995 694 738 162 319 36;
  • 43) 0.813 476 562 499 999 995 694 738 162 319 36 × 2 = 1 + 0.626 953 124 999 999 991 389 476 324 638 72;
  • 44) 0.626 953 124 999 999 991 389 476 324 638 72 × 2 = 1 + 0.253 906 249 999 999 982 778 952 649 277 44;
  • 45) 0.253 906 249 999 999 982 778 952 649 277 44 × 2 = 0 + 0.507 812 499 999 999 965 557 905 298 554 88;
  • 46) 0.507 812 499 999 999 965 557 905 298 554 88 × 2 = 1 + 0.015 624 999 999 999 931 115 810 597 109 76;
  • 47) 0.015 624 999 999 999 931 115 810 597 109 76 × 2 = 0 + 0.031 249 999 999 999 862 231 621 194 219 52;
  • 48) 0.031 249 999 999 999 862 231 621 194 219 52 × 2 = 0 + 0.062 499 999 999 999 724 463 242 388 439 04;
  • 49) 0.062 499 999 999 999 724 463 242 388 439 04 × 2 = 0 + 0.124 999 999 999 999 448 926 484 776 878 08;
  • 50) 0.124 999 999 999 999 448 926 484 776 878 08 × 2 = 0 + 0.249 999 999 999 998 897 852 969 553 756 16;
  • 51) 0.249 999 999 999 998 897 852 969 553 756 16 × 2 = 0 + 0.499 999 999 999 997 795 705 939 107 512 32;
  • 52) 0.499 999 999 999 997 795 705 939 107 512 32 × 2 = 0 + 0.999 999 999 999 995 591 411 878 215 024 64;
  • 53) 0.999 999 999 999 995 591 411 878 215 024 64 × 2 = 1 + 0.999 999 999 999 991 182 823 756 430 049 28;

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:


0.745 459 324 169 999 826 281 696 186 923 84(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 923 84(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.745 459 324 169 999 826 281 696 186 923 84(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1(2) × 20


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): 0

Mantissa (not normalized):
1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

0 + 2(11-1) - 1 =

(0 + 1 023)(10) =

1 023(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;
  • 1 023 ÷ 2 = 511 + 1;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

1023(10) =

011 1111 1111(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 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000 1 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


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) =
011 1111 1111

Mantissa (52 bits) =
1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000


Decimal number 1.745 459 324 169 999 826 281 696 186 923 84 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0000

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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