1.745 459 324 169 999 826 281 696 186 924 794 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 924 794 6(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
1.745 459 324 169 999 826 281 696 186 924 794 6(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: 0.745 459 324 169 999 826 281 696 186 924 794 6.

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) 0.745 459 324 169 999 826 281 696 186 924 794 6 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 849 589 2;
  • 2) 0.490 918 648 339 999 652 563 392 373 849 589 2 × 2 = 0 + 0.981 837 296 679 999 305 126 784 747 699 178 4;
  • 3) 0.981 837 296 679 999 305 126 784 747 699 178 4 × 2 = 1 + 0.963 674 593 359 998 610 253 569 495 398 356 8;
  • 4) 0.963 674 593 359 998 610 253 569 495 398 356 8 × 2 = 1 + 0.927 349 186 719 997 220 507 138 990 796 713 6;
  • 5) 0.927 349 186 719 997 220 507 138 990 796 713 6 × 2 = 1 + 0.854 698 373 439 994 441 014 277 981 593 427 2;
  • 6) 0.854 698 373 439 994 441 014 277 981 593 427 2 × 2 = 1 + 0.709 396 746 879 988 882 028 555 963 186 854 4;
  • 7) 0.709 396 746 879 988 882 028 555 963 186 854 4 × 2 = 1 + 0.418 793 493 759 977 764 057 111 926 373 708 8;
  • 8) 0.418 793 493 759 977 764 057 111 926 373 708 8 × 2 = 0 + 0.837 586 987 519 955 528 114 223 852 747 417 6;
  • 9) 0.837 586 987 519 955 528 114 223 852 747 417 6 × 2 = 1 + 0.675 173 975 039 911 056 228 447 705 494 835 2;
  • 10) 0.675 173 975 039 911 056 228 447 705 494 835 2 × 2 = 1 + 0.350 347 950 079 822 112 456 895 410 989 670 4;
  • 11) 0.350 347 950 079 822 112 456 895 410 989 670 4 × 2 = 0 + 0.700 695 900 159 644 224 913 790 821 979 340 8;
  • 12) 0.700 695 900 159 644 224 913 790 821 979 340 8 × 2 = 1 + 0.401 391 800 319 288 449 827 581 643 958 681 6;
  • 13) 0.401 391 800 319 288 449 827 581 643 958 681 6 × 2 = 0 + 0.802 783 600 638 576 899 655 163 287 917 363 2;
  • 14) 0.802 783 600 638 576 899 655 163 287 917 363 2 × 2 = 1 + 0.605 567 201 277 153 799 310 326 575 834 726 4;
  • 15) 0.605 567 201 277 153 799 310 326 575 834 726 4 × 2 = 1 + 0.211 134 402 554 307 598 620 653 151 669 452 8;
  • 16) 0.211 134 402 554 307 598 620 653 151 669 452 8 × 2 = 0 + 0.422 268 805 108 615 197 241 306 303 338 905 6;
  • 17) 0.422 268 805 108 615 197 241 306 303 338 905 6 × 2 = 0 + 0.844 537 610 217 230 394 482 612 606 677 811 2;
  • 18) 0.844 537 610 217 230 394 482 612 606 677 811 2 × 2 = 1 + 0.689 075 220 434 460 788 965 225 213 355 622 4;
  • 19) 0.689 075 220 434 460 788 965 225 213 355 622 4 × 2 = 1 + 0.378 150 440 868 921 577 930 450 426 711 244 8;
  • 20) 0.378 150 440 868 921 577 930 450 426 711 244 8 × 2 = 0 + 0.756 300 881 737 843 155 860 900 853 422 489 6;
  • 21) 0.756 300 881 737 843 155 860 900 853 422 489 6 × 2 = 1 + 0.512 601 763 475 686 311 721 801 706 844 979 2;
  • 22) 0.512 601 763 475 686 311 721 801 706 844 979 2 × 2 = 1 + 0.025 203 526 951 372 623 443 603 413 689 958 4;
  • 23) 0.025 203 526 951 372 623 443 603 413 689 958 4 × 2 = 0 + 0.050 407 053 902 745 246 887 206 827 379 916 8;
  • 24) 0.050 407 053 902 745 246 887 206 827 379 916 8 × 2 = 0 + 0.100 814 107 805 490 493 774 413 654 759 833 6;
  • 25) 0.100 814 107 805 490 493 774 413 654 759 833 6 × 2 = 0 + 0.201 628 215 610 980 987 548 827 309 519 667 2;
  • 26) 0.201 628 215 610 980 987 548 827 309 519 667 2 × 2 = 0 + 0.403 256 431 221 961 975 097 654 619 039 334 4;
  • 27) 0.403 256 431 221 961 975 097 654 619 039 334 4 × 2 = 0 + 0.806 512 862 443 923 950 195 309 238 078 668 8;
  • 28) 0.806 512 862 443 923 950 195 309 238 078 668 8 × 2 = 1 + 0.613 025 724 887 847 900 390 618 476 157 337 6;
  • 29) 0.613 025 724 887 847 900 390 618 476 157 337 6 × 2 = 1 + 0.226 051 449 775 695 800 781 236 952 314 675 2;
  • 30) 0.226 051 449 775 695 800 781 236 952 314 675 2 × 2 = 0 + 0.452 102 899 551 391 601 562 473 904 629 350 4;
  • 31) 0.452 102 899 551 391 601 562 473 904 629 350 4 × 2 = 0 + 0.904 205 799 102 783 203 124 947 809 258 700 8;
  • 32) 0.904 205 799 102 783 203 124 947 809 258 700 8 × 2 = 1 + 0.808 411 598 205 566 406 249 895 618 517 401 6;
  • 33) 0.808 411 598 205 566 406 249 895 618 517 401 6 × 2 = 1 + 0.616 823 196 411 132 812 499 791 237 034 803 2;
  • 34) 0.616 823 196 411 132 812 499 791 237 034 803 2 × 2 = 1 + 0.233 646 392 822 265 624 999 582 474 069 606 4;
  • 35) 0.233 646 392 822 265 624 999 582 474 069 606 4 × 2 = 0 + 0.467 292 785 644 531 249 999 164 948 139 212 8;
  • 36) 0.467 292 785 644 531 249 999 164 948 139 212 8 × 2 = 0 + 0.934 585 571 289 062 499 998 329 896 278 425 6;
  • 37) 0.934 585 571 289 062 499 998 329 896 278 425 6 × 2 = 1 + 0.869 171 142 578 124 999 996 659 792 556 851 2;
  • 38) 0.869 171 142 578 124 999 996 659 792 556 851 2 × 2 = 1 + 0.738 342 285 156 249 999 993 319 585 113 702 4;
  • 39) 0.738 342 285 156 249 999 993 319 585 113 702 4 × 2 = 1 + 0.476 684 570 312 499 999 986 639 170 227 404 8;
  • 40) 0.476 684 570 312 499 999 986 639 170 227 404 8 × 2 = 0 + 0.953 369 140 624 999 999 973 278 340 454 809 6;
  • 41) 0.953 369 140 624 999 999 973 278 340 454 809 6 × 2 = 1 + 0.906 738 281 249 999 999 946 556 680 909 619 2;
  • 42) 0.906 738 281 249 999 999 946 556 680 909 619 2 × 2 = 1 + 0.813 476 562 499 999 999 893 113 361 819 238 4;
  • 43) 0.813 476 562 499 999 999 893 113 361 819 238 4 × 2 = 1 + 0.626 953 124 999 999 999 786 226 723 638 476 8;
  • 44) 0.626 953 124 999 999 999 786 226 723 638 476 8 × 2 = 1 + 0.253 906 249 999 999 999 572 453 447 276 953 6;
  • 45) 0.253 906 249 999 999 999 572 453 447 276 953 6 × 2 = 0 + 0.507 812 499 999 999 999 144 906 894 553 907 2;
  • 46) 0.507 812 499 999 999 999 144 906 894 553 907 2 × 2 = 1 + 0.015 624 999 999 999 998 289 813 789 107 814 4;
  • 47) 0.015 624 999 999 999 998 289 813 789 107 814 4 × 2 = 0 + 0.031 249 999 999 999 996 579 627 578 215 628 8;
  • 48) 0.031 249 999 999 999 996 579 627 578 215 628 8 × 2 = 0 + 0.062 499 999 999 999 993 159 255 156 431 257 6;
  • 49) 0.062 499 999 999 999 993 159 255 156 431 257 6 × 2 = 0 + 0.124 999 999 999 999 986 318 510 312 862 515 2;
  • 50) 0.124 999 999 999 999 986 318 510 312 862 515 2 × 2 = 0 + 0.249 999 999 999 999 972 637 020 625 725 030 4;
  • 51) 0.249 999 999 999 999 972 637 020 625 725 030 4 × 2 = 0 + 0.499 999 999 999 999 945 274 041 251 450 060 8;
  • 52) 0.499 999 999 999 999 945 274 041 251 450 060 8 × 2 = 0 + 0.999 999 999 999 999 890 548 082 502 900 121 6;
  • 53) 0.999 999 999 999 999 890 548 082 502 900 121 6 × 2 = 1 + 0.999 999 999 999 999 781 096 165 005 800 243 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:


0.745 459 324 169 999 826 281 696 186 924 794 6(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 924 794 6(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:


1.745 459 324 169 999 826 281 696 186 924 794 6(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 1.745 459 324 169 999 826 281 696 186 924 794 6 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