1.797 693 134 862 315 708 145 274 35 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.797 693 134 862 315 708 145 274 35(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
1.797 693 134 862 315 708 145 274 35(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: 0.797 693 134 862 315 708 145 274 35.

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) 0.797 693 134 862 315 708 145 274 35 × 2 = 1 + 0.595 386 269 724 631 416 290 548 7;
  • 2) 0.595 386 269 724 631 416 290 548 7 × 2 = 1 + 0.190 772 539 449 262 832 581 097 4;
  • 3) 0.190 772 539 449 262 832 581 097 4 × 2 = 0 + 0.381 545 078 898 525 665 162 194 8;
  • 4) 0.381 545 078 898 525 665 162 194 8 × 2 = 0 + 0.763 090 157 797 051 330 324 389 6;
  • 5) 0.763 090 157 797 051 330 324 389 6 × 2 = 1 + 0.526 180 315 594 102 660 648 779 2;
  • 6) 0.526 180 315 594 102 660 648 779 2 × 2 = 1 + 0.052 360 631 188 205 321 297 558 4;
  • 7) 0.052 360 631 188 205 321 297 558 4 × 2 = 0 + 0.104 721 262 376 410 642 595 116 8;
  • 8) 0.104 721 262 376 410 642 595 116 8 × 2 = 0 + 0.209 442 524 752 821 285 190 233 6;
  • 9) 0.209 442 524 752 821 285 190 233 6 × 2 = 0 + 0.418 885 049 505 642 570 380 467 2;
  • 10) 0.418 885 049 505 642 570 380 467 2 × 2 = 0 + 0.837 770 099 011 285 140 760 934 4;
  • 11) 0.837 770 099 011 285 140 760 934 4 × 2 = 1 + 0.675 540 198 022 570 281 521 868 8;
  • 12) 0.675 540 198 022 570 281 521 868 8 × 2 = 1 + 0.351 080 396 045 140 563 043 737 6;
  • 13) 0.351 080 396 045 140 563 043 737 6 × 2 = 0 + 0.702 160 792 090 281 126 087 475 2;
  • 14) 0.702 160 792 090 281 126 087 475 2 × 2 = 1 + 0.404 321 584 180 562 252 174 950 4;
  • 15) 0.404 321 584 180 562 252 174 950 4 × 2 = 0 + 0.808 643 168 361 124 504 349 900 8;
  • 16) 0.808 643 168 361 124 504 349 900 8 × 2 = 1 + 0.617 286 336 722 249 008 699 801 6;
  • 17) 0.617 286 336 722 249 008 699 801 6 × 2 = 1 + 0.234 572 673 444 498 017 399 603 2;
  • 18) 0.234 572 673 444 498 017 399 603 2 × 2 = 0 + 0.469 145 346 888 996 034 799 206 4;
  • 19) 0.469 145 346 888 996 034 799 206 4 × 2 = 0 + 0.938 290 693 777 992 069 598 412 8;
  • 20) 0.938 290 693 777 992 069 598 412 8 × 2 = 1 + 0.876 581 387 555 984 139 196 825 6;
  • 21) 0.876 581 387 555 984 139 196 825 6 × 2 = 1 + 0.753 162 775 111 968 278 393 651 2;
  • 22) 0.753 162 775 111 968 278 393 651 2 × 2 = 1 + 0.506 325 550 223 936 556 787 302 4;
  • 23) 0.506 325 550 223 936 556 787 302 4 × 2 = 1 + 0.012 651 100 447 873 113 574 604 8;
  • 24) 0.012 651 100 447 873 113 574 604 8 × 2 = 0 + 0.025 302 200 895 746 227 149 209 6;
  • 25) 0.025 302 200 895 746 227 149 209 6 × 2 = 0 + 0.050 604 401 791 492 454 298 419 2;
  • 26) 0.050 604 401 791 492 454 298 419 2 × 2 = 0 + 0.101 208 803 582 984 908 596 838 4;
  • 27) 0.101 208 803 582 984 908 596 838 4 × 2 = 0 + 0.202 417 607 165 969 817 193 676 8;
  • 28) 0.202 417 607 165 969 817 193 676 8 × 2 = 0 + 0.404 835 214 331 939 634 387 353 6;
  • 29) 0.404 835 214 331 939 634 387 353 6 × 2 = 0 + 0.809 670 428 663 879 268 774 707 2;
  • 30) 0.809 670 428 663 879 268 774 707 2 × 2 = 1 + 0.619 340 857 327 758 537 549 414 4;
  • 31) 0.619 340 857 327 758 537 549 414 4 × 2 = 1 + 0.238 681 714 655 517 075 098 828 8;
  • 32) 0.238 681 714 655 517 075 098 828 8 × 2 = 0 + 0.477 363 429 311 034 150 197 657 6;
  • 33) 0.477 363 429 311 034 150 197 657 6 × 2 = 0 + 0.954 726 858 622 068 300 395 315 2;
  • 34) 0.954 726 858 622 068 300 395 315 2 × 2 = 1 + 0.909 453 717 244 136 600 790 630 4;
  • 35) 0.909 453 717 244 136 600 790 630 4 × 2 = 1 + 0.818 907 434 488 273 201 581 260 8;
  • 36) 0.818 907 434 488 273 201 581 260 8 × 2 = 1 + 0.637 814 868 976 546 403 162 521 6;
  • 37) 0.637 814 868 976 546 403 162 521 6 × 2 = 1 + 0.275 629 737 953 092 806 325 043 2;
  • 38) 0.275 629 737 953 092 806 325 043 2 × 2 = 0 + 0.551 259 475 906 185 612 650 086 4;
  • 39) 0.551 259 475 906 185 612 650 086 4 × 2 = 1 + 0.102 518 951 812 371 225 300 172 8;
  • 40) 0.102 518 951 812 371 225 300 172 8 × 2 = 0 + 0.205 037 903 624 742 450 600 345 6;
  • 41) 0.205 037 903 624 742 450 600 345 6 × 2 = 0 + 0.410 075 807 249 484 901 200 691 2;
  • 42) 0.410 075 807 249 484 901 200 691 2 × 2 = 0 + 0.820 151 614 498 969 802 401 382 4;
  • 43) 0.820 151 614 498 969 802 401 382 4 × 2 = 1 + 0.640 303 228 997 939 604 802 764 8;
  • 44) 0.640 303 228 997 939 604 802 764 8 × 2 = 1 + 0.280 606 457 995 879 209 605 529 6;
  • 45) 0.280 606 457 995 879 209 605 529 6 × 2 = 0 + 0.561 212 915 991 758 419 211 059 2;
  • 46) 0.561 212 915 991 758 419 211 059 2 × 2 = 1 + 0.122 425 831 983 516 838 422 118 4;
  • 47) 0.122 425 831 983 516 838 422 118 4 × 2 = 0 + 0.244 851 663 967 033 676 844 236 8;
  • 48) 0.244 851 663 967 033 676 844 236 8 × 2 = 0 + 0.489 703 327 934 067 353 688 473 6;
  • 49) 0.489 703 327 934 067 353 688 473 6 × 2 = 0 + 0.979 406 655 868 134 707 376 947 2;
  • 50) 0.979 406 655 868 134 707 376 947 2 × 2 = 1 + 0.958 813 311 736 269 414 753 894 4;
  • 51) 0.958 813 311 736 269 414 753 894 4 × 2 = 1 + 0.917 626 623 472 538 829 507 788 8;
  • 52) 0.917 626 623 472 538 829 507 788 8 × 2 = 1 + 0.835 253 246 945 077 659 015 577 6;
  • 53) 0.835 253 246 945 077 659 015 577 6 × 2 = 1 + 0.670 506 493 890 155 318 031 155 2;

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:


0.797 693 134 862 315 708 145 274 35(10) =

0.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111 1(2)

5. Positive number before normalization:

1.797 693 134 862 315 708 145 274 35(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111 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:


1.797 693 134 862 315 708 145 274 35(10) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111 1(2) =

1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111 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):
1.1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111 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. 1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111 1 =

1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111


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) =
1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111


Decimal number 1.797 693 134 862 315 708 145 274 35 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1111 - 1100 1100 0011 0101 1001 1110 0000 0110 0111 1010 0011 0100 0111

» Convert decimal 1.797 693 134 862 315 708 145 275 19 to 64 bit double precision IEEE 754 binary floating point representation standard

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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