1.745 459 324 169 999 826 281 696 186 924 734 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 924 734(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
1.745 459 324 169 999 826 281 696 186 924 734(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: 0.745 459 324 169 999 826 281 696 186 924 734.

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) 0.745 459 324 169 999 826 281 696 186 924 734 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 849 468;
  • 2) 0.490 918 648 339 999 652 563 392 373 849 468 × 2 = 0 + 0.981 837 296 679 999 305 126 784 747 698 936;
  • 3) 0.981 837 296 679 999 305 126 784 747 698 936 × 2 = 1 + 0.963 674 593 359 998 610 253 569 495 397 872;
  • 4) 0.963 674 593 359 998 610 253 569 495 397 872 × 2 = 1 + 0.927 349 186 719 997 220 507 138 990 795 744;
  • 5) 0.927 349 186 719 997 220 507 138 990 795 744 × 2 = 1 + 0.854 698 373 439 994 441 014 277 981 591 488;
  • 6) 0.854 698 373 439 994 441 014 277 981 591 488 × 2 = 1 + 0.709 396 746 879 988 882 028 555 963 182 976;
  • 7) 0.709 396 746 879 988 882 028 555 963 182 976 × 2 = 1 + 0.418 793 493 759 977 764 057 111 926 365 952;
  • 8) 0.418 793 493 759 977 764 057 111 926 365 952 × 2 = 0 + 0.837 586 987 519 955 528 114 223 852 731 904;
  • 9) 0.837 586 987 519 955 528 114 223 852 731 904 × 2 = 1 + 0.675 173 975 039 911 056 228 447 705 463 808;
  • 10) 0.675 173 975 039 911 056 228 447 705 463 808 × 2 = 1 + 0.350 347 950 079 822 112 456 895 410 927 616;
  • 11) 0.350 347 950 079 822 112 456 895 410 927 616 × 2 = 0 + 0.700 695 900 159 644 224 913 790 821 855 232;
  • 12) 0.700 695 900 159 644 224 913 790 821 855 232 × 2 = 1 + 0.401 391 800 319 288 449 827 581 643 710 464;
  • 13) 0.401 391 800 319 288 449 827 581 643 710 464 × 2 = 0 + 0.802 783 600 638 576 899 655 163 287 420 928;
  • 14) 0.802 783 600 638 576 899 655 163 287 420 928 × 2 = 1 + 0.605 567 201 277 153 799 310 326 574 841 856;
  • 15) 0.605 567 201 277 153 799 310 326 574 841 856 × 2 = 1 + 0.211 134 402 554 307 598 620 653 149 683 712;
  • 16) 0.211 134 402 554 307 598 620 653 149 683 712 × 2 = 0 + 0.422 268 805 108 615 197 241 306 299 367 424;
  • 17) 0.422 268 805 108 615 197 241 306 299 367 424 × 2 = 0 + 0.844 537 610 217 230 394 482 612 598 734 848;
  • 18) 0.844 537 610 217 230 394 482 612 598 734 848 × 2 = 1 + 0.689 075 220 434 460 788 965 225 197 469 696;
  • 19) 0.689 075 220 434 460 788 965 225 197 469 696 × 2 = 1 + 0.378 150 440 868 921 577 930 450 394 939 392;
  • 20) 0.378 150 440 868 921 577 930 450 394 939 392 × 2 = 0 + 0.756 300 881 737 843 155 860 900 789 878 784;
  • 21) 0.756 300 881 737 843 155 860 900 789 878 784 × 2 = 1 + 0.512 601 763 475 686 311 721 801 579 757 568;
  • 22) 0.512 601 763 475 686 311 721 801 579 757 568 × 2 = 1 + 0.025 203 526 951 372 623 443 603 159 515 136;
  • 23) 0.025 203 526 951 372 623 443 603 159 515 136 × 2 = 0 + 0.050 407 053 902 745 246 887 206 319 030 272;
  • 24) 0.050 407 053 902 745 246 887 206 319 030 272 × 2 = 0 + 0.100 814 107 805 490 493 774 412 638 060 544;
  • 25) 0.100 814 107 805 490 493 774 412 638 060 544 × 2 = 0 + 0.201 628 215 610 980 987 548 825 276 121 088;
  • 26) 0.201 628 215 610 980 987 548 825 276 121 088 × 2 = 0 + 0.403 256 431 221 961 975 097 650 552 242 176;
  • 27) 0.403 256 431 221 961 975 097 650 552 242 176 × 2 = 0 + 0.806 512 862 443 923 950 195 301 104 484 352;
  • 28) 0.806 512 862 443 923 950 195 301 104 484 352 × 2 = 1 + 0.613 025 724 887 847 900 390 602 208 968 704;
  • 29) 0.613 025 724 887 847 900 390 602 208 968 704 × 2 = 1 + 0.226 051 449 775 695 800 781 204 417 937 408;
  • 30) 0.226 051 449 775 695 800 781 204 417 937 408 × 2 = 0 + 0.452 102 899 551 391 601 562 408 835 874 816;
  • 31) 0.452 102 899 551 391 601 562 408 835 874 816 × 2 = 0 + 0.904 205 799 102 783 203 124 817 671 749 632;
  • 32) 0.904 205 799 102 783 203 124 817 671 749 632 × 2 = 1 + 0.808 411 598 205 566 406 249 635 343 499 264;
  • 33) 0.808 411 598 205 566 406 249 635 343 499 264 × 2 = 1 + 0.616 823 196 411 132 812 499 270 686 998 528;
  • 34) 0.616 823 196 411 132 812 499 270 686 998 528 × 2 = 1 + 0.233 646 392 822 265 624 998 541 373 997 056;
  • 35) 0.233 646 392 822 265 624 998 541 373 997 056 × 2 = 0 + 0.467 292 785 644 531 249 997 082 747 994 112;
  • 36) 0.467 292 785 644 531 249 997 082 747 994 112 × 2 = 0 + 0.934 585 571 289 062 499 994 165 495 988 224;
  • 37) 0.934 585 571 289 062 499 994 165 495 988 224 × 2 = 1 + 0.869 171 142 578 124 999 988 330 991 976 448;
  • 38) 0.869 171 142 578 124 999 988 330 991 976 448 × 2 = 1 + 0.738 342 285 156 249 999 976 661 983 952 896;
  • 39) 0.738 342 285 156 249 999 976 661 983 952 896 × 2 = 1 + 0.476 684 570 312 499 999 953 323 967 905 792;
  • 40) 0.476 684 570 312 499 999 953 323 967 905 792 × 2 = 0 + 0.953 369 140 624 999 999 906 647 935 811 584;
  • 41) 0.953 369 140 624 999 999 906 647 935 811 584 × 2 = 1 + 0.906 738 281 249 999 999 813 295 871 623 168;
  • 42) 0.906 738 281 249 999 999 813 295 871 623 168 × 2 = 1 + 0.813 476 562 499 999 999 626 591 743 246 336;
  • 43) 0.813 476 562 499 999 999 626 591 743 246 336 × 2 = 1 + 0.626 953 124 999 999 999 253 183 486 492 672;
  • 44) 0.626 953 124 999 999 999 253 183 486 492 672 × 2 = 1 + 0.253 906 249 999 999 998 506 366 972 985 344;
  • 45) 0.253 906 249 999 999 998 506 366 972 985 344 × 2 = 0 + 0.507 812 499 999 999 997 012 733 945 970 688;
  • 46) 0.507 812 499 999 999 997 012 733 945 970 688 × 2 = 1 + 0.015 624 999 999 999 994 025 467 891 941 376;
  • 47) 0.015 624 999 999 999 994 025 467 891 941 376 × 2 = 0 + 0.031 249 999 999 999 988 050 935 783 882 752;
  • 48) 0.031 249 999 999 999 988 050 935 783 882 752 × 2 = 0 + 0.062 499 999 999 999 976 101 871 567 765 504;
  • 49) 0.062 499 999 999 999 976 101 871 567 765 504 × 2 = 0 + 0.124 999 999 999 999 952 203 743 135 531 008;
  • 50) 0.124 999 999 999 999 952 203 743 135 531 008 × 2 = 0 + 0.249 999 999 999 999 904 407 486 271 062 016;
  • 51) 0.249 999 999 999 999 904 407 486 271 062 016 × 2 = 0 + 0.499 999 999 999 999 808 814 972 542 124 032;
  • 52) 0.499 999 999 999 999 808 814 972 542 124 032 × 2 = 0 + 0.999 999 999 999 999 617 629 945 084 248 064;
  • 53) 0.999 999 999 999 999 617 629 945 084 248 064 × 2 = 1 + 0.999 999 999 999 999 235 259 890 168 496 128;

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:


0.745 459 324 169 999 826 281 696 186 924 734(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 924 734(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:


1.745 459 324 169 999 826 281 696 186 924 734(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 1.745 459 324 169 999 826 281 696 186 924 734 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