1.745 459 324 169 999 826 281 696 186 924 818 748 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 924 818 748(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
1.745 459 324 169 999 826 281 696 186 924 818 748(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: 0.745 459 324 169 999 826 281 696 186 924 818 748.

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) 0.745 459 324 169 999 826 281 696 186 924 818 748 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 849 637 496;
  • 2) 0.490 918 648 339 999 652 563 392 373 849 637 496 × 2 = 0 + 0.981 837 296 679 999 305 126 784 747 699 274 992;
  • 3) 0.981 837 296 679 999 305 126 784 747 699 274 992 × 2 = 1 + 0.963 674 593 359 998 610 253 569 495 398 549 984;
  • 4) 0.963 674 593 359 998 610 253 569 495 398 549 984 × 2 = 1 + 0.927 349 186 719 997 220 507 138 990 797 099 968;
  • 5) 0.927 349 186 719 997 220 507 138 990 797 099 968 × 2 = 1 + 0.854 698 373 439 994 441 014 277 981 594 199 936;
  • 6) 0.854 698 373 439 994 441 014 277 981 594 199 936 × 2 = 1 + 0.709 396 746 879 988 882 028 555 963 188 399 872;
  • 7) 0.709 396 746 879 988 882 028 555 963 188 399 872 × 2 = 1 + 0.418 793 493 759 977 764 057 111 926 376 799 744;
  • 8) 0.418 793 493 759 977 764 057 111 926 376 799 744 × 2 = 0 + 0.837 586 987 519 955 528 114 223 852 753 599 488;
  • 9) 0.837 586 987 519 955 528 114 223 852 753 599 488 × 2 = 1 + 0.675 173 975 039 911 056 228 447 705 507 198 976;
  • 10) 0.675 173 975 039 911 056 228 447 705 507 198 976 × 2 = 1 + 0.350 347 950 079 822 112 456 895 411 014 397 952;
  • 11) 0.350 347 950 079 822 112 456 895 411 014 397 952 × 2 = 0 + 0.700 695 900 159 644 224 913 790 822 028 795 904;
  • 12) 0.700 695 900 159 644 224 913 790 822 028 795 904 × 2 = 1 + 0.401 391 800 319 288 449 827 581 644 057 591 808;
  • 13) 0.401 391 800 319 288 449 827 581 644 057 591 808 × 2 = 0 + 0.802 783 600 638 576 899 655 163 288 115 183 616;
  • 14) 0.802 783 600 638 576 899 655 163 288 115 183 616 × 2 = 1 + 0.605 567 201 277 153 799 310 326 576 230 367 232;
  • 15) 0.605 567 201 277 153 799 310 326 576 230 367 232 × 2 = 1 + 0.211 134 402 554 307 598 620 653 152 460 734 464;
  • 16) 0.211 134 402 554 307 598 620 653 152 460 734 464 × 2 = 0 + 0.422 268 805 108 615 197 241 306 304 921 468 928;
  • 17) 0.422 268 805 108 615 197 241 306 304 921 468 928 × 2 = 0 + 0.844 537 610 217 230 394 482 612 609 842 937 856;
  • 18) 0.844 537 610 217 230 394 482 612 609 842 937 856 × 2 = 1 + 0.689 075 220 434 460 788 965 225 219 685 875 712;
  • 19) 0.689 075 220 434 460 788 965 225 219 685 875 712 × 2 = 1 + 0.378 150 440 868 921 577 930 450 439 371 751 424;
  • 20) 0.378 150 440 868 921 577 930 450 439 371 751 424 × 2 = 0 + 0.756 300 881 737 843 155 860 900 878 743 502 848;
  • 21) 0.756 300 881 737 843 155 860 900 878 743 502 848 × 2 = 1 + 0.512 601 763 475 686 311 721 801 757 487 005 696;
  • 22) 0.512 601 763 475 686 311 721 801 757 487 005 696 × 2 = 1 + 0.025 203 526 951 372 623 443 603 514 974 011 392;
  • 23) 0.025 203 526 951 372 623 443 603 514 974 011 392 × 2 = 0 + 0.050 407 053 902 745 246 887 207 029 948 022 784;
  • 24) 0.050 407 053 902 745 246 887 207 029 948 022 784 × 2 = 0 + 0.100 814 107 805 490 493 774 414 059 896 045 568;
  • 25) 0.100 814 107 805 490 493 774 414 059 896 045 568 × 2 = 0 + 0.201 628 215 610 980 987 548 828 119 792 091 136;
  • 26) 0.201 628 215 610 980 987 548 828 119 792 091 136 × 2 = 0 + 0.403 256 431 221 961 975 097 656 239 584 182 272;
  • 27) 0.403 256 431 221 961 975 097 656 239 584 182 272 × 2 = 0 + 0.806 512 862 443 923 950 195 312 479 168 364 544;
  • 28) 0.806 512 862 443 923 950 195 312 479 168 364 544 × 2 = 1 + 0.613 025 724 887 847 900 390 624 958 336 729 088;
  • 29) 0.613 025 724 887 847 900 390 624 958 336 729 088 × 2 = 1 + 0.226 051 449 775 695 800 781 249 916 673 458 176;
  • 30) 0.226 051 449 775 695 800 781 249 916 673 458 176 × 2 = 0 + 0.452 102 899 551 391 601 562 499 833 346 916 352;
  • 31) 0.452 102 899 551 391 601 562 499 833 346 916 352 × 2 = 0 + 0.904 205 799 102 783 203 124 999 666 693 832 704;
  • 32) 0.904 205 799 102 783 203 124 999 666 693 832 704 × 2 = 1 + 0.808 411 598 205 566 406 249 999 333 387 665 408;
  • 33) 0.808 411 598 205 566 406 249 999 333 387 665 408 × 2 = 1 + 0.616 823 196 411 132 812 499 998 666 775 330 816;
  • 34) 0.616 823 196 411 132 812 499 998 666 775 330 816 × 2 = 1 + 0.233 646 392 822 265 624 999 997 333 550 661 632;
  • 35) 0.233 646 392 822 265 624 999 997 333 550 661 632 × 2 = 0 + 0.467 292 785 644 531 249 999 994 667 101 323 264;
  • 36) 0.467 292 785 644 531 249 999 994 667 101 323 264 × 2 = 0 + 0.934 585 571 289 062 499 999 989 334 202 646 528;
  • 37) 0.934 585 571 289 062 499 999 989 334 202 646 528 × 2 = 1 + 0.869 171 142 578 124 999 999 978 668 405 293 056;
  • 38) 0.869 171 142 578 124 999 999 978 668 405 293 056 × 2 = 1 + 0.738 342 285 156 249 999 999 957 336 810 586 112;
  • 39) 0.738 342 285 156 249 999 999 957 336 810 586 112 × 2 = 1 + 0.476 684 570 312 499 999 999 914 673 621 172 224;
  • 40) 0.476 684 570 312 499 999 999 914 673 621 172 224 × 2 = 0 + 0.953 369 140 624 999 999 999 829 347 242 344 448;
  • 41) 0.953 369 140 624 999 999 999 829 347 242 344 448 × 2 = 1 + 0.906 738 281 249 999 999 999 658 694 484 688 896;
  • 42) 0.906 738 281 249 999 999 999 658 694 484 688 896 × 2 = 1 + 0.813 476 562 499 999 999 999 317 388 969 377 792;
  • 43) 0.813 476 562 499 999 999 999 317 388 969 377 792 × 2 = 1 + 0.626 953 124 999 999 999 998 634 777 938 755 584;
  • 44) 0.626 953 124 999 999 999 998 634 777 938 755 584 × 2 = 1 + 0.253 906 249 999 999 999 997 269 555 877 511 168;
  • 45) 0.253 906 249 999 999 999 997 269 555 877 511 168 × 2 = 0 + 0.507 812 499 999 999 999 994 539 111 755 022 336;
  • 46) 0.507 812 499 999 999 999 994 539 111 755 022 336 × 2 = 1 + 0.015 624 999 999 999 999 989 078 223 510 044 672;
  • 47) 0.015 624 999 999 999 999 989 078 223 510 044 672 × 2 = 0 + 0.031 249 999 999 999 999 978 156 447 020 089 344;
  • 48) 0.031 249 999 999 999 999 978 156 447 020 089 344 × 2 = 0 + 0.062 499 999 999 999 999 956 312 894 040 178 688;
  • 49) 0.062 499 999 999 999 999 956 312 894 040 178 688 × 2 = 0 + 0.124 999 999 999 999 999 912 625 788 080 357 376;
  • 50) 0.124 999 999 999 999 999 912 625 788 080 357 376 × 2 = 0 + 0.249 999 999 999 999 999 825 251 576 160 714 752;
  • 51) 0.249 999 999 999 999 999 825 251 576 160 714 752 × 2 = 0 + 0.499 999 999 999 999 999 650 503 152 321 429 504;
  • 52) 0.499 999 999 999 999 999 650 503 152 321 429 504 × 2 = 0 + 0.999 999 999 999 999 999 301 006 304 642 859 008;
  • 53) 0.999 999 999 999 999 999 301 006 304 642 859 008 × 2 = 1 + 0.999 999 999 999 999 998 602 012 609 285 718 016;

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:


0.745 459 324 169 999 826 281 696 186 924 818 748(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 924 818 748(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:


1.745 459 324 169 999 826 281 696 186 924 818 748(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 1.745 459 324 169 999 826 281 696 186 924 818 748 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