-0.000 000 000 742 147 674 51 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 674 51(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
-0.000 000 000 742 147 674 51(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:

|-0.000 000 000 742 147 674 51| = 0.000 000 000 742 147 674 51


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: 0.000 000 000 742 147 674 51.

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) 0.000 000 000 742 147 674 51 × 2 = 0 + 0.000 000 001 484 295 349 02;
  • 2) 0.000 000 001 484 295 349 02 × 2 = 0 + 0.000 000 002 968 590 698 04;
  • 3) 0.000 000 002 968 590 698 04 × 2 = 0 + 0.000 000 005 937 181 396 08;
  • 4) 0.000 000 005 937 181 396 08 × 2 = 0 + 0.000 000 011 874 362 792 16;
  • 5) 0.000 000 011 874 362 792 16 × 2 = 0 + 0.000 000 023 748 725 584 32;
  • 6) 0.000 000 023 748 725 584 32 × 2 = 0 + 0.000 000 047 497 451 168 64;
  • 7) 0.000 000 047 497 451 168 64 × 2 = 0 + 0.000 000 094 994 902 337 28;
  • 8) 0.000 000 094 994 902 337 28 × 2 = 0 + 0.000 000 189 989 804 674 56;
  • 9) 0.000 000 189 989 804 674 56 × 2 = 0 + 0.000 000 379 979 609 349 12;
  • 10) 0.000 000 379 979 609 349 12 × 2 = 0 + 0.000 000 759 959 218 698 24;
  • 11) 0.000 000 759 959 218 698 24 × 2 = 0 + 0.000 001 519 918 437 396 48;
  • 12) 0.000 001 519 918 437 396 48 × 2 = 0 + 0.000 003 039 836 874 792 96;
  • 13) 0.000 003 039 836 874 792 96 × 2 = 0 + 0.000 006 079 673 749 585 92;
  • 14) 0.000 006 079 673 749 585 92 × 2 = 0 + 0.000 012 159 347 499 171 84;
  • 15) 0.000 012 159 347 499 171 84 × 2 = 0 + 0.000 024 318 694 998 343 68;
  • 16) 0.000 024 318 694 998 343 68 × 2 = 0 + 0.000 048 637 389 996 687 36;
  • 17) 0.000 048 637 389 996 687 36 × 2 = 0 + 0.000 097 274 779 993 374 72;
  • 18) 0.000 097 274 779 993 374 72 × 2 = 0 + 0.000 194 549 559 986 749 44;
  • 19) 0.000 194 549 559 986 749 44 × 2 = 0 + 0.000 389 099 119 973 498 88;
  • 20) 0.000 389 099 119 973 498 88 × 2 = 0 + 0.000 778 198 239 946 997 76;
  • 21) 0.000 778 198 239 946 997 76 × 2 = 0 + 0.001 556 396 479 893 995 52;
  • 22) 0.001 556 396 479 893 995 52 × 2 = 0 + 0.003 112 792 959 787 991 04;
  • 23) 0.003 112 792 959 787 991 04 × 2 = 0 + 0.006 225 585 919 575 982 08;
  • 24) 0.006 225 585 919 575 982 08 × 2 = 0 + 0.012 451 171 839 151 964 16;
  • 25) 0.012 451 171 839 151 964 16 × 2 = 0 + 0.024 902 343 678 303 928 32;
  • 26) 0.024 902 343 678 303 928 32 × 2 = 0 + 0.049 804 687 356 607 856 64;
  • 27) 0.049 804 687 356 607 856 64 × 2 = 0 + 0.099 609 374 713 215 713 28;
  • 28) 0.099 609 374 713 215 713 28 × 2 = 0 + 0.199 218 749 426 431 426 56;
  • 29) 0.199 218 749 426 431 426 56 × 2 = 0 + 0.398 437 498 852 862 853 12;
  • 30) 0.398 437 498 852 862 853 12 × 2 = 0 + 0.796 874 997 705 725 706 24;
  • 31) 0.796 874 997 705 725 706 24 × 2 = 1 + 0.593 749 995 411 451 412 48;
  • 32) 0.593 749 995 411 451 412 48 × 2 = 1 + 0.187 499 990 822 902 824 96;
  • 33) 0.187 499 990 822 902 824 96 × 2 = 0 + 0.374 999 981 645 805 649 92;
  • 34) 0.374 999 981 645 805 649 92 × 2 = 0 + 0.749 999 963 291 611 299 84;
  • 35) 0.749 999 963 291 611 299 84 × 2 = 1 + 0.499 999 926 583 222 599 68;
  • 36) 0.499 999 926 583 222 599 68 × 2 = 0 + 0.999 999 853 166 445 199 36;
  • 37) 0.999 999 853 166 445 199 36 × 2 = 1 + 0.999 999 706 332 890 398 72;
  • 38) 0.999 999 706 332 890 398 72 × 2 = 1 + 0.999 999 412 665 780 797 44;
  • 39) 0.999 999 412 665 780 797 44 × 2 = 1 + 0.999 998 825 331 561 594 88;
  • 40) 0.999 998 825 331 561 594 88 × 2 = 1 + 0.999 997 650 663 123 189 76;
  • 41) 0.999 997 650 663 123 189 76 × 2 = 1 + 0.999 995 301 326 246 379 52;
  • 42) 0.999 995 301 326 246 379 52 × 2 = 1 + 0.999 990 602 652 492 759 04;
  • 43) 0.999 990 602 652 492 759 04 × 2 = 1 + 0.999 981 205 304 985 518 08;
  • 44) 0.999 981 205 304 985 518 08 × 2 = 1 + 0.999 962 410 609 971 036 16;
  • 45) 0.999 962 410 609 971 036 16 × 2 = 1 + 0.999 924 821 219 942 072 32;
  • 46) 0.999 924 821 219 942 072 32 × 2 = 1 + 0.999 849 642 439 884 144 64;
  • 47) 0.999 849 642 439 884 144 64 × 2 = 1 + 0.999 699 284 879 768 289 28;
  • 48) 0.999 699 284 879 768 289 28 × 2 = 1 + 0.999 398 569 759 536 578 56;
  • 49) 0.999 398 569 759 536 578 56 × 2 = 1 + 0.998 797 139 519 073 157 12;
  • 50) 0.998 797 139 519 073 157 12 × 2 = 1 + 0.997 594 279 038 146 314 24;
  • 51) 0.997 594 279 038 146 314 24 × 2 = 1 + 0.995 188 558 076 292 628 48;
  • 52) 0.995 188 558 076 292 628 48 × 2 = 1 + 0.990 377 116 152 585 256 96;
  • 53) 0.990 377 116 152 585 256 96 × 2 = 1 + 0.980 754 232 305 170 513 92;
  • 54) 0.980 754 232 305 170 513 92 × 2 = 1 + 0.961 508 464 610 341 027 84;

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:


0.000 000 000 742 147 674 51(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 674 51(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:


0.000 000 000 742 147 674 51(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 -0.000 000 000 742 147 674 51 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