1.745 459 324 169 999 826 281 673 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 673(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
1.745 459 324 169 999 826 281 673(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: 0.745 459 324 169 999 826 281 673.

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) 0.745 459 324 169 999 826 281 673 × 2 = 1 + 0.490 918 648 339 999 652 563 346;
  • 2) 0.490 918 648 339 999 652 563 346 × 2 = 0 + 0.981 837 296 679 999 305 126 692;
  • 3) 0.981 837 296 679 999 305 126 692 × 2 = 1 + 0.963 674 593 359 998 610 253 384;
  • 4) 0.963 674 593 359 998 610 253 384 × 2 = 1 + 0.927 349 186 719 997 220 506 768;
  • 5) 0.927 349 186 719 997 220 506 768 × 2 = 1 + 0.854 698 373 439 994 441 013 536;
  • 6) 0.854 698 373 439 994 441 013 536 × 2 = 1 + 0.709 396 746 879 988 882 027 072;
  • 7) 0.709 396 746 879 988 882 027 072 × 2 = 1 + 0.418 793 493 759 977 764 054 144;
  • 8) 0.418 793 493 759 977 764 054 144 × 2 = 0 + 0.837 586 987 519 955 528 108 288;
  • 9) 0.837 586 987 519 955 528 108 288 × 2 = 1 + 0.675 173 975 039 911 056 216 576;
  • 10) 0.675 173 975 039 911 056 216 576 × 2 = 1 + 0.350 347 950 079 822 112 433 152;
  • 11) 0.350 347 950 079 822 112 433 152 × 2 = 0 + 0.700 695 900 159 644 224 866 304;
  • 12) 0.700 695 900 159 644 224 866 304 × 2 = 1 + 0.401 391 800 319 288 449 732 608;
  • 13) 0.401 391 800 319 288 449 732 608 × 2 = 0 + 0.802 783 600 638 576 899 465 216;
  • 14) 0.802 783 600 638 576 899 465 216 × 2 = 1 + 0.605 567 201 277 153 798 930 432;
  • 15) 0.605 567 201 277 153 798 930 432 × 2 = 1 + 0.211 134 402 554 307 597 860 864;
  • 16) 0.211 134 402 554 307 597 860 864 × 2 = 0 + 0.422 268 805 108 615 195 721 728;
  • 17) 0.422 268 805 108 615 195 721 728 × 2 = 0 + 0.844 537 610 217 230 391 443 456;
  • 18) 0.844 537 610 217 230 391 443 456 × 2 = 1 + 0.689 075 220 434 460 782 886 912;
  • 19) 0.689 075 220 434 460 782 886 912 × 2 = 1 + 0.378 150 440 868 921 565 773 824;
  • 20) 0.378 150 440 868 921 565 773 824 × 2 = 0 + 0.756 300 881 737 843 131 547 648;
  • 21) 0.756 300 881 737 843 131 547 648 × 2 = 1 + 0.512 601 763 475 686 263 095 296;
  • 22) 0.512 601 763 475 686 263 095 296 × 2 = 1 + 0.025 203 526 951 372 526 190 592;
  • 23) 0.025 203 526 951 372 526 190 592 × 2 = 0 + 0.050 407 053 902 745 052 381 184;
  • 24) 0.050 407 053 902 745 052 381 184 × 2 = 0 + 0.100 814 107 805 490 104 762 368;
  • 25) 0.100 814 107 805 490 104 762 368 × 2 = 0 + 0.201 628 215 610 980 209 524 736;
  • 26) 0.201 628 215 610 980 209 524 736 × 2 = 0 + 0.403 256 431 221 960 419 049 472;
  • 27) 0.403 256 431 221 960 419 049 472 × 2 = 0 + 0.806 512 862 443 920 838 098 944;
  • 28) 0.806 512 862 443 920 838 098 944 × 2 = 1 + 0.613 025 724 887 841 676 197 888;
  • 29) 0.613 025 724 887 841 676 197 888 × 2 = 1 + 0.226 051 449 775 683 352 395 776;
  • 30) 0.226 051 449 775 683 352 395 776 × 2 = 0 + 0.452 102 899 551 366 704 791 552;
  • 31) 0.452 102 899 551 366 704 791 552 × 2 = 0 + 0.904 205 799 102 733 409 583 104;
  • 32) 0.904 205 799 102 733 409 583 104 × 2 = 1 + 0.808 411 598 205 466 819 166 208;
  • 33) 0.808 411 598 205 466 819 166 208 × 2 = 1 + 0.616 823 196 410 933 638 332 416;
  • 34) 0.616 823 196 410 933 638 332 416 × 2 = 1 + 0.233 646 392 821 867 276 664 832;
  • 35) 0.233 646 392 821 867 276 664 832 × 2 = 0 + 0.467 292 785 643 734 553 329 664;
  • 36) 0.467 292 785 643 734 553 329 664 × 2 = 0 + 0.934 585 571 287 469 106 659 328;
  • 37) 0.934 585 571 287 469 106 659 328 × 2 = 1 + 0.869 171 142 574 938 213 318 656;
  • 38) 0.869 171 142 574 938 213 318 656 × 2 = 1 + 0.738 342 285 149 876 426 637 312;
  • 39) 0.738 342 285 149 876 426 637 312 × 2 = 1 + 0.476 684 570 299 752 853 274 624;
  • 40) 0.476 684 570 299 752 853 274 624 × 2 = 0 + 0.953 369 140 599 505 706 549 248;
  • 41) 0.953 369 140 599 505 706 549 248 × 2 = 1 + 0.906 738 281 199 011 413 098 496;
  • 42) 0.906 738 281 199 011 413 098 496 × 2 = 1 + 0.813 476 562 398 022 826 196 992;
  • 43) 0.813 476 562 398 022 826 196 992 × 2 = 1 + 0.626 953 124 796 045 652 393 984;
  • 44) 0.626 953 124 796 045 652 393 984 × 2 = 1 + 0.253 906 249 592 091 304 787 968;
  • 45) 0.253 906 249 592 091 304 787 968 × 2 = 0 + 0.507 812 499 184 182 609 575 936;
  • 46) 0.507 812 499 184 182 609 575 936 × 2 = 1 + 0.015 624 998 368 365 219 151 872;
  • 47) 0.015 624 998 368 365 219 151 872 × 2 = 0 + 0.031 249 996 736 730 438 303 744;
  • 48) 0.031 249 996 736 730 438 303 744 × 2 = 0 + 0.062 499 993 473 460 876 607 488;
  • 49) 0.062 499 993 473 460 876 607 488 × 2 = 0 + 0.124 999 986 946 921 753 214 976;
  • 50) 0.124 999 986 946 921 753 214 976 × 2 = 0 + 0.249 999 973 893 843 506 429 952;
  • 51) 0.249 999 973 893 843 506 429 952 × 2 = 0 + 0.499 999 947 787 687 012 859 904;
  • 52) 0.499 999 947 787 687 012 859 904 × 2 = 0 + 0.999 999 895 575 374 025 719 808;
  • 53) 0.999 999 895 575 374 025 719 808 × 2 = 1 + 0.999 999 791 150 748 051 439 616;

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:


0.745 459 324 169 999 826 281 673(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 673(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:


1.745 459 324 169 999 826 281 673(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 1.745 459 324 169 999 826 281 673 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