1.745 459 324 169 999 826 281 696 153 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 696 153(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
1.745 459 324 169 999 826 281 696 153(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: 0.745 459 324 169 999 826 281 696 153.

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) 0.745 459 324 169 999 826 281 696 153 × 2 = 1 + 0.490 918 648 339 999 652 563 392 306;
  • 2) 0.490 918 648 339 999 652 563 392 306 × 2 = 0 + 0.981 837 296 679 999 305 126 784 612;
  • 3) 0.981 837 296 679 999 305 126 784 612 × 2 = 1 + 0.963 674 593 359 998 610 253 569 224;
  • 4) 0.963 674 593 359 998 610 253 569 224 × 2 = 1 + 0.927 349 186 719 997 220 507 138 448;
  • 5) 0.927 349 186 719 997 220 507 138 448 × 2 = 1 + 0.854 698 373 439 994 441 014 276 896;
  • 6) 0.854 698 373 439 994 441 014 276 896 × 2 = 1 + 0.709 396 746 879 988 882 028 553 792;
  • 7) 0.709 396 746 879 988 882 028 553 792 × 2 = 1 + 0.418 793 493 759 977 764 057 107 584;
  • 8) 0.418 793 493 759 977 764 057 107 584 × 2 = 0 + 0.837 586 987 519 955 528 114 215 168;
  • 9) 0.837 586 987 519 955 528 114 215 168 × 2 = 1 + 0.675 173 975 039 911 056 228 430 336;
  • 10) 0.675 173 975 039 911 056 228 430 336 × 2 = 1 + 0.350 347 950 079 822 112 456 860 672;
  • 11) 0.350 347 950 079 822 112 456 860 672 × 2 = 0 + 0.700 695 900 159 644 224 913 721 344;
  • 12) 0.700 695 900 159 644 224 913 721 344 × 2 = 1 + 0.401 391 800 319 288 449 827 442 688;
  • 13) 0.401 391 800 319 288 449 827 442 688 × 2 = 0 + 0.802 783 600 638 576 899 654 885 376;
  • 14) 0.802 783 600 638 576 899 654 885 376 × 2 = 1 + 0.605 567 201 277 153 799 309 770 752;
  • 15) 0.605 567 201 277 153 799 309 770 752 × 2 = 1 + 0.211 134 402 554 307 598 619 541 504;
  • 16) 0.211 134 402 554 307 598 619 541 504 × 2 = 0 + 0.422 268 805 108 615 197 239 083 008;
  • 17) 0.422 268 805 108 615 197 239 083 008 × 2 = 0 + 0.844 537 610 217 230 394 478 166 016;
  • 18) 0.844 537 610 217 230 394 478 166 016 × 2 = 1 + 0.689 075 220 434 460 788 956 332 032;
  • 19) 0.689 075 220 434 460 788 956 332 032 × 2 = 1 + 0.378 150 440 868 921 577 912 664 064;
  • 20) 0.378 150 440 868 921 577 912 664 064 × 2 = 0 + 0.756 300 881 737 843 155 825 328 128;
  • 21) 0.756 300 881 737 843 155 825 328 128 × 2 = 1 + 0.512 601 763 475 686 311 650 656 256;
  • 22) 0.512 601 763 475 686 311 650 656 256 × 2 = 1 + 0.025 203 526 951 372 623 301 312 512;
  • 23) 0.025 203 526 951 372 623 301 312 512 × 2 = 0 + 0.050 407 053 902 745 246 602 625 024;
  • 24) 0.050 407 053 902 745 246 602 625 024 × 2 = 0 + 0.100 814 107 805 490 493 205 250 048;
  • 25) 0.100 814 107 805 490 493 205 250 048 × 2 = 0 + 0.201 628 215 610 980 986 410 500 096;
  • 26) 0.201 628 215 610 980 986 410 500 096 × 2 = 0 + 0.403 256 431 221 961 972 821 000 192;
  • 27) 0.403 256 431 221 961 972 821 000 192 × 2 = 0 + 0.806 512 862 443 923 945 642 000 384;
  • 28) 0.806 512 862 443 923 945 642 000 384 × 2 = 1 + 0.613 025 724 887 847 891 284 000 768;
  • 29) 0.613 025 724 887 847 891 284 000 768 × 2 = 1 + 0.226 051 449 775 695 782 568 001 536;
  • 30) 0.226 051 449 775 695 782 568 001 536 × 2 = 0 + 0.452 102 899 551 391 565 136 003 072;
  • 31) 0.452 102 899 551 391 565 136 003 072 × 2 = 0 + 0.904 205 799 102 783 130 272 006 144;
  • 32) 0.904 205 799 102 783 130 272 006 144 × 2 = 1 + 0.808 411 598 205 566 260 544 012 288;
  • 33) 0.808 411 598 205 566 260 544 012 288 × 2 = 1 + 0.616 823 196 411 132 521 088 024 576;
  • 34) 0.616 823 196 411 132 521 088 024 576 × 2 = 1 + 0.233 646 392 822 265 042 176 049 152;
  • 35) 0.233 646 392 822 265 042 176 049 152 × 2 = 0 + 0.467 292 785 644 530 084 352 098 304;
  • 36) 0.467 292 785 644 530 084 352 098 304 × 2 = 0 + 0.934 585 571 289 060 168 704 196 608;
  • 37) 0.934 585 571 289 060 168 704 196 608 × 2 = 1 + 0.869 171 142 578 120 337 408 393 216;
  • 38) 0.869 171 142 578 120 337 408 393 216 × 2 = 1 + 0.738 342 285 156 240 674 816 786 432;
  • 39) 0.738 342 285 156 240 674 816 786 432 × 2 = 1 + 0.476 684 570 312 481 349 633 572 864;
  • 40) 0.476 684 570 312 481 349 633 572 864 × 2 = 0 + 0.953 369 140 624 962 699 267 145 728;
  • 41) 0.953 369 140 624 962 699 267 145 728 × 2 = 1 + 0.906 738 281 249 925 398 534 291 456;
  • 42) 0.906 738 281 249 925 398 534 291 456 × 2 = 1 + 0.813 476 562 499 850 797 068 582 912;
  • 43) 0.813 476 562 499 850 797 068 582 912 × 2 = 1 + 0.626 953 124 999 701 594 137 165 824;
  • 44) 0.626 953 124 999 701 594 137 165 824 × 2 = 1 + 0.253 906 249 999 403 188 274 331 648;
  • 45) 0.253 906 249 999 403 188 274 331 648 × 2 = 0 + 0.507 812 499 998 806 376 548 663 296;
  • 46) 0.507 812 499 998 806 376 548 663 296 × 2 = 1 + 0.015 624 999 997 612 753 097 326 592;
  • 47) 0.015 624 999 997 612 753 097 326 592 × 2 = 0 + 0.031 249 999 995 225 506 194 653 184;
  • 48) 0.031 249 999 995 225 506 194 653 184 × 2 = 0 + 0.062 499 999 990 451 012 389 306 368;
  • 49) 0.062 499 999 990 451 012 389 306 368 × 2 = 0 + 0.124 999 999 980 902 024 778 612 736;
  • 50) 0.124 999 999 980 902 024 778 612 736 × 2 = 0 + 0.249 999 999 961 804 049 557 225 472;
  • 51) 0.249 999 999 961 804 049 557 225 472 × 2 = 0 + 0.499 999 999 923 608 099 114 450 944;
  • 52) 0.499 999 999 923 608 099 114 450 944 × 2 = 0 + 0.999 999 999 847 216 198 228 901 888;
  • 53) 0.999 999 999 847 216 198 228 901 888 × 2 = 1 + 0.999 999 999 694 432 396 457 803 776;

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:


0.745 459 324 169 999 826 281 696 153(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 696 153(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:


1.745 459 324 169 999 826 281 696 153(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 1.745 459 324 169 999 826 281 696 153 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