1.745 459 324 169 999 826 281 779 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 779(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
1.745 459 324 169 999 826 281 779(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: 0.745 459 324 169 999 826 281 779.

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) 0.745 459 324 169 999 826 281 779 × 2 = 1 + 0.490 918 648 339 999 652 563 558;
  • 2) 0.490 918 648 339 999 652 563 558 × 2 = 0 + 0.981 837 296 679 999 305 127 116;
  • 3) 0.981 837 296 679 999 305 127 116 × 2 = 1 + 0.963 674 593 359 998 610 254 232;
  • 4) 0.963 674 593 359 998 610 254 232 × 2 = 1 + 0.927 349 186 719 997 220 508 464;
  • 5) 0.927 349 186 719 997 220 508 464 × 2 = 1 + 0.854 698 373 439 994 441 016 928;
  • 6) 0.854 698 373 439 994 441 016 928 × 2 = 1 + 0.709 396 746 879 988 882 033 856;
  • 7) 0.709 396 746 879 988 882 033 856 × 2 = 1 + 0.418 793 493 759 977 764 067 712;
  • 8) 0.418 793 493 759 977 764 067 712 × 2 = 0 + 0.837 586 987 519 955 528 135 424;
  • 9) 0.837 586 987 519 955 528 135 424 × 2 = 1 + 0.675 173 975 039 911 056 270 848;
  • 10) 0.675 173 975 039 911 056 270 848 × 2 = 1 + 0.350 347 950 079 822 112 541 696;
  • 11) 0.350 347 950 079 822 112 541 696 × 2 = 0 + 0.700 695 900 159 644 225 083 392;
  • 12) 0.700 695 900 159 644 225 083 392 × 2 = 1 + 0.401 391 800 319 288 450 166 784;
  • 13) 0.401 391 800 319 288 450 166 784 × 2 = 0 + 0.802 783 600 638 576 900 333 568;
  • 14) 0.802 783 600 638 576 900 333 568 × 2 = 1 + 0.605 567 201 277 153 800 667 136;
  • 15) 0.605 567 201 277 153 800 667 136 × 2 = 1 + 0.211 134 402 554 307 601 334 272;
  • 16) 0.211 134 402 554 307 601 334 272 × 2 = 0 + 0.422 268 805 108 615 202 668 544;
  • 17) 0.422 268 805 108 615 202 668 544 × 2 = 0 + 0.844 537 610 217 230 405 337 088;
  • 18) 0.844 537 610 217 230 405 337 088 × 2 = 1 + 0.689 075 220 434 460 810 674 176;
  • 19) 0.689 075 220 434 460 810 674 176 × 2 = 1 + 0.378 150 440 868 921 621 348 352;
  • 20) 0.378 150 440 868 921 621 348 352 × 2 = 0 + 0.756 300 881 737 843 242 696 704;
  • 21) 0.756 300 881 737 843 242 696 704 × 2 = 1 + 0.512 601 763 475 686 485 393 408;
  • 22) 0.512 601 763 475 686 485 393 408 × 2 = 1 + 0.025 203 526 951 372 970 786 816;
  • 23) 0.025 203 526 951 372 970 786 816 × 2 = 0 + 0.050 407 053 902 745 941 573 632;
  • 24) 0.050 407 053 902 745 941 573 632 × 2 = 0 + 0.100 814 107 805 491 883 147 264;
  • 25) 0.100 814 107 805 491 883 147 264 × 2 = 0 + 0.201 628 215 610 983 766 294 528;
  • 26) 0.201 628 215 610 983 766 294 528 × 2 = 0 + 0.403 256 431 221 967 532 589 056;
  • 27) 0.403 256 431 221 967 532 589 056 × 2 = 0 + 0.806 512 862 443 935 065 178 112;
  • 28) 0.806 512 862 443 935 065 178 112 × 2 = 1 + 0.613 025 724 887 870 130 356 224;
  • 29) 0.613 025 724 887 870 130 356 224 × 2 = 1 + 0.226 051 449 775 740 260 712 448;
  • 30) 0.226 051 449 775 740 260 712 448 × 2 = 0 + 0.452 102 899 551 480 521 424 896;
  • 31) 0.452 102 899 551 480 521 424 896 × 2 = 0 + 0.904 205 799 102 961 042 849 792;
  • 32) 0.904 205 799 102 961 042 849 792 × 2 = 1 + 0.808 411 598 205 922 085 699 584;
  • 33) 0.808 411 598 205 922 085 699 584 × 2 = 1 + 0.616 823 196 411 844 171 399 168;
  • 34) 0.616 823 196 411 844 171 399 168 × 2 = 1 + 0.233 646 392 823 688 342 798 336;
  • 35) 0.233 646 392 823 688 342 798 336 × 2 = 0 + 0.467 292 785 647 376 685 596 672;
  • 36) 0.467 292 785 647 376 685 596 672 × 2 = 0 + 0.934 585 571 294 753 371 193 344;
  • 37) 0.934 585 571 294 753 371 193 344 × 2 = 1 + 0.869 171 142 589 506 742 386 688;
  • 38) 0.869 171 142 589 506 742 386 688 × 2 = 1 + 0.738 342 285 179 013 484 773 376;
  • 39) 0.738 342 285 179 013 484 773 376 × 2 = 1 + 0.476 684 570 358 026 969 546 752;
  • 40) 0.476 684 570 358 026 969 546 752 × 2 = 0 + 0.953 369 140 716 053 939 093 504;
  • 41) 0.953 369 140 716 053 939 093 504 × 2 = 1 + 0.906 738 281 432 107 878 187 008;
  • 42) 0.906 738 281 432 107 878 187 008 × 2 = 1 + 0.813 476 562 864 215 756 374 016;
  • 43) 0.813 476 562 864 215 756 374 016 × 2 = 1 + 0.626 953 125 728 431 512 748 032;
  • 44) 0.626 953 125 728 431 512 748 032 × 2 = 1 + 0.253 906 251 456 863 025 496 064;
  • 45) 0.253 906 251 456 863 025 496 064 × 2 = 0 + 0.507 812 502 913 726 050 992 128;
  • 46) 0.507 812 502 913 726 050 992 128 × 2 = 1 + 0.015 625 005 827 452 101 984 256;
  • 47) 0.015 625 005 827 452 101 984 256 × 2 = 0 + 0.031 250 011 654 904 203 968 512;
  • 48) 0.031 250 011 654 904 203 968 512 × 2 = 0 + 0.062 500 023 309 808 407 937 024;
  • 49) 0.062 500 023 309 808 407 937 024 × 2 = 0 + 0.125 000 046 619 616 815 874 048;
  • 50) 0.125 000 046 619 616 815 874 048 × 2 = 0 + 0.250 000 093 239 233 631 748 096;
  • 51) 0.250 000 093 239 233 631 748 096 × 2 = 0 + 0.500 000 186 478 467 263 496 192;
  • 52) 0.500 000 186 478 467 263 496 192 × 2 = 1 + 0.000 000 372 956 934 526 992 384;
  • 53) 0.000 000 372 956 934 526 992 384 × 2 = 0 + 0.000 000 745 913 869 053 984 768;

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:


0.745 459 324 169 999 826 281 779(10) =

0.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(2)

5. Positive number before normalization:

1.745 459 324 169 999 826 281 779(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(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:


1.745 459 324 169 999 826 281 779(10) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(2) =

1.1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001 0(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 0001 0


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 0001 0 =

1011 1110 1101 0110 0110 1100 0001 1001 1100 1110 1111 0100 0001


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 0001


Decimal number 1.745 459 324 169 999 826 281 779 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 0001

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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