1.745 459 324 169 999 826 281 696 175 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 175 3(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
1.745 459 324 169 999 826 281 696 175 3(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: 0.745 459 324 169 999 826 281 696 175 3.

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) 0.745 459 324 169 999 826 281 696 175 3 × 2 = 1 + 0.490 918 648 339 999 652 563 392 350 6;
  • 2) 0.490 918 648 339 999 652 563 392 350 6 × 2 = 0 + 0.981 837 296 679 999 305 126 784 701 2;
  • 3) 0.981 837 296 679 999 305 126 784 701 2 × 2 = 1 + 0.963 674 593 359 998 610 253 569 402 4;
  • 4) 0.963 674 593 359 998 610 253 569 402 4 × 2 = 1 + 0.927 349 186 719 997 220 507 138 804 8;
  • 5) 0.927 349 186 719 997 220 507 138 804 8 × 2 = 1 + 0.854 698 373 439 994 441 014 277 609 6;
  • 6) 0.854 698 373 439 994 441 014 277 609 6 × 2 = 1 + 0.709 396 746 879 988 882 028 555 219 2;
  • 7) 0.709 396 746 879 988 882 028 555 219 2 × 2 = 1 + 0.418 793 493 759 977 764 057 110 438 4;
  • 8) 0.418 793 493 759 977 764 057 110 438 4 × 2 = 0 + 0.837 586 987 519 955 528 114 220 876 8;
  • 9) 0.837 586 987 519 955 528 114 220 876 8 × 2 = 1 + 0.675 173 975 039 911 056 228 441 753 6;
  • 10) 0.675 173 975 039 911 056 228 441 753 6 × 2 = 1 + 0.350 347 950 079 822 112 456 883 507 2;
  • 11) 0.350 347 950 079 822 112 456 883 507 2 × 2 = 0 + 0.700 695 900 159 644 224 913 767 014 4;
  • 12) 0.700 695 900 159 644 224 913 767 014 4 × 2 = 1 + 0.401 391 800 319 288 449 827 534 028 8;
  • 13) 0.401 391 800 319 288 449 827 534 028 8 × 2 = 0 + 0.802 783 600 638 576 899 655 068 057 6;
  • 14) 0.802 783 600 638 576 899 655 068 057 6 × 2 = 1 + 0.605 567 201 277 153 799 310 136 115 2;
  • 15) 0.605 567 201 277 153 799 310 136 115 2 × 2 = 1 + 0.211 134 402 554 307 598 620 272 230 4;
  • 16) 0.211 134 402 554 307 598 620 272 230 4 × 2 = 0 + 0.422 268 805 108 615 197 240 544 460 8;
  • 17) 0.422 268 805 108 615 197 240 544 460 8 × 2 = 0 + 0.844 537 610 217 230 394 481 088 921 6;
  • 18) 0.844 537 610 217 230 394 481 088 921 6 × 2 = 1 + 0.689 075 220 434 460 788 962 177 843 2;
  • 19) 0.689 075 220 434 460 788 962 177 843 2 × 2 = 1 + 0.378 150 440 868 921 577 924 355 686 4;
  • 20) 0.378 150 440 868 921 577 924 355 686 4 × 2 = 0 + 0.756 300 881 737 843 155 848 711 372 8;
  • 21) 0.756 300 881 737 843 155 848 711 372 8 × 2 = 1 + 0.512 601 763 475 686 311 697 422 745 6;
  • 22) 0.512 601 763 475 686 311 697 422 745 6 × 2 = 1 + 0.025 203 526 951 372 623 394 845 491 2;
  • 23) 0.025 203 526 951 372 623 394 845 491 2 × 2 = 0 + 0.050 407 053 902 745 246 789 690 982 4;
  • 24) 0.050 407 053 902 745 246 789 690 982 4 × 2 = 0 + 0.100 814 107 805 490 493 579 381 964 8;
  • 25) 0.100 814 107 805 490 493 579 381 964 8 × 2 = 0 + 0.201 628 215 610 980 987 158 763 929 6;
  • 26) 0.201 628 215 610 980 987 158 763 929 6 × 2 = 0 + 0.403 256 431 221 961 974 317 527 859 2;
  • 27) 0.403 256 431 221 961 974 317 527 859 2 × 2 = 0 + 0.806 512 862 443 923 948 635 055 718 4;
  • 28) 0.806 512 862 443 923 948 635 055 718 4 × 2 = 1 + 0.613 025 724 887 847 897 270 111 436 8;
  • 29) 0.613 025 724 887 847 897 270 111 436 8 × 2 = 1 + 0.226 051 449 775 695 794 540 222 873 6;
  • 30) 0.226 051 449 775 695 794 540 222 873 6 × 2 = 0 + 0.452 102 899 551 391 589 080 445 747 2;
  • 31) 0.452 102 899 551 391 589 080 445 747 2 × 2 = 0 + 0.904 205 799 102 783 178 160 891 494 4;
  • 32) 0.904 205 799 102 783 178 160 891 494 4 × 2 = 1 + 0.808 411 598 205 566 356 321 782 988 8;
  • 33) 0.808 411 598 205 566 356 321 782 988 8 × 2 = 1 + 0.616 823 196 411 132 712 643 565 977 6;
  • 34) 0.616 823 196 411 132 712 643 565 977 6 × 2 = 1 + 0.233 646 392 822 265 425 287 131 955 2;
  • 35) 0.233 646 392 822 265 425 287 131 955 2 × 2 = 0 + 0.467 292 785 644 530 850 574 263 910 4;
  • 36) 0.467 292 785 644 530 850 574 263 910 4 × 2 = 0 + 0.934 585 571 289 061 701 148 527 820 8;
  • 37) 0.934 585 571 289 061 701 148 527 820 8 × 2 = 1 + 0.869 171 142 578 123 402 297 055 641 6;
  • 38) 0.869 171 142 578 123 402 297 055 641 6 × 2 = 1 + 0.738 342 285 156 246 804 594 111 283 2;
  • 39) 0.738 342 285 156 246 804 594 111 283 2 × 2 = 1 + 0.476 684 570 312 493 609 188 222 566 4;
  • 40) 0.476 684 570 312 493 609 188 222 566 4 × 2 = 0 + 0.953 369 140 624 987 218 376 445 132 8;
  • 41) 0.953 369 140 624 987 218 376 445 132 8 × 2 = 1 + 0.906 738 281 249 974 436 752 890 265 6;
  • 42) 0.906 738 281 249 974 436 752 890 265 6 × 2 = 1 + 0.813 476 562 499 948 873 505 780 531 2;
  • 43) 0.813 476 562 499 948 873 505 780 531 2 × 2 = 1 + 0.626 953 124 999 897 747 011 561 062 4;
  • 44) 0.626 953 124 999 897 747 011 561 062 4 × 2 = 1 + 0.253 906 249 999 795 494 023 122 124 8;
  • 45) 0.253 906 249 999 795 494 023 122 124 8 × 2 = 0 + 0.507 812 499 999 590 988 046 244 249 6;
  • 46) 0.507 812 499 999 590 988 046 244 249 6 × 2 = 1 + 0.015 624 999 999 181 976 092 488 499 2;
  • 47) 0.015 624 999 999 181 976 092 488 499 2 × 2 = 0 + 0.031 249 999 998 363 952 184 976 998 4;
  • 48) 0.031 249 999 998 363 952 184 976 998 4 × 2 = 0 + 0.062 499 999 996 727 904 369 953 996 8;
  • 49) 0.062 499 999 996 727 904 369 953 996 8 × 2 = 0 + 0.124 999 999 993 455 808 739 907 993 6;
  • 50) 0.124 999 999 993 455 808 739 907 993 6 × 2 = 0 + 0.249 999 999 986 911 617 479 815 987 2;
  • 51) 0.249 999 999 986 911 617 479 815 987 2 × 2 = 0 + 0.499 999 999 973 823 234 959 631 974 4;
  • 52) 0.499 999 999 973 823 234 959 631 974 4 × 2 = 0 + 0.999 999 999 947 646 469 919 263 948 8;
  • 53) 0.999 999 999 947 646 469 919 263 948 8 × 2 = 1 + 0.999 999 999 895 292 939 838 527 897 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:


0.745 459 324 169 999 826 281 696 175 3(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 175 3(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:


1.745 459 324 169 999 826 281 696 175 3(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 1.745 459 324 169 999 826 281 696 175 3 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 1.745 459 324 169 999 826 281 696 181 5 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: Are You Using Communication by Design, or by Default? NotebookLM YouTube Video of 6.33 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