-0.145 067 813 487 901 050 860 48 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.145 067 813 487 901 050 860 48(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
-0.145 067 813 487 901 050 860 48(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. Start with the positive version of the number:

|-0.145 067 813 487 901 050 860 48| = 0.145 067 813 487 901 050 860 48


2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

3. 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.

0(10) =

0(2)


4. Convert to binary (base 2) the fractional part: 0.145 067 813 487 901 050 860 48.

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) 0.145 067 813 487 901 050 860 48 × 2 = 0 + 0.290 135 626 975 802 101 720 96;
  • 2) 0.290 135 626 975 802 101 720 96 × 2 = 0 + 0.580 271 253 951 604 203 441 92;
  • 3) 0.580 271 253 951 604 203 441 92 × 2 = 1 + 0.160 542 507 903 208 406 883 84;
  • 4) 0.160 542 507 903 208 406 883 84 × 2 = 0 + 0.321 085 015 806 416 813 767 68;
  • 5) 0.321 085 015 806 416 813 767 68 × 2 = 0 + 0.642 170 031 612 833 627 535 36;
  • 6) 0.642 170 031 612 833 627 535 36 × 2 = 1 + 0.284 340 063 225 667 255 070 72;
  • 7) 0.284 340 063 225 667 255 070 72 × 2 = 0 + 0.568 680 126 451 334 510 141 44;
  • 8) 0.568 680 126 451 334 510 141 44 × 2 = 1 + 0.137 360 252 902 669 020 282 88;
  • 9) 0.137 360 252 902 669 020 282 88 × 2 = 0 + 0.274 720 505 805 338 040 565 76;
  • 10) 0.274 720 505 805 338 040 565 76 × 2 = 0 + 0.549 441 011 610 676 081 131 52;
  • 11) 0.549 441 011 610 676 081 131 52 × 2 = 1 + 0.098 882 023 221 352 162 263 04;
  • 12) 0.098 882 023 221 352 162 263 04 × 2 = 0 + 0.197 764 046 442 704 324 526 08;
  • 13) 0.197 764 046 442 704 324 526 08 × 2 = 0 + 0.395 528 092 885 408 649 052 16;
  • 14) 0.395 528 092 885 408 649 052 16 × 2 = 0 + 0.791 056 185 770 817 298 104 32;
  • 15) 0.791 056 185 770 817 298 104 32 × 2 = 1 + 0.582 112 371 541 634 596 208 64;
  • 16) 0.582 112 371 541 634 596 208 64 × 2 = 1 + 0.164 224 743 083 269 192 417 28;
  • 17) 0.164 224 743 083 269 192 417 28 × 2 = 0 + 0.328 449 486 166 538 384 834 56;
  • 18) 0.328 449 486 166 538 384 834 56 × 2 = 0 + 0.656 898 972 333 076 769 669 12;
  • 19) 0.656 898 972 333 076 769 669 12 × 2 = 1 + 0.313 797 944 666 153 539 338 24;
  • 20) 0.313 797 944 666 153 539 338 24 × 2 = 0 + 0.627 595 889 332 307 078 676 48;
  • 21) 0.627 595 889 332 307 078 676 48 × 2 = 1 + 0.255 191 778 664 614 157 352 96;
  • 22) 0.255 191 778 664 614 157 352 96 × 2 = 0 + 0.510 383 557 329 228 314 705 92;
  • 23) 0.510 383 557 329 228 314 705 92 × 2 = 1 + 0.020 767 114 658 456 629 411 84;
  • 24) 0.020 767 114 658 456 629 411 84 × 2 = 0 + 0.041 534 229 316 913 258 823 68;
  • 25) 0.041 534 229 316 913 258 823 68 × 2 = 0 + 0.083 068 458 633 826 517 647 36;
  • 26) 0.083 068 458 633 826 517 647 36 × 2 = 0 + 0.166 136 917 267 653 035 294 72;
  • 27) 0.166 136 917 267 653 035 294 72 × 2 = 0 + 0.332 273 834 535 306 070 589 44;
  • 28) 0.332 273 834 535 306 070 589 44 × 2 = 0 + 0.664 547 669 070 612 141 178 88;
  • 29) 0.664 547 669 070 612 141 178 88 × 2 = 1 + 0.329 095 338 141 224 282 357 76;
  • 30) 0.329 095 338 141 224 282 357 76 × 2 = 0 + 0.658 190 676 282 448 564 715 52;
  • 31) 0.658 190 676 282 448 564 715 52 × 2 = 1 + 0.316 381 352 564 897 129 431 04;
  • 32) 0.316 381 352 564 897 129 431 04 × 2 = 0 + 0.632 762 705 129 794 258 862 08;
  • 33) 0.632 762 705 129 794 258 862 08 × 2 = 1 + 0.265 525 410 259 588 517 724 16;
  • 34) 0.265 525 410 259 588 517 724 16 × 2 = 0 + 0.531 050 820 519 177 035 448 32;
  • 35) 0.531 050 820 519 177 035 448 32 × 2 = 1 + 0.062 101 641 038 354 070 896 64;
  • 36) 0.062 101 641 038 354 070 896 64 × 2 = 0 + 0.124 203 282 076 708 141 793 28;
  • 37) 0.124 203 282 076 708 141 793 28 × 2 = 0 + 0.248 406 564 153 416 283 586 56;
  • 38) 0.248 406 564 153 416 283 586 56 × 2 = 0 + 0.496 813 128 306 832 567 173 12;
  • 39) 0.496 813 128 306 832 567 173 12 × 2 = 0 + 0.993 626 256 613 665 134 346 24;
  • 40) 0.993 626 256 613 665 134 346 24 × 2 = 1 + 0.987 252 513 227 330 268 692 48;
  • 41) 0.987 252 513 227 330 268 692 48 × 2 = 1 + 0.974 505 026 454 660 537 384 96;
  • 42) 0.974 505 026 454 660 537 384 96 × 2 = 1 + 0.949 010 052 909 321 074 769 92;
  • 43) 0.949 010 052 909 321 074 769 92 × 2 = 1 + 0.898 020 105 818 642 149 539 84;
  • 44) 0.898 020 105 818 642 149 539 84 × 2 = 1 + 0.796 040 211 637 284 299 079 68;
  • 45) 0.796 040 211 637 284 299 079 68 × 2 = 1 + 0.592 080 423 274 568 598 159 36;
  • 46) 0.592 080 423 274 568 598 159 36 × 2 = 1 + 0.184 160 846 549 137 196 318 72;
  • 47) 0.184 160 846 549 137 196 318 72 × 2 = 0 + 0.368 321 693 098 274 392 637 44;
  • 48) 0.368 321 693 098 274 392 637 44 × 2 = 0 + 0.736 643 386 196 548 785 274 88;
  • 49) 0.736 643 386 196 548 785 274 88 × 2 = 1 + 0.473 286 772 393 097 570 549 76;
  • 50) 0.473 286 772 393 097 570 549 76 × 2 = 0 + 0.946 573 544 786 195 141 099 52;
  • 51) 0.946 573 544 786 195 141 099 52 × 2 = 1 + 0.893 147 089 572 390 282 199 04;
  • 52) 0.893 147 089 572 390 282 199 04 × 2 = 1 + 0.786 294 179 144 780 564 398 08;
  • 53) 0.786 294 179 144 780 564 398 08 × 2 = 1 + 0.572 588 358 289 561 128 796 16;
  • 54) 0.572 588 358 289 561 128 796 16 × 2 = 1 + 0.145 176 716 579 122 257 592 32;
  • 55) 0.145 176 716 579 122 257 592 32 × 2 = 0 + 0.290 353 433 158 244 515 184 64;

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).


5. 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:


0.145 067 813 487 901 050 860 48(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2)

6. Positive number before normalization:

0.145 067 813 487 901 050 860 48(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right, so that only one non zero digit remains to the left of it:


0.145 067 813 487 901 050 860 48(10) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2) =

0.0010 0101 0010 0011 0010 1010 0000 1010 1010 0001 1111 1100 1011 110(2) × 20 =

1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110(2) × 2-3


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): -3

Mantissa (not normalized):
1.0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-3 + 2(11-1) - 1 =

(-3 + 1 023)(10) =

1 020(10)


10. 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;
  • 1 020 ÷ 2 = 510 + 0;
  • 510 ÷ 2 = 255 + 0;
  • 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;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1020(10) =

011 1111 1100(2)


12. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110 =

0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)

Exponent (11 bits) =
011 1111 1100

Mantissa (52 bits) =
0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110


Decimal number -0.145 067 813 487 901 050 860 48 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1100 - 0010 1001 0001 1001 0101 0000 0101 0101 0000 1111 1110 0101 1110

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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