1.745 459 324 169 999 826 281 697 91 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 697 91(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
1.745 459 324 169 999 826 281 697 91(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: 0.745 459 324 169 999 826 281 697 91.

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) 0.745 459 324 169 999 826 281 697 91 × 2 = 1 + 0.490 918 648 339 999 652 563 395 82;
  • 2) 0.490 918 648 339 999 652 563 395 82 × 2 = 0 + 0.981 837 296 679 999 305 126 791 64;
  • 3) 0.981 837 296 679 999 305 126 791 64 × 2 = 1 + 0.963 674 593 359 998 610 253 583 28;
  • 4) 0.963 674 593 359 998 610 253 583 28 × 2 = 1 + 0.927 349 186 719 997 220 507 166 56;
  • 5) 0.927 349 186 719 997 220 507 166 56 × 2 = 1 + 0.854 698 373 439 994 441 014 333 12;
  • 6) 0.854 698 373 439 994 441 014 333 12 × 2 = 1 + 0.709 396 746 879 988 882 028 666 24;
  • 7) 0.709 396 746 879 988 882 028 666 24 × 2 = 1 + 0.418 793 493 759 977 764 057 332 48;
  • 8) 0.418 793 493 759 977 764 057 332 48 × 2 = 0 + 0.837 586 987 519 955 528 114 664 96;
  • 9) 0.837 586 987 519 955 528 114 664 96 × 2 = 1 + 0.675 173 975 039 911 056 229 329 92;
  • 10) 0.675 173 975 039 911 056 229 329 92 × 2 = 1 + 0.350 347 950 079 822 112 458 659 84;
  • 11) 0.350 347 950 079 822 112 458 659 84 × 2 = 0 + 0.700 695 900 159 644 224 917 319 68;
  • 12) 0.700 695 900 159 644 224 917 319 68 × 2 = 1 + 0.401 391 800 319 288 449 834 639 36;
  • 13) 0.401 391 800 319 288 449 834 639 36 × 2 = 0 + 0.802 783 600 638 576 899 669 278 72;
  • 14) 0.802 783 600 638 576 899 669 278 72 × 2 = 1 + 0.605 567 201 277 153 799 338 557 44;
  • 15) 0.605 567 201 277 153 799 338 557 44 × 2 = 1 + 0.211 134 402 554 307 598 677 114 88;
  • 16) 0.211 134 402 554 307 598 677 114 88 × 2 = 0 + 0.422 268 805 108 615 197 354 229 76;
  • 17) 0.422 268 805 108 615 197 354 229 76 × 2 = 0 + 0.844 537 610 217 230 394 708 459 52;
  • 18) 0.844 537 610 217 230 394 708 459 52 × 2 = 1 + 0.689 075 220 434 460 789 416 919 04;
  • 19) 0.689 075 220 434 460 789 416 919 04 × 2 = 1 + 0.378 150 440 868 921 578 833 838 08;
  • 20) 0.378 150 440 868 921 578 833 838 08 × 2 = 0 + 0.756 300 881 737 843 157 667 676 16;
  • 21) 0.756 300 881 737 843 157 667 676 16 × 2 = 1 + 0.512 601 763 475 686 315 335 352 32;
  • 22) 0.512 601 763 475 686 315 335 352 32 × 2 = 1 + 0.025 203 526 951 372 630 670 704 64;
  • 23) 0.025 203 526 951 372 630 670 704 64 × 2 = 0 + 0.050 407 053 902 745 261 341 409 28;
  • 24) 0.050 407 053 902 745 261 341 409 28 × 2 = 0 + 0.100 814 107 805 490 522 682 818 56;
  • 25) 0.100 814 107 805 490 522 682 818 56 × 2 = 0 + 0.201 628 215 610 981 045 365 637 12;
  • 26) 0.201 628 215 610 981 045 365 637 12 × 2 = 0 + 0.403 256 431 221 962 090 731 274 24;
  • 27) 0.403 256 431 221 962 090 731 274 24 × 2 = 0 + 0.806 512 862 443 924 181 462 548 48;
  • 28) 0.806 512 862 443 924 181 462 548 48 × 2 = 1 + 0.613 025 724 887 848 362 925 096 96;
  • 29) 0.613 025 724 887 848 362 925 096 96 × 2 = 1 + 0.226 051 449 775 696 725 850 193 92;
  • 30) 0.226 051 449 775 696 725 850 193 92 × 2 = 0 + 0.452 102 899 551 393 451 700 387 84;
  • 31) 0.452 102 899 551 393 451 700 387 84 × 2 = 0 + 0.904 205 799 102 786 903 400 775 68;
  • 32) 0.904 205 799 102 786 903 400 775 68 × 2 = 1 + 0.808 411 598 205 573 806 801 551 36;
  • 33) 0.808 411 598 205 573 806 801 551 36 × 2 = 1 + 0.616 823 196 411 147 613 603 102 72;
  • 34) 0.616 823 196 411 147 613 603 102 72 × 2 = 1 + 0.233 646 392 822 295 227 206 205 44;
  • 35) 0.233 646 392 822 295 227 206 205 44 × 2 = 0 + 0.467 292 785 644 590 454 412 410 88;
  • 36) 0.467 292 785 644 590 454 412 410 88 × 2 = 0 + 0.934 585 571 289 180 908 824 821 76;
  • 37) 0.934 585 571 289 180 908 824 821 76 × 2 = 1 + 0.869 171 142 578 361 817 649 643 52;
  • 38) 0.869 171 142 578 361 817 649 643 52 × 2 = 1 + 0.738 342 285 156 723 635 299 287 04;
  • 39) 0.738 342 285 156 723 635 299 287 04 × 2 = 1 + 0.476 684 570 313 447 270 598 574 08;
  • 40) 0.476 684 570 313 447 270 598 574 08 × 2 = 0 + 0.953 369 140 626 894 541 197 148 16;
  • 41) 0.953 369 140 626 894 541 197 148 16 × 2 = 1 + 0.906 738 281 253 789 082 394 296 32;
  • 42) 0.906 738 281 253 789 082 394 296 32 × 2 = 1 + 0.813 476 562 507 578 164 788 592 64;
  • 43) 0.813 476 562 507 578 164 788 592 64 × 2 = 1 + 0.626 953 125 015 156 329 577 185 28;
  • 44) 0.626 953 125 015 156 329 577 185 28 × 2 = 1 + 0.253 906 250 030 312 659 154 370 56;
  • 45) 0.253 906 250 030 312 659 154 370 56 × 2 = 0 + 0.507 812 500 060 625 318 308 741 12;
  • 46) 0.507 812 500 060 625 318 308 741 12 × 2 = 1 + 0.015 625 000 121 250 636 617 482 24;
  • 47) 0.015 625 000 121 250 636 617 482 24 × 2 = 0 + 0.031 250 000 242 501 273 234 964 48;
  • 48) 0.031 250 000 242 501 273 234 964 48 × 2 = 0 + 0.062 500 000 485 002 546 469 928 96;
  • 49) 0.062 500 000 485 002 546 469 928 96 × 2 = 0 + 0.125 000 000 970 005 092 939 857 92;
  • 50) 0.125 000 000 970 005 092 939 857 92 × 2 = 0 + 0.250 000 001 940 010 185 879 715 84;
  • 51) 0.250 000 001 940 010 185 879 715 84 × 2 = 0 + 0.500 000 003 880 020 371 759 431 68;
  • 52) 0.500 000 003 880 020 371 759 431 68 × 2 = 1 + 0.000 000 007 760 040 743 518 863 36;
  • 53) 0.000 000 007 760 040 743 518 863 36 × 2 = 0 + 0.000 000 015 520 081 487 037 726 72;

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:


0.745 459 324 169 999 826 281 697 91(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 697 91(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:


1.745 459 324 169 999 826 281 697 91(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 1.745 459 324 169 999 826 281 697 91 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