1.745 459 324 169 999 826 281 696 186 578 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 186 578(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
1.745 459 324 169 999 826 281 696 186 578(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: 0.745 459 324 169 999 826 281 696 186 578.

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) 0.745 459 324 169 999 826 281 696 186 578 × 2 = 1 + 0.490 918 648 339 999 652 563 392 373 156;
  • 2) 0.490 918 648 339 999 652 563 392 373 156 × 2 = 0 + 0.981 837 296 679 999 305 126 784 746 312;
  • 3) 0.981 837 296 679 999 305 126 784 746 312 × 2 = 1 + 0.963 674 593 359 998 610 253 569 492 624;
  • 4) 0.963 674 593 359 998 610 253 569 492 624 × 2 = 1 + 0.927 349 186 719 997 220 507 138 985 248;
  • 5) 0.927 349 186 719 997 220 507 138 985 248 × 2 = 1 + 0.854 698 373 439 994 441 014 277 970 496;
  • 6) 0.854 698 373 439 994 441 014 277 970 496 × 2 = 1 + 0.709 396 746 879 988 882 028 555 940 992;
  • 7) 0.709 396 746 879 988 882 028 555 940 992 × 2 = 1 + 0.418 793 493 759 977 764 057 111 881 984;
  • 8) 0.418 793 493 759 977 764 057 111 881 984 × 2 = 0 + 0.837 586 987 519 955 528 114 223 763 968;
  • 9) 0.837 586 987 519 955 528 114 223 763 968 × 2 = 1 + 0.675 173 975 039 911 056 228 447 527 936;
  • 10) 0.675 173 975 039 911 056 228 447 527 936 × 2 = 1 + 0.350 347 950 079 822 112 456 895 055 872;
  • 11) 0.350 347 950 079 822 112 456 895 055 872 × 2 = 0 + 0.700 695 900 159 644 224 913 790 111 744;
  • 12) 0.700 695 900 159 644 224 913 790 111 744 × 2 = 1 + 0.401 391 800 319 288 449 827 580 223 488;
  • 13) 0.401 391 800 319 288 449 827 580 223 488 × 2 = 0 + 0.802 783 600 638 576 899 655 160 446 976;
  • 14) 0.802 783 600 638 576 899 655 160 446 976 × 2 = 1 + 0.605 567 201 277 153 799 310 320 893 952;
  • 15) 0.605 567 201 277 153 799 310 320 893 952 × 2 = 1 + 0.211 134 402 554 307 598 620 641 787 904;
  • 16) 0.211 134 402 554 307 598 620 641 787 904 × 2 = 0 + 0.422 268 805 108 615 197 241 283 575 808;
  • 17) 0.422 268 805 108 615 197 241 283 575 808 × 2 = 0 + 0.844 537 610 217 230 394 482 567 151 616;
  • 18) 0.844 537 610 217 230 394 482 567 151 616 × 2 = 1 + 0.689 075 220 434 460 788 965 134 303 232;
  • 19) 0.689 075 220 434 460 788 965 134 303 232 × 2 = 1 + 0.378 150 440 868 921 577 930 268 606 464;
  • 20) 0.378 150 440 868 921 577 930 268 606 464 × 2 = 0 + 0.756 300 881 737 843 155 860 537 212 928;
  • 21) 0.756 300 881 737 843 155 860 537 212 928 × 2 = 1 + 0.512 601 763 475 686 311 721 074 425 856;
  • 22) 0.512 601 763 475 686 311 721 074 425 856 × 2 = 1 + 0.025 203 526 951 372 623 442 148 851 712;
  • 23) 0.025 203 526 951 372 623 442 148 851 712 × 2 = 0 + 0.050 407 053 902 745 246 884 297 703 424;
  • 24) 0.050 407 053 902 745 246 884 297 703 424 × 2 = 0 + 0.100 814 107 805 490 493 768 595 406 848;
  • 25) 0.100 814 107 805 490 493 768 595 406 848 × 2 = 0 + 0.201 628 215 610 980 987 537 190 813 696;
  • 26) 0.201 628 215 610 980 987 537 190 813 696 × 2 = 0 + 0.403 256 431 221 961 975 074 381 627 392;
  • 27) 0.403 256 431 221 961 975 074 381 627 392 × 2 = 0 + 0.806 512 862 443 923 950 148 763 254 784;
  • 28) 0.806 512 862 443 923 950 148 763 254 784 × 2 = 1 + 0.613 025 724 887 847 900 297 526 509 568;
  • 29) 0.613 025 724 887 847 900 297 526 509 568 × 2 = 1 + 0.226 051 449 775 695 800 595 053 019 136;
  • 30) 0.226 051 449 775 695 800 595 053 019 136 × 2 = 0 + 0.452 102 899 551 391 601 190 106 038 272;
  • 31) 0.452 102 899 551 391 601 190 106 038 272 × 2 = 0 + 0.904 205 799 102 783 202 380 212 076 544;
  • 32) 0.904 205 799 102 783 202 380 212 076 544 × 2 = 1 + 0.808 411 598 205 566 404 760 424 153 088;
  • 33) 0.808 411 598 205 566 404 760 424 153 088 × 2 = 1 + 0.616 823 196 411 132 809 520 848 306 176;
  • 34) 0.616 823 196 411 132 809 520 848 306 176 × 2 = 1 + 0.233 646 392 822 265 619 041 696 612 352;
  • 35) 0.233 646 392 822 265 619 041 696 612 352 × 2 = 0 + 0.467 292 785 644 531 238 083 393 224 704;
  • 36) 0.467 292 785 644 531 238 083 393 224 704 × 2 = 0 + 0.934 585 571 289 062 476 166 786 449 408;
  • 37) 0.934 585 571 289 062 476 166 786 449 408 × 2 = 1 + 0.869 171 142 578 124 952 333 572 898 816;
  • 38) 0.869 171 142 578 124 952 333 572 898 816 × 2 = 1 + 0.738 342 285 156 249 904 667 145 797 632;
  • 39) 0.738 342 285 156 249 904 667 145 797 632 × 2 = 1 + 0.476 684 570 312 499 809 334 291 595 264;
  • 40) 0.476 684 570 312 499 809 334 291 595 264 × 2 = 0 + 0.953 369 140 624 999 618 668 583 190 528;
  • 41) 0.953 369 140 624 999 618 668 583 190 528 × 2 = 1 + 0.906 738 281 249 999 237 337 166 381 056;
  • 42) 0.906 738 281 249 999 237 337 166 381 056 × 2 = 1 + 0.813 476 562 499 998 474 674 332 762 112;
  • 43) 0.813 476 562 499 998 474 674 332 762 112 × 2 = 1 + 0.626 953 124 999 996 949 348 665 524 224;
  • 44) 0.626 953 124 999 996 949 348 665 524 224 × 2 = 1 + 0.253 906 249 999 993 898 697 331 048 448;
  • 45) 0.253 906 249 999 993 898 697 331 048 448 × 2 = 0 + 0.507 812 499 999 987 797 394 662 096 896;
  • 46) 0.507 812 499 999 987 797 394 662 096 896 × 2 = 1 + 0.015 624 999 999 975 594 789 324 193 792;
  • 47) 0.015 624 999 999 975 594 789 324 193 792 × 2 = 0 + 0.031 249 999 999 951 189 578 648 387 584;
  • 48) 0.031 249 999 999 951 189 578 648 387 584 × 2 = 0 + 0.062 499 999 999 902 379 157 296 775 168;
  • 49) 0.062 499 999 999 902 379 157 296 775 168 × 2 = 0 + 0.124 999 999 999 804 758 314 593 550 336;
  • 50) 0.124 999 999 999 804 758 314 593 550 336 × 2 = 0 + 0.249 999 999 999 609 516 629 187 100 672;
  • 51) 0.249 999 999 999 609 516 629 187 100 672 × 2 = 0 + 0.499 999 999 999 219 033 258 374 201 344;
  • 52) 0.499 999 999 999 219 033 258 374 201 344 × 2 = 0 + 0.999 999 999 998 438 066 516 748 402 688;
  • 53) 0.999 999 999 998 438 066 516 748 402 688 × 2 = 1 + 0.999 999 999 996 876 133 033 496 805 376;

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:


0.745 459 324 169 999 826 281 696 186 578(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 186 578(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:


1.745 459 324 169 999 826 281 696 186 578(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 1.745 459 324 169 999 826 281 696 186 578 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