-0.000 000 000 742 147 675 985 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 675 985(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
-0.000 000 000 742 147 675 985(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:

|-0.000 000 000 742 147 675 985| = 0.000 000 000 742 147 675 985


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: 0.000 000 000 742 147 675 985.

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) 0.000 000 000 742 147 675 985 × 2 = 0 + 0.000 000 001 484 295 351 97;
  • 2) 0.000 000 001 484 295 351 97 × 2 = 0 + 0.000 000 002 968 590 703 94;
  • 3) 0.000 000 002 968 590 703 94 × 2 = 0 + 0.000 000 005 937 181 407 88;
  • 4) 0.000 000 005 937 181 407 88 × 2 = 0 + 0.000 000 011 874 362 815 76;
  • 5) 0.000 000 011 874 362 815 76 × 2 = 0 + 0.000 000 023 748 725 631 52;
  • 6) 0.000 000 023 748 725 631 52 × 2 = 0 + 0.000 000 047 497 451 263 04;
  • 7) 0.000 000 047 497 451 263 04 × 2 = 0 + 0.000 000 094 994 902 526 08;
  • 8) 0.000 000 094 994 902 526 08 × 2 = 0 + 0.000 000 189 989 805 052 16;
  • 9) 0.000 000 189 989 805 052 16 × 2 = 0 + 0.000 000 379 979 610 104 32;
  • 10) 0.000 000 379 979 610 104 32 × 2 = 0 + 0.000 000 759 959 220 208 64;
  • 11) 0.000 000 759 959 220 208 64 × 2 = 0 + 0.000 001 519 918 440 417 28;
  • 12) 0.000 001 519 918 440 417 28 × 2 = 0 + 0.000 003 039 836 880 834 56;
  • 13) 0.000 003 039 836 880 834 56 × 2 = 0 + 0.000 006 079 673 761 669 12;
  • 14) 0.000 006 079 673 761 669 12 × 2 = 0 + 0.000 012 159 347 523 338 24;
  • 15) 0.000 012 159 347 523 338 24 × 2 = 0 + 0.000 024 318 695 046 676 48;
  • 16) 0.000 024 318 695 046 676 48 × 2 = 0 + 0.000 048 637 390 093 352 96;
  • 17) 0.000 048 637 390 093 352 96 × 2 = 0 + 0.000 097 274 780 186 705 92;
  • 18) 0.000 097 274 780 186 705 92 × 2 = 0 + 0.000 194 549 560 373 411 84;
  • 19) 0.000 194 549 560 373 411 84 × 2 = 0 + 0.000 389 099 120 746 823 68;
  • 20) 0.000 389 099 120 746 823 68 × 2 = 0 + 0.000 778 198 241 493 647 36;
  • 21) 0.000 778 198 241 493 647 36 × 2 = 0 + 0.001 556 396 482 987 294 72;
  • 22) 0.001 556 396 482 987 294 72 × 2 = 0 + 0.003 112 792 965 974 589 44;
  • 23) 0.003 112 792 965 974 589 44 × 2 = 0 + 0.006 225 585 931 949 178 88;
  • 24) 0.006 225 585 931 949 178 88 × 2 = 0 + 0.012 451 171 863 898 357 76;
  • 25) 0.012 451 171 863 898 357 76 × 2 = 0 + 0.024 902 343 727 796 715 52;
  • 26) 0.024 902 343 727 796 715 52 × 2 = 0 + 0.049 804 687 455 593 431 04;
  • 27) 0.049 804 687 455 593 431 04 × 2 = 0 + 0.099 609 374 911 186 862 08;
  • 28) 0.099 609 374 911 186 862 08 × 2 = 0 + 0.199 218 749 822 373 724 16;
  • 29) 0.199 218 749 822 373 724 16 × 2 = 0 + 0.398 437 499 644 747 448 32;
  • 30) 0.398 437 499 644 747 448 32 × 2 = 0 + 0.796 874 999 289 494 896 64;
  • 31) 0.796 874 999 289 494 896 64 × 2 = 1 + 0.593 749 998 578 989 793 28;
  • 32) 0.593 749 998 578 989 793 28 × 2 = 1 + 0.187 499 997 157 979 586 56;
  • 33) 0.187 499 997 157 979 586 56 × 2 = 0 + 0.374 999 994 315 959 173 12;
  • 34) 0.374 999 994 315 959 173 12 × 2 = 0 + 0.749 999 988 631 918 346 24;
  • 35) 0.749 999 988 631 918 346 24 × 2 = 1 + 0.499 999 977 263 836 692 48;
  • 36) 0.499 999 977 263 836 692 48 × 2 = 0 + 0.999 999 954 527 673 384 96;
  • 37) 0.999 999 954 527 673 384 96 × 2 = 1 + 0.999 999 909 055 346 769 92;
  • 38) 0.999 999 909 055 346 769 92 × 2 = 1 + 0.999 999 818 110 693 539 84;
  • 39) 0.999 999 818 110 693 539 84 × 2 = 1 + 0.999 999 636 221 387 079 68;
  • 40) 0.999 999 636 221 387 079 68 × 2 = 1 + 0.999 999 272 442 774 159 36;
  • 41) 0.999 999 272 442 774 159 36 × 2 = 1 + 0.999 998 544 885 548 318 72;
  • 42) 0.999 998 544 885 548 318 72 × 2 = 1 + 0.999 997 089 771 096 637 44;
  • 43) 0.999 997 089 771 096 637 44 × 2 = 1 + 0.999 994 179 542 193 274 88;
  • 44) 0.999 994 179 542 193 274 88 × 2 = 1 + 0.999 988 359 084 386 549 76;
  • 45) 0.999 988 359 084 386 549 76 × 2 = 1 + 0.999 976 718 168 773 099 52;
  • 46) 0.999 976 718 168 773 099 52 × 2 = 1 + 0.999 953 436 337 546 199 04;
  • 47) 0.999 953 436 337 546 199 04 × 2 = 1 + 0.999 906 872 675 092 398 08;
  • 48) 0.999 906 872 675 092 398 08 × 2 = 1 + 0.999 813 745 350 184 796 16;
  • 49) 0.999 813 745 350 184 796 16 × 2 = 1 + 0.999 627 490 700 369 592 32;
  • 50) 0.999 627 490 700 369 592 32 × 2 = 1 + 0.999 254 981 400 739 184 64;
  • 51) 0.999 254 981 400 739 184 64 × 2 = 1 + 0.998 509 962 801 478 369 28;
  • 52) 0.998 509 962 801 478 369 28 × 2 = 1 + 0.997 019 925 602 956 738 56;
  • 53) 0.997 019 925 602 956 738 56 × 2 = 1 + 0.994 039 851 205 913 477 12;
  • 54) 0.994 039 851 205 913 477 12 × 2 = 1 + 0.988 079 702 411 826 954 24;

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:


0.000 000 000 742 147 675 985(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 675 985(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:


0.000 000 000 742 147 675 985(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 -0.000 000 000 742 147 675 985 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