1.745 459 324 169 999 826 281 695 41 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.745 459 324 169 999 826 281 695 41(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
1.745 459 324 169 999 826 281 695 41(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: 0.745 459 324 169 999 826 281 695 41.

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) 0.745 459 324 169 999 826 281 695 41 × 2 = 1 + 0.490 918 648 339 999 652 563 390 82;
  • 2) 0.490 918 648 339 999 652 563 390 82 × 2 = 0 + 0.981 837 296 679 999 305 126 781 64;
  • 3) 0.981 837 296 679 999 305 126 781 64 × 2 = 1 + 0.963 674 593 359 998 610 253 563 28;
  • 4) 0.963 674 593 359 998 610 253 563 28 × 2 = 1 + 0.927 349 186 719 997 220 507 126 56;
  • 5) 0.927 349 186 719 997 220 507 126 56 × 2 = 1 + 0.854 698 373 439 994 441 014 253 12;
  • 6) 0.854 698 373 439 994 441 014 253 12 × 2 = 1 + 0.709 396 746 879 988 882 028 506 24;
  • 7) 0.709 396 746 879 988 882 028 506 24 × 2 = 1 + 0.418 793 493 759 977 764 057 012 48;
  • 8) 0.418 793 493 759 977 764 057 012 48 × 2 = 0 + 0.837 586 987 519 955 528 114 024 96;
  • 9) 0.837 586 987 519 955 528 114 024 96 × 2 = 1 + 0.675 173 975 039 911 056 228 049 92;
  • 10) 0.675 173 975 039 911 056 228 049 92 × 2 = 1 + 0.350 347 950 079 822 112 456 099 84;
  • 11) 0.350 347 950 079 822 112 456 099 84 × 2 = 0 + 0.700 695 900 159 644 224 912 199 68;
  • 12) 0.700 695 900 159 644 224 912 199 68 × 2 = 1 + 0.401 391 800 319 288 449 824 399 36;
  • 13) 0.401 391 800 319 288 449 824 399 36 × 2 = 0 + 0.802 783 600 638 576 899 648 798 72;
  • 14) 0.802 783 600 638 576 899 648 798 72 × 2 = 1 + 0.605 567 201 277 153 799 297 597 44;
  • 15) 0.605 567 201 277 153 799 297 597 44 × 2 = 1 + 0.211 134 402 554 307 598 595 194 88;
  • 16) 0.211 134 402 554 307 598 595 194 88 × 2 = 0 + 0.422 268 805 108 615 197 190 389 76;
  • 17) 0.422 268 805 108 615 197 190 389 76 × 2 = 0 + 0.844 537 610 217 230 394 380 779 52;
  • 18) 0.844 537 610 217 230 394 380 779 52 × 2 = 1 + 0.689 075 220 434 460 788 761 559 04;
  • 19) 0.689 075 220 434 460 788 761 559 04 × 2 = 1 + 0.378 150 440 868 921 577 523 118 08;
  • 20) 0.378 150 440 868 921 577 523 118 08 × 2 = 0 + 0.756 300 881 737 843 155 046 236 16;
  • 21) 0.756 300 881 737 843 155 046 236 16 × 2 = 1 + 0.512 601 763 475 686 310 092 472 32;
  • 22) 0.512 601 763 475 686 310 092 472 32 × 2 = 1 + 0.025 203 526 951 372 620 184 944 64;
  • 23) 0.025 203 526 951 372 620 184 944 64 × 2 = 0 + 0.050 407 053 902 745 240 369 889 28;
  • 24) 0.050 407 053 902 745 240 369 889 28 × 2 = 0 + 0.100 814 107 805 490 480 739 778 56;
  • 25) 0.100 814 107 805 490 480 739 778 56 × 2 = 0 + 0.201 628 215 610 980 961 479 557 12;
  • 26) 0.201 628 215 610 980 961 479 557 12 × 2 = 0 + 0.403 256 431 221 961 922 959 114 24;
  • 27) 0.403 256 431 221 961 922 959 114 24 × 2 = 0 + 0.806 512 862 443 923 845 918 228 48;
  • 28) 0.806 512 862 443 923 845 918 228 48 × 2 = 1 + 0.613 025 724 887 847 691 836 456 96;
  • 29) 0.613 025 724 887 847 691 836 456 96 × 2 = 1 + 0.226 051 449 775 695 383 672 913 92;
  • 30) 0.226 051 449 775 695 383 672 913 92 × 2 = 0 + 0.452 102 899 551 390 767 345 827 84;
  • 31) 0.452 102 899 551 390 767 345 827 84 × 2 = 0 + 0.904 205 799 102 781 534 691 655 68;
  • 32) 0.904 205 799 102 781 534 691 655 68 × 2 = 1 + 0.808 411 598 205 563 069 383 311 36;
  • 33) 0.808 411 598 205 563 069 383 311 36 × 2 = 1 + 0.616 823 196 411 126 138 766 622 72;
  • 34) 0.616 823 196 411 126 138 766 622 72 × 2 = 1 + 0.233 646 392 822 252 277 533 245 44;
  • 35) 0.233 646 392 822 252 277 533 245 44 × 2 = 0 + 0.467 292 785 644 504 555 066 490 88;
  • 36) 0.467 292 785 644 504 555 066 490 88 × 2 = 0 + 0.934 585 571 289 009 110 132 981 76;
  • 37) 0.934 585 571 289 009 110 132 981 76 × 2 = 1 + 0.869 171 142 578 018 220 265 963 52;
  • 38) 0.869 171 142 578 018 220 265 963 52 × 2 = 1 + 0.738 342 285 156 036 440 531 927 04;
  • 39) 0.738 342 285 156 036 440 531 927 04 × 2 = 1 + 0.476 684 570 312 072 881 063 854 08;
  • 40) 0.476 684 570 312 072 881 063 854 08 × 2 = 0 + 0.953 369 140 624 145 762 127 708 16;
  • 41) 0.953 369 140 624 145 762 127 708 16 × 2 = 1 + 0.906 738 281 248 291 524 255 416 32;
  • 42) 0.906 738 281 248 291 524 255 416 32 × 2 = 1 + 0.813 476 562 496 583 048 510 832 64;
  • 43) 0.813 476 562 496 583 048 510 832 64 × 2 = 1 + 0.626 953 124 993 166 097 021 665 28;
  • 44) 0.626 953 124 993 166 097 021 665 28 × 2 = 1 + 0.253 906 249 986 332 194 043 330 56;
  • 45) 0.253 906 249 986 332 194 043 330 56 × 2 = 0 + 0.507 812 499 972 664 388 086 661 12;
  • 46) 0.507 812 499 972 664 388 086 661 12 × 2 = 1 + 0.015 624 999 945 328 776 173 322 24;
  • 47) 0.015 624 999 945 328 776 173 322 24 × 2 = 0 + 0.031 249 999 890 657 552 346 644 48;
  • 48) 0.031 249 999 890 657 552 346 644 48 × 2 = 0 + 0.062 499 999 781 315 104 693 288 96;
  • 49) 0.062 499 999 781 315 104 693 288 96 × 2 = 0 + 0.124 999 999 562 630 209 386 577 92;
  • 50) 0.124 999 999 562 630 209 386 577 92 × 2 = 0 + 0.249 999 999 125 260 418 773 155 84;
  • 51) 0.249 999 999 125 260 418 773 155 84 × 2 = 0 + 0.499 999 998 250 520 837 546 311 68;
  • 52) 0.499 999 998 250 520 837 546 311 68 × 2 = 0 + 0.999 999 996 501 041 675 092 623 36;
  • 53) 0.999 999 996 501 041 675 092 623 36 × 2 = 1 + 0.999 999 993 002 083 350 185 246 72;

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:


0.745 459 324 169 999 826 281 695 41(10) =

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

5. Positive number before normalization:

1.745 459 324 169 999 826 281 695 41(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:


1.745 459 324 169 999 826 281 695 41(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 1.745 459 324 169 999 826 281 695 41 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