1.745 459 324 169 999 826 281 696 182 81 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 182 81(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
1.745 459 324 169 999 826 281 696 182 81(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: 0.745 459 324 169 999 826 281 696 182 81.

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) 0.745 459 324 169 999 826 281 696 182 81 × 2 = 1 + 0.490 918 648 339 999 652 563 392 365 62;
  • 2) 0.490 918 648 339 999 652 563 392 365 62 × 2 = 0 + 0.981 837 296 679 999 305 126 784 731 24;
  • 3) 0.981 837 296 679 999 305 126 784 731 24 × 2 = 1 + 0.963 674 593 359 998 610 253 569 462 48;
  • 4) 0.963 674 593 359 998 610 253 569 462 48 × 2 = 1 + 0.927 349 186 719 997 220 507 138 924 96;
  • 5) 0.927 349 186 719 997 220 507 138 924 96 × 2 = 1 + 0.854 698 373 439 994 441 014 277 849 92;
  • 6) 0.854 698 373 439 994 441 014 277 849 92 × 2 = 1 + 0.709 396 746 879 988 882 028 555 699 84;
  • 7) 0.709 396 746 879 988 882 028 555 699 84 × 2 = 1 + 0.418 793 493 759 977 764 057 111 399 68;
  • 8) 0.418 793 493 759 977 764 057 111 399 68 × 2 = 0 + 0.837 586 987 519 955 528 114 222 799 36;
  • 9) 0.837 586 987 519 955 528 114 222 799 36 × 2 = 1 + 0.675 173 975 039 911 056 228 445 598 72;
  • 10) 0.675 173 975 039 911 056 228 445 598 72 × 2 = 1 + 0.350 347 950 079 822 112 456 891 197 44;
  • 11) 0.350 347 950 079 822 112 456 891 197 44 × 2 = 0 + 0.700 695 900 159 644 224 913 782 394 88;
  • 12) 0.700 695 900 159 644 224 913 782 394 88 × 2 = 1 + 0.401 391 800 319 288 449 827 564 789 76;
  • 13) 0.401 391 800 319 288 449 827 564 789 76 × 2 = 0 + 0.802 783 600 638 576 899 655 129 579 52;
  • 14) 0.802 783 600 638 576 899 655 129 579 52 × 2 = 1 + 0.605 567 201 277 153 799 310 259 159 04;
  • 15) 0.605 567 201 277 153 799 310 259 159 04 × 2 = 1 + 0.211 134 402 554 307 598 620 518 318 08;
  • 16) 0.211 134 402 554 307 598 620 518 318 08 × 2 = 0 + 0.422 268 805 108 615 197 241 036 636 16;
  • 17) 0.422 268 805 108 615 197 241 036 636 16 × 2 = 0 + 0.844 537 610 217 230 394 482 073 272 32;
  • 18) 0.844 537 610 217 230 394 482 073 272 32 × 2 = 1 + 0.689 075 220 434 460 788 964 146 544 64;
  • 19) 0.689 075 220 434 460 788 964 146 544 64 × 2 = 1 + 0.378 150 440 868 921 577 928 293 089 28;
  • 20) 0.378 150 440 868 921 577 928 293 089 28 × 2 = 0 + 0.756 300 881 737 843 155 856 586 178 56;
  • 21) 0.756 300 881 737 843 155 856 586 178 56 × 2 = 1 + 0.512 601 763 475 686 311 713 172 357 12;
  • 22) 0.512 601 763 475 686 311 713 172 357 12 × 2 = 1 + 0.025 203 526 951 372 623 426 344 714 24;
  • 23) 0.025 203 526 951 372 623 426 344 714 24 × 2 = 0 + 0.050 407 053 902 745 246 852 689 428 48;
  • 24) 0.050 407 053 902 745 246 852 689 428 48 × 2 = 0 + 0.100 814 107 805 490 493 705 378 856 96;
  • 25) 0.100 814 107 805 490 493 705 378 856 96 × 2 = 0 + 0.201 628 215 610 980 987 410 757 713 92;
  • 26) 0.201 628 215 610 980 987 410 757 713 92 × 2 = 0 + 0.403 256 431 221 961 974 821 515 427 84;
  • 27) 0.403 256 431 221 961 974 821 515 427 84 × 2 = 0 + 0.806 512 862 443 923 949 643 030 855 68;
  • 28) 0.806 512 862 443 923 949 643 030 855 68 × 2 = 1 + 0.613 025 724 887 847 899 286 061 711 36;
  • 29) 0.613 025 724 887 847 899 286 061 711 36 × 2 = 1 + 0.226 051 449 775 695 798 572 123 422 72;
  • 30) 0.226 051 449 775 695 798 572 123 422 72 × 2 = 0 + 0.452 102 899 551 391 597 144 246 845 44;
  • 31) 0.452 102 899 551 391 597 144 246 845 44 × 2 = 0 + 0.904 205 799 102 783 194 288 493 690 88;
  • 32) 0.904 205 799 102 783 194 288 493 690 88 × 2 = 1 + 0.808 411 598 205 566 388 576 987 381 76;
  • 33) 0.808 411 598 205 566 388 576 987 381 76 × 2 = 1 + 0.616 823 196 411 132 777 153 974 763 52;
  • 34) 0.616 823 196 411 132 777 153 974 763 52 × 2 = 1 + 0.233 646 392 822 265 554 307 949 527 04;
  • 35) 0.233 646 392 822 265 554 307 949 527 04 × 2 = 0 + 0.467 292 785 644 531 108 615 899 054 08;
  • 36) 0.467 292 785 644 531 108 615 899 054 08 × 2 = 0 + 0.934 585 571 289 062 217 231 798 108 16;
  • 37) 0.934 585 571 289 062 217 231 798 108 16 × 2 = 1 + 0.869 171 142 578 124 434 463 596 216 32;
  • 38) 0.869 171 142 578 124 434 463 596 216 32 × 2 = 1 + 0.738 342 285 156 248 868 927 192 432 64;
  • 39) 0.738 342 285 156 248 868 927 192 432 64 × 2 = 1 + 0.476 684 570 312 497 737 854 384 865 28;
  • 40) 0.476 684 570 312 497 737 854 384 865 28 × 2 = 0 + 0.953 369 140 624 995 475 708 769 730 56;
  • 41) 0.953 369 140 624 995 475 708 769 730 56 × 2 = 1 + 0.906 738 281 249 990 951 417 539 461 12;
  • 42) 0.906 738 281 249 990 951 417 539 461 12 × 2 = 1 + 0.813 476 562 499 981 902 835 078 922 24;
  • 43) 0.813 476 562 499 981 902 835 078 922 24 × 2 = 1 + 0.626 953 124 999 963 805 670 157 844 48;
  • 44) 0.626 953 124 999 963 805 670 157 844 48 × 2 = 1 + 0.253 906 249 999 927 611 340 315 688 96;
  • 45) 0.253 906 249 999 927 611 340 315 688 96 × 2 = 0 + 0.507 812 499 999 855 222 680 631 377 92;
  • 46) 0.507 812 499 999 855 222 680 631 377 92 × 2 = 1 + 0.015 624 999 999 710 445 361 262 755 84;
  • 47) 0.015 624 999 999 710 445 361 262 755 84 × 2 = 0 + 0.031 249 999 999 420 890 722 525 511 68;
  • 48) 0.031 249 999 999 420 890 722 525 511 68 × 2 = 0 + 0.062 499 999 998 841 781 445 051 023 36;
  • 49) 0.062 499 999 998 841 781 445 051 023 36 × 2 = 0 + 0.124 999 999 997 683 562 890 102 046 72;
  • 50) 0.124 999 999 997 683 562 890 102 046 72 × 2 = 0 + 0.249 999 999 995 367 125 780 204 093 44;
  • 51) 0.249 999 999 995 367 125 780 204 093 44 × 2 = 0 + 0.499 999 999 990 734 251 560 408 186 88;
  • 52) 0.499 999 999 990 734 251 560 408 186 88 × 2 = 0 + 0.999 999 999 981 468 503 120 816 373 76;
  • 53) 0.999 999 999 981 468 503 120 816 373 76 × 2 = 1 + 0.999 999 999 962 937 006 241 632 747 52;

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:


0.745 459 324 169 999 826 281 696 182 81(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 182 81(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:


1.745 459 324 169 999 826 281 696 182 81(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 1.745 459 324 169 999 826 281 696 182 81 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