1.745 459 324 169 999 826 281 647 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 647(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
1.745 459 324 169 999 826 281 647(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: 0.745 459 324 169 999 826 281 647.

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) 0.745 459 324 169 999 826 281 647 × 2 = 1 + 0.490 918 648 339 999 652 563 294;
  • 2) 0.490 918 648 339 999 652 563 294 × 2 = 0 + 0.981 837 296 679 999 305 126 588;
  • 3) 0.981 837 296 679 999 305 126 588 × 2 = 1 + 0.963 674 593 359 998 610 253 176;
  • 4) 0.963 674 593 359 998 610 253 176 × 2 = 1 + 0.927 349 186 719 997 220 506 352;
  • 5) 0.927 349 186 719 997 220 506 352 × 2 = 1 + 0.854 698 373 439 994 441 012 704;
  • 6) 0.854 698 373 439 994 441 012 704 × 2 = 1 + 0.709 396 746 879 988 882 025 408;
  • 7) 0.709 396 746 879 988 882 025 408 × 2 = 1 + 0.418 793 493 759 977 764 050 816;
  • 8) 0.418 793 493 759 977 764 050 816 × 2 = 0 + 0.837 586 987 519 955 528 101 632;
  • 9) 0.837 586 987 519 955 528 101 632 × 2 = 1 + 0.675 173 975 039 911 056 203 264;
  • 10) 0.675 173 975 039 911 056 203 264 × 2 = 1 + 0.350 347 950 079 822 112 406 528;
  • 11) 0.350 347 950 079 822 112 406 528 × 2 = 0 + 0.700 695 900 159 644 224 813 056;
  • 12) 0.700 695 900 159 644 224 813 056 × 2 = 1 + 0.401 391 800 319 288 449 626 112;
  • 13) 0.401 391 800 319 288 449 626 112 × 2 = 0 + 0.802 783 600 638 576 899 252 224;
  • 14) 0.802 783 600 638 576 899 252 224 × 2 = 1 + 0.605 567 201 277 153 798 504 448;
  • 15) 0.605 567 201 277 153 798 504 448 × 2 = 1 + 0.211 134 402 554 307 597 008 896;
  • 16) 0.211 134 402 554 307 597 008 896 × 2 = 0 + 0.422 268 805 108 615 194 017 792;
  • 17) 0.422 268 805 108 615 194 017 792 × 2 = 0 + 0.844 537 610 217 230 388 035 584;
  • 18) 0.844 537 610 217 230 388 035 584 × 2 = 1 + 0.689 075 220 434 460 776 071 168;
  • 19) 0.689 075 220 434 460 776 071 168 × 2 = 1 + 0.378 150 440 868 921 552 142 336;
  • 20) 0.378 150 440 868 921 552 142 336 × 2 = 0 + 0.756 300 881 737 843 104 284 672;
  • 21) 0.756 300 881 737 843 104 284 672 × 2 = 1 + 0.512 601 763 475 686 208 569 344;
  • 22) 0.512 601 763 475 686 208 569 344 × 2 = 1 + 0.025 203 526 951 372 417 138 688;
  • 23) 0.025 203 526 951 372 417 138 688 × 2 = 0 + 0.050 407 053 902 744 834 277 376;
  • 24) 0.050 407 053 902 744 834 277 376 × 2 = 0 + 0.100 814 107 805 489 668 554 752;
  • 25) 0.100 814 107 805 489 668 554 752 × 2 = 0 + 0.201 628 215 610 979 337 109 504;
  • 26) 0.201 628 215 610 979 337 109 504 × 2 = 0 + 0.403 256 431 221 958 674 219 008;
  • 27) 0.403 256 431 221 958 674 219 008 × 2 = 0 + 0.806 512 862 443 917 348 438 016;
  • 28) 0.806 512 862 443 917 348 438 016 × 2 = 1 + 0.613 025 724 887 834 696 876 032;
  • 29) 0.613 025 724 887 834 696 876 032 × 2 = 1 + 0.226 051 449 775 669 393 752 064;
  • 30) 0.226 051 449 775 669 393 752 064 × 2 = 0 + 0.452 102 899 551 338 787 504 128;
  • 31) 0.452 102 899 551 338 787 504 128 × 2 = 0 + 0.904 205 799 102 677 575 008 256;
  • 32) 0.904 205 799 102 677 575 008 256 × 2 = 1 + 0.808 411 598 205 355 150 016 512;
  • 33) 0.808 411 598 205 355 150 016 512 × 2 = 1 + 0.616 823 196 410 710 300 033 024;
  • 34) 0.616 823 196 410 710 300 033 024 × 2 = 1 + 0.233 646 392 821 420 600 066 048;
  • 35) 0.233 646 392 821 420 600 066 048 × 2 = 0 + 0.467 292 785 642 841 200 132 096;
  • 36) 0.467 292 785 642 841 200 132 096 × 2 = 0 + 0.934 585 571 285 682 400 264 192;
  • 37) 0.934 585 571 285 682 400 264 192 × 2 = 1 + 0.869 171 142 571 364 800 528 384;
  • 38) 0.869 171 142 571 364 800 528 384 × 2 = 1 + 0.738 342 285 142 729 601 056 768;
  • 39) 0.738 342 285 142 729 601 056 768 × 2 = 1 + 0.476 684 570 285 459 202 113 536;
  • 40) 0.476 684 570 285 459 202 113 536 × 2 = 0 + 0.953 369 140 570 918 404 227 072;
  • 41) 0.953 369 140 570 918 404 227 072 × 2 = 1 + 0.906 738 281 141 836 808 454 144;
  • 42) 0.906 738 281 141 836 808 454 144 × 2 = 1 + 0.813 476 562 283 673 616 908 288;
  • 43) 0.813 476 562 283 673 616 908 288 × 2 = 1 + 0.626 953 124 567 347 233 816 576;
  • 44) 0.626 953 124 567 347 233 816 576 × 2 = 1 + 0.253 906 249 134 694 467 633 152;
  • 45) 0.253 906 249 134 694 467 633 152 × 2 = 0 + 0.507 812 498 269 388 935 266 304;
  • 46) 0.507 812 498 269 388 935 266 304 × 2 = 1 + 0.015 624 996 538 777 870 532 608;
  • 47) 0.015 624 996 538 777 870 532 608 × 2 = 0 + 0.031 249 993 077 555 741 065 216;
  • 48) 0.031 249 993 077 555 741 065 216 × 2 = 0 + 0.062 499 986 155 111 482 130 432;
  • 49) 0.062 499 986 155 111 482 130 432 × 2 = 0 + 0.124 999 972 310 222 964 260 864;
  • 50) 0.124 999 972 310 222 964 260 864 × 2 = 0 + 0.249 999 944 620 445 928 521 728;
  • 51) 0.249 999 944 620 445 928 521 728 × 2 = 0 + 0.499 999 889 240 891 857 043 456;
  • 52) 0.499 999 889 240 891 857 043 456 × 2 = 0 + 0.999 999 778 481 783 714 086 912;
  • 53) 0.999 999 778 481 783 714 086 912 × 2 = 1 + 0.999 999 556 963 567 428 173 824;

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:


0.745 459 324 169 999 826 281 647(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 647(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:


1.745 459 324 169 999 826 281 647(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 1.745 459 324 169 999 826 281 647 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