-0.000 000 000 742 147 676 646 681 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 681 6(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
-0.000 000 000 742 147 676 646 681 6(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. Start with the positive version of the number:

|-0.000 000 000 742 147 676 646 681 6| = 0.000 000 000 742 147 676 646 681 6


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: 0.000 000 000 742 147 676 646 681 6.

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) 0.000 000 000 742 147 676 646 681 6 × 2 = 0 + 0.000 000 001 484 295 353 293 363 2;
  • 2) 0.000 000 001 484 295 353 293 363 2 × 2 = 0 + 0.000 000 002 968 590 706 586 726 4;
  • 3) 0.000 000 002 968 590 706 586 726 4 × 2 = 0 + 0.000 000 005 937 181 413 173 452 8;
  • 4) 0.000 000 005 937 181 413 173 452 8 × 2 = 0 + 0.000 000 011 874 362 826 346 905 6;
  • 5) 0.000 000 011 874 362 826 346 905 6 × 2 = 0 + 0.000 000 023 748 725 652 693 811 2;
  • 6) 0.000 000 023 748 725 652 693 811 2 × 2 = 0 + 0.000 000 047 497 451 305 387 622 4;
  • 7) 0.000 000 047 497 451 305 387 622 4 × 2 = 0 + 0.000 000 094 994 902 610 775 244 8;
  • 8) 0.000 000 094 994 902 610 775 244 8 × 2 = 0 + 0.000 000 189 989 805 221 550 489 6;
  • 9) 0.000 000 189 989 805 221 550 489 6 × 2 = 0 + 0.000 000 379 979 610 443 100 979 2;
  • 10) 0.000 000 379 979 610 443 100 979 2 × 2 = 0 + 0.000 000 759 959 220 886 201 958 4;
  • 11) 0.000 000 759 959 220 886 201 958 4 × 2 = 0 + 0.000 001 519 918 441 772 403 916 8;
  • 12) 0.000 001 519 918 441 772 403 916 8 × 2 = 0 + 0.000 003 039 836 883 544 807 833 6;
  • 13) 0.000 003 039 836 883 544 807 833 6 × 2 = 0 + 0.000 006 079 673 767 089 615 667 2;
  • 14) 0.000 006 079 673 767 089 615 667 2 × 2 = 0 + 0.000 012 159 347 534 179 231 334 4;
  • 15) 0.000 012 159 347 534 179 231 334 4 × 2 = 0 + 0.000 024 318 695 068 358 462 668 8;
  • 16) 0.000 024 318 695 068 358 462 668 8 × 2 = 0 + 0.000 048 637 390 136 716 925 337 6;
  • 17) 0.000 048 637 390 136 716 925 337 6 × 2 = 0 + 0.000 097 274 780 273 433 850 675 2;
  • 18) 0.000 097 274 780 273 433 850 675 2 × 2 = 0 + 0.000 194 549 560 546 867 701 350 4;
  • 19) 0.000 194 549 560 546 867 701 350 4 × 2 = 0 + 0.000 389 099 121 093 735 402 700 8;
  • 20) 0.000 389 099 121 093 735 402 700 8 × 2 = 0 + 0.000 778 198 242 187 470 805 401 6;
  • 21) 0.000 778 198 242 187 470 805 401 6 × 2 = 0 + 0.001 556 396 484 374 941 610 803 2;
  • 22) 0.001 556 396 484 374 941 610 803 2 × 2 = 0 + 0.003 112 792 968 749 883 221 606 4;
  • 23) 0.003 112 792 968 749 883 221 606 4 × 2 = 0 + 0.006 225 585 937 499 766 443 212 8;
  • 24) 0.006 225 585 937 499 766 443 212 8 × 2 = 0 + 0.012 451 171 874 999 532 886 425 6;
  • 25) 0.012 451 171 874 999 532 886 425 6 × 2 = 0 + 0.024 902 343 749 999 065 772 851 2;
  • 26) 0.024 902 343 749 999 065 772 851 2 × 2 = 0 + 0.049 804 687 499 998 131 545 702 4;
  • 27) 0.049 804 687 499 998 131 545 702 4 × 2 = 0 + 0.099 609 374 999 996 263 091 404 8;
  • 28) 0.099 609 374 999 996 263 091 404 8 × 2 = 0 + 0.199 218 749 999 992 526 182 809 6;
  • 29) 0.199 218 749 999 992 526 182 809 6 × 2 = 0 + 0.398 437 499 999 985 052 365 619 2;
  • 30) 0.398 437 499 999 985 052 365 619 2 × 2 = 0 + 0.796 874 999 999 970 104 731 238 4;
  • 31) 0.796 874 999 999 970 104 731 238 4 × 2 = 1 + 0.593 749 999 999 940 209 462 476 8;
  • 32) 0.593 749 999 999 940 209 462 476 8 × 2 = 1 + 0.187 499 999 999 880 418 924 953 6;
  • 33) 0.187 499 999 999 880 418 924 953 6 × 2 = 0 + 0.374 999 999 999 760 837 849 907 2;
  • 34) 0.374 999 999 999 760 837 849 907 2 × 2 = 0 + 0.749 999 999 999 521 675 699 814 4;
  • 35) 0.749 999 999 999 521 675 699 814 4 × 2 = 1 + 0.499 999 999 999 043 351 399 628 8;
  • 36) 0.499 999 999 999 043 351 399 628 8 × 2 = 0 + 0.999 999 999 998 086 702 799 257 6;
  • 37) 0.999 999 999 998 086 702 799 257 6 × 2 = 1 + 0.999 999 999 996 173 405 598 515 2;
  • 38) 0.999 999 999 996 173 405 598 515 2 × 2 = 1 + 0.999 999 999 992 346 811 197 030 4;
  • 39) 0.999 999 999 992 346 811 197 030 4 × 2 = 1 + 0.999 999 999 984 693 622 394 060 8;
  • 40) 0.999 999 999 984 693 622 394 060 8 × 2 = 1 + 0.999 999 999 969 387 244 788 121 6;
  • 41) 0.999 999 999 969 387 244 788 121 6 × 2 = 1 + 0.999 999 999 938 774 489 576 243 2;
  • 42) 0.999 999 999 938 774 489 576 243 2 × 2 = 1 + 0.999 999 999 877 548 979 152 486 4;
  • 43) 0.999 999 999 877 548 979 152 486 4 × 2 = 1 + 0.999 999 999 755 097 958 304 972 8;
  • 44) 0.999 999 999 755 097 958 304 972 8 × 2 = 1 + 0.999 999 999 510 195 916 609 945 6;
  • 45) 0.999 999 999 510 195 916 609 945 6 × 2 = 1 + 0.999 999 999 020 391 833 219 891 2;
  • 46) 0.999 999 999 020 391 833 219 891 2 × 2 = 1 + 0.999 999 998 040 783 666 439 782 4;
  • 47) 0.999 999 998 040 783 666 439 782 4 × 2 = 1 + 0.999 999 996 081 567 332 879 564 8;
  • 48) 0.999 999 996 081 567 332 879 564 8 × 2 = 1 + 0.999 999 992 163 134 665 759 129 6;
  • 49) 0.999 999 992 163 134 665 759 129 6 × 2 = 1 + 0.999 999 984 326 269 331 518 259 2;
  • 50) 0.999 999 984 326 269 331 518 259 2 × 2 = 1 + 0.999 999 968 652 538 663 036 518 4;
  • 51) 0.999 999 968 652 538 663 036 518 4 × 2 = 1 + 0.999 999 937 305 077 326 073 036 8;
  • 52) 0.999 999 937 305 077 326 073 036 8 × 2 = 1 + 0.999 999 874 610 154 652 146 073 6;
  • 53) 0.999 999 874 610 154 652 146 073 6 × 2 = 1 + 0.999 999 749 220 309 304 292 147 2;
  • 54) 0.999 999 749 220 309 304 292 147 2 × 2 = 1 + 0.999 999 498 440 618 608 584 294 4;

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:


0.000 000 000 742 147 676 646 681 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 681 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 646 681 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 1 (a negative number)

Exponent (unadjusted): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(10)


10. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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) =

96(10) =

0110 0000(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


13. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

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

Exponent (8 bits) =
0110 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 646 681 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

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

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111