1.745 459 324 169 999 826 281 696 186 924 818 903 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 924 818 903 19(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
1.745 459 324 169 999 826 281 696 186 924 818 903 19(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: 0.745 459 324 169 999 826 281 696 186 924 818 903 19.

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) 0.745 459 324 169 999 826 281 696 186 924 818 903 19 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 849 637 806 38;
  • 2) 0.490 918 648 339 999 652 563 392 373 849 637 806 38 × 2 = 0 + 0.981 837 296 679 999 305 126 784 747 699 275 612 76;
  • 3) 0.981 837 296 679 999 305 126 784 747 699 275 612 76 × 2 = 1 + 0.963 674 593 359 998 610 253 569 495 398 551 225 52;
  • 4) 0.963 674 593 359 998 610 253 569 495 398 551 225 52 × 2 = 1 + 0.927 349 186 719 997 220 507 138 990 797 102 451 04;
  • 5) 0.927 349 186 719 997 220 507 138 990 797 102 451 04 × 2 = 1 + 0.854 698 373 439 994 441 014 277 981 594 204 902 08;
  • 6) 0.854 698 373 439 994 441 014 277 981 594 204 902 08 × 2 = 1 + 0.709 396 746 879 988 882 028 555 963 188 409 804 16;
  • 7) 0.709 396 746 879 988 882 028 555 963 188 409 804 16 × 2 = 1 + 0.418 793 493 759 977 764 057 111 926 376 819 608 32;
  • 8) 0.418 793 493 759 977 764 057 111 926 376 819 608 32 × 2 = 0 + 0.837 586 987 519 955 528 114 223 852 753 639 216 64;
  • 9) 0.837 586 987 519 955 528 114 223 852 753 639 216 64 × 2 = 1 + 0.675 173 975 039 911 056 228 447 705 507 278 433 28;
  • 10) 0.675 173 975 039 911 056 228 447 705 507 278 433 28 × 2 = 1 + 0.350 347 950 079 822 112 456 895 411 014 556 866 56;
  • 11) 0.350 347 950 079 822 112 456 895 411 014 556 866 56 × 2 = 0 + 0.700 695 900 159 644 224 913 790 822 029 113 733 12;
  • 12) 0.700 695 900 159 644 224 913 790 822 029 113 733 12 × 2 = 1 + 0.401 391 800 319 288 449 827 581 644 058 227 466 24;
  • 13) 0.401 391 800 319 288 449 827 581 644 058 227 466 24 × 2 = 0 + 0.802 783 600 638 576 899 655 163 288 116 454 932 48;
  • 14) 0.802 783 600 638 576 899 655 163 288 116 454 932 48 × 2 = 1 + 0.605 567 201 277 153 799 310 326 576 232 909 864 96;
  • 15) 0.605 567 201 277 153 799 310 326 576 232 909 864 96 × 2 = 1 + 0.211 134 402 554 307 598 620 653 152 465 819 729 92;
  • 16) 0.211 134 402 554 307 598 620 653 152 465 819 729 92 × 2 = 0 + 0.422 268 805 108 615 197 241 306 304 931 639 459 84;
  • 17) 0.422 268 805 108 615 197 241 306 304 931 639 459 84 × 2 = 0 + 0.844 537 610 217 230 394 482 612 609 863 278 919 68;
  • 18) 0.844 537 610 217 230 394 482 612 609 863 278 919 68 × 2 = 1 + 0.689 075 220 434 460 788 965 225 219 726 557 839 36;
  • 19) 0.689 075 220 434 460 788 965 225 219 726 557 839 36 × 2 = 1 + 0.378 150 440 868 921 577 930 450 439 453 115 678 72;
  • 20) 0.378 150 440 868 921 577 930 450 439 453 115 678 72 × 2 = 0 + 0.756 300 881 737 843 155 860 900 878 906 231 357 44;
  • 21) 0.756 300 881 737 843 155 860 900 878 906 231 357 44 × 2 = 1 + 0.512 601 763 475 686 311 721 801 757 812 462 714 88;
  • 22) 0.512 601 763 475 686 311 721 801 757 812 462 714 88 × 2 = 1 + 0.025 203 526 951 372 623 443 603 515 624 925 429 76;
  • 23) 0.025 203 526 951 372 623 443 603 515 624 925 429 76 × 2 = 0 + 0.050 407 053 902 745 246 887 207 031 249 850 859 52;
  • 24) 0.050 407 053 902 745 246 887 207 031 249 850 859 52 × 2 = 0 + 0.100 814 107 805 490 493 774 414 062 499 701 719 04;
  • 25) 0.100 814 107 805 490 493 774 414 062 499 701 719 04 × 2 = 0 + 0.201 628 215 610 980 987 548 828 124 999 403 438 08;
  • 26) 0.201 628 215 610 980 987 548 828 124 999 403 438 08 × 2 = 0 + 0.403 256 431 221 961 975 097 656 249 998 806 876 16;
  • 27) 0.403 256 431 221 961 975 097 656 249 998 806 876 16 × 2 = 0 + 0.806 512 862 443 923 950 195 312 499 997 613 752 32;
  • 28) 0.806 512 862 443 923 950 195 312 499 997 613 752 32 × 2 = 1 + 0.613 025 724 887 847 900 390 624 999 995 227 504 64;
  • 29) 0.613 025 724 887 847 900 390 624 999 995 227 504 64 × 2 = 1 + 0.226 051 449 775 695 800 781 249 999 990 455 009 28;
  • 30) 0.226 051 449 775 695 800 781 249 999 990 455 009 28 × 2 = 0 + 0.452 102 899 551 391 601 562 499 999 980 910 018 56;
  • 31) 0.452 102 899 551 391 601 562 499 999 980 910 018 56 × 2 = 0 + 0.904 205 799 102 783 203 124 999 999 961 820 037 12;
  • 32) 0.904 205 799 102 783 203 124 999 999 961 820 037 12 × 2 = 1 + 0.808 411 598 205 566 406 249 999 999 923 640 074 24;
  • 33) 0.808 411 598 205 566 406 249 999 999 923 640 074 24 × 2 = 1 + 0.616 823 196 411 132 812 499 999 999 847 280 148 48;
  • 34) 0.616 823 196 411 132 812 499 999 999 847 280 148 48 × 2 = 1 + 0.233 646 392 822 265 624 999 999 999 694 560 296 96;
  • 35) 0.233 646 392 822 265 624 999 999 999 694 560 296 96 × 2 = 0 + 0.467 292 785 644 531 249 999 999 999 389 120 593 92;
  • 36) 0.467 292 785 644 531 249 999 999 999 389 120 593 92 × 2 = 0 + 0.934 585 571 289 062 499 999 999 998 778 241 187 84;
  • 37) 0.934 585 571 289 062 499 999 999 998 778 241 187 84 × 2 = 1 + 0.869 171 142 578 124 999 999 999 997 556 482 375 68;
  • 38) 0.869 171 142 578 124 999 999 999 997 556 482 375 68 × 2 = 1 + 0.738 342 285 156 249 999 999 999 995 112 964 751 36;
  • 39) 0.738 342 285 156 249 999 999 999 995 112 964 751 36 × 2 = 1 + 0.476 684 570 312 499 999 999 999 990 225 929 502 72;
  • 40) 0.476 684 570 312 499 999 999 999 990 225 929 502 72 × 2 = 0 + 0.953 369 140 624 999 999 999 999 980 451 859 005 44;
  • 41) 0.953 369 140 624 999 999 999 999 980 451 859 005 44 × 2 = 1 + 0.906 738 281 249 999 999 999 999 960 903 718 010 88;
  • 42) 0.906 738 281 249 999 999 999 999 960 903 718 010 88 × 2 = 1 + 0.813 476 562 499 999 999 999 999 921 807 436 021 76;
  • 43) 0.813 476 562 499 999 999 999 999 921 807 436 021 76 × 2 = 1 + 0.626 953 124 999 999 999 999 999 843 614 872 043 52;
  • 44) 0.626 953 124 999 999 999 999 999 843 614 872 043 52 × 2 = 1 + 0.253 906 249 999 999 999 999 999 687 229 744 087 04;
  • 45) 0.253 906 249 999 999 999 999 999 687 229 744 087 04 × 2 = 0 + 0.507 812 499 999 999 999 999 999 374 459 488 174 08;
  • 46) 0.507 812 499 999 999 999 999 999 374 459 488 174 08 × 2 = 1 + 0.015 624 999 999 999 999 999 998 748 918 976 348 16;
  • 47) 0.015 624 999 999 999 999 999 998 748 918 976 348 16 × 2 = 0 + 0.031 249 999 999 999 999 999 997 497 837 952 696 32;
  • 48) 0.031 249 999 999 999 999 999 997 497 837 952 696 32 × 2 = 0 + 0.062 499 999 999 999 999 999 994 995 675 905 392 64;
  • 49) 0.062 499 999 999 999 999 999 994 995 675 905 392 64 × 2 = 0 + 0.124 999 999 999 999 999 999 989 991 351 810 785 28;
  • 50) 0.124 999 999 999 999 999 999 989 991 351 810 785 28 × 2 = 0 + 0.249 999 999 999 999 999 999 979 982 703 621 570 56;
  • 51) 0.249 999 999 999 999 999 999 979 982 703 621 570 56 × 2 = 0 + 0.499 999 999 999 999 999 999 959 965 407 243 141 12;
  • 52) 0.499 999 999 999 999 999 999 959 965 407 243 141 12 × 2 = 0 + 0.999 999 999 999 999 999 999 919 930 814 486 282 24;
  • 53) 0.999 999 999 999 999 999 999 919 930 814 486 282 24 × 2 = 1 + 0.999 999 999 999 999 999 999 839 861 628 972 564 48;

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:


0.745 459 324 169 999 826 281 696 186 924 818 903 19(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 924 818 903 19(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:


1.745 459 324 169 999 826 281 696 186 924 818 903 19(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 1.745 459 324 169 999 826 281 696 186 924 818 903 19 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

Share

» Convert decimal 1.745 459 324 169 999 826 281 696 186 924 818 903 3 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: Reputation: Bridge Between Company's Actions & Market Value. NotebookLM YouTube Video of 6.55 minutes

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