1.745 459 324 169 999 826 281 506 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 506(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
1.745 459 324 169 999 826 281 506(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: 0.745 459 324 169 999 826 281 506.

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) 0.745 459 324 169 999 826 281 506 × 2 = 1 + 0.490 918 648 339 999 652 563 012;
  • 2) 0.490 918 648 339 999 652 563 012 × 2 = 0 + 0.981 837 296 679 999 305 126 024;
  • 3) 0.981 837 296 679 999 305 126 024 × 2 = 1 + 0.963 674 593 359 998 610 252 048;
  • 4) 0.963 674 593 359 998 610 252 048 × 2 = 1 + 0.927 349 186 719 997 220 504 096;
  • 5) 0.927 349 186 719 997 220 504 096 × 2 = 1 + 0.854 698 373 439 994 441 008 192;
  • 6) 0.854 698 373 439 994 441 008 192 × 2 = 1 + 0.709 396 746 879 988 882 016 384;
  • 7) 0.709 396 746 879 988 882 016 384 × 2 = 1 + 0.418 793 493 759 977 764 032 768;
  • 8) 0.418 793 493 759 977 764 032 768 × 2 = 0 + 0.837 586 987 519 955 528 065 536;
  • 9) 0.837 586 987 519 955 528 065 536 × 2 = 1 + 0.675 173 975 039 911 056 131 072;
  • 10) 0.675 173 975 039 911 056 131 072 × 2 = 1 + 0.350 347 950 079 822 112 262 144;
  • 11) 0.350 347 950 079 822 112 262 144 × 2 = 0 + 0.700 695 900 159 644 224 524 288;
  • 12) 0.700 695 900 159 644 224 524 288 × 2 = 1 + 0.401 391 800 319 288 449 048 576;
  • 13) 0.401 391 800 319 288 449 048 576 × 2 = 0 + 0.802 783 600 638 576 898 097 152;
  • 14) 0.802 783 600 638 576 898 097 152 × 2 = 1 + 0.605 567 201 277 153 796 194 304;
  • 15) 0.605 567 201 277 153 796 194 304 × 2 = 1 + 0.211 134 402 554 307 592 388 608;
  • 16) 0.211 134 402 554 307 592 388 608 × 2 = 0 + 0.422 268 805 108 615 184 777 216;
  • 17) 0.422 268 805 108 615 184 777 216 × 2 = 0 + 0.844 537 610 217 230 369 554 432;
  • 18) 0.844 537 610 217 230 369 554 432 × 2 = 1 + 0.689 075 220 434 460 739 108 864;
  • 19) 0.689 075 220 434 460 739 108 864 × 2 = 1 + 0.378 150 440 868 921 478 217 728;
  • 20) 0.378 150 440 868 921 478 217 728 × 2 = 0 + 0.756 300 881 737 842 956 435 456;
  • 21) 0.756 300 881 737 842 956 435 456 × 2 = 1 + 0.512 601 763 475 685 912 870 912;
  • 22) 0.512 601 763 475 685 912 870 912 × 2 = 1 + 0.025 203 526 951 371 825 741 824;
  • 23) 0.025 203 526 951 371 825 741 824 × 2 = 0 + 0.050 407 053 902 743 651 483 648;
  • 24) 0.050 407 053 902 743 651 483 648 × 2 = 0 + 0.100 814 107 805 487 302 967 296;
  • 25) 0.100 814 107 805 487 302 967 296 × 2 = 0 + 0.201 628 215 610 974 605 934 592;
  • 26) 0.201 628 215 610 974 605 934 592 × 2 = 0 + 0.403 256 431 221 949 211 869 184;
  • 27) 0.403 256 431 221 949 211 869 184 × 2 = 0 + 0.806 512 862 443 898 423 738 368;
  • 28) 0.806 512 862 443 898 423 738 368 × 2 = 1 + 0.613 025 724 887 796 847 476 736;
  • 29) 0.613 025 724 887 796 847 476 736 × 2 = 1 + 0.226 051 449 775 593 694 953 472;
  • 30) 0.226 051 449 775 593 694 953 472 × 2 = 0 + 0.452 102 899 551 187 389 906 944;
  • 31) 0.452 102 899 551 187 389 906 944 × 2 = 0 + 0.904 205 799 102 374 779 813 888;
  • 32) 0.904 205 799 102 374 779 813 888 × 2 = 1 + 0.808 411 598 204 749 559 627 776;
  • 33) 0.808 411 598 204 749 559 627 776 × 2 = 1 + 0.616 823 196 409 499 119 255 552;
  • 34) 0.616 823 196 409 499 119 255 552 × 2 = 1 + 0.233 646 392 818 998 238 511 104;
  • 35) 0.233 646 392 818 998 238 511 104 × 2 = 0 + 0.467 292 785 637 996 477 022 208;
  • 36) 0.467 292 785 637 996 477 022 208 × 2 = 0 + 0.934 585 571 275 992 954 044 416;
  • 37) 0.934 585 571 275 992 954 044 416 × 2 = 1 + 0.869 171 142 551 985 908 088 832;
  • 38) 0.869 171 142 551 985 908 088 832 × 2 = 1 + 0.738 342 285 103 971 816 177 664;
  • 39) 0.738 342 285 103 971 816 177 664 × 2 = 1 + 0.476 684 570 207 943 632 355 328;
  • 40) 0.476 684 570 207 943 632 355 328 × 2 = 0 + 0.953 369 140 415 887 264 710 656;
  • 41) 0.953 369 140 415 887 264 710 656 × 2 = 1 + 0.906 738 280 831 774 529 421 312;
  • 42) 0.906 738 280 831 774 529 421 312 × 2 = 1 + 0.813 476 561 663 549 058 842 624;
  • 43) 0.813 476 561 663 549 058 842 624 × 2 = 1 + 0.626 953 123 327 098 117 685 248;
  • 44) 0.626 953 123 327 098 117 685 248 × 2 = 1 + 0.253 906 246 654 196 235 370 496;
  • 45) 0.253 906 246 654 196 235 370 496 × 2 = 0 + 0.507 812 493 308 392 470 740 992;
  • 46) 0.507 812 493 308 392 470 740 992 × 2 = 1 + 0.015 624 986 616 784 941 481 984;
  • 47) 0.015 624 986 616 784 941 481 984 × 2 = 0 + 0.031 249 973 233 569 882 963 968;
  • 48) 0.031 249 973 233 569 882 963 968 × 2 = 0 + 0.062 499 946 467 139 765 927 936;
  • 49) 0.062 499 946 467 139 765 927 936 × 2 = 0 + 0.124 999 892 934 279 531 855 872;
  • 50) 0.124 999 892 934 279 531 855 872 × 2 = 0 + 0.249 999 785 868 559 063 711 744;
  • 51) 0.249 999 785 868 559 063 711 744 × 2 = 0 + 0.499 999 571 737 118 127 423 488;
  • 52) 0.499 999 571 737 118 127 423 488 × 2 = 0 + 0.999 999 143 474 236 254 846 976;
  • 53) 0.999 999 143 474 236 254 846 976 × 2 = 1 + 0.999 998 286 948 472 509 693 952;

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:


0.745 459 324 169 999 826 281 506(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 506(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:


1.745 459 324 169 999 826 281 506(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 1.745 459 324 169 999 826 281 506 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