-0.000 000 000 742 146 72 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 72(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
-0.000 000 000 742 146 72(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:

|-0.000 000 000 742 146 72| = 0.000 000 000 742 146 72


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: 0.000 000 000 742 146 72.

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) 0.000 000 000 742 146 72 × 2 = 0 + 0.000 000 001 484 293 44;
  • 2) 0.000 000 001 484 293 44 × 2 = 0 + 0.000 000 002 968 586 88;
  • 3) 0.000 000 002 968 586 88 × 2 = 0 + 0.000 000 005 937 173 76;
  • 4) 0.000 000 005 937 173 76 × 2 = 0 + 0.000 000 011 874 347 52;
  • 5) 0.000 000 011 874 347 52 × 2 = 0 + 0.000 000 023 748 695 04;
  • 6) 0.000 000 023 748 695 04 × 2 = 0 + 0.000 000 047 497 390 08;
  • 7) 0.000 000 047 497 390 08 × 2 = 0 + 0.000 000 094 994 780 16;
  • 8) 0.000 000 094 994 780 16 × 2 = 0 + 0.000 000 189 989 560 32;
  • 9) 0.000 000 189 989 560 32 × 2 = 0 + 0.000 000 379 979 120 64;
  • 10) 0.000 000 379 979 120 64 × 2 = 0 + 0.000 000 759 958 241 28;
  • 11) 0.000 000 759 958 241 28 × 2 = 0 + 0.000 001 519 916 482 56;
  • 12) 0.000 001 519 916 482 56 × 2 = 0 + 0.000 003 039 832 965 12;
  • 13) 0.000 003 039 832 965 12 × 2 = 0 + 0.000 006 079 665 930 24;
  • 14) 0.000 006 079 665 930 24 × 2 = 0 + 0.000 012 159 331 860 48;
  • 15) 0.000 012 159 331 860 48 × 2 = 0 + 0.000 024 318 663 720 96;
  • 16) 0.000 024 318 663 720 96 × 2 = 0 + 0.000 048 637 327 441 92;
  • 17) 0.000 048 637 327 441 92 × 2 = 0 + 0.000 097 274 654 883 84;
  • 18) 0.000 097 274 654 883 84 × 2 = 0 + 0.000 194 549 309 767 68;
  • 19) 0.000 194 549 309 767 68 × 2 = 0 + 0.000 389 098 619 535 36;
  • 20) 0.000 389 098 619 535 36 × 2 = 0 + 0.000 778 197 239 070 72;
  • 21) 0.000 778 197 239 070 72 × 2 = 0 + 0.001 556 394 478 141 44;
  • 22) 0.001 556 394 478 141 44 × 2 = 0 + 0.003 112 788 956 282 88;
  • 23) 0.003 112 788 956 282 88 × 2 = 0 + 0.006 225 577 912 565 76;
  • 24) 0.006 225 577 912 565 76 × 2 = 0 + 0.012 451 155 825 131 52;
  • 25) 0.012 451 155 825 131 52 × 2 = 0 + 0.024 902 311 650 263 04;
  • 26) 0.024 902 311 650 263 04 × 2 = 0 + 0.049 804 623 300 526 08;
  • 27) 0.049 804 623 300 526 08 × 2 = 0 + 0.099 609 246 601 052 16;
  • 28) 0.099 609 246 601 052 16 × 2 = 0 + 0.199 218 493 202 104 32;
  • 29) 0.199 218 493 202 104 32 × 2 = 0 + 0.398 436 986 404 208 64;
  • 30) 0.398 436 986 404 208 64 × 2 = 0 + 0.796 873 972 808 417 28;
  • 31) 0.796 873 972 808 417 28 × 2 = 1 + 0.593 747 945 616 834 56;
  • 32) 0.593 747 945 616 834 56 × 2 = 1 + 0.187 495 891 233 669 12;
  • 33) 0.187 495 891 233 669 12 × 2 = 0 + 0.374 991 782 467 338 24;
  • 34) 0.374 991 782 467 338 24 × 2 = 0 + 0.749 983 564 934 676 48;
  • 35) 0.749 983 564 934 676 48 × 2 = 1 + 0.499 967 129 869 352 96;
  • 36) 0.499 967 129 869 352 96 × 2 = 0 + 0.999 934 259 738 705 92;
  • 37) 0.999 934 259 738 705 92 × 2 = 1 + 0.999 868 519 477 411 84;
  • 38) 0.999 868 519 477 411 84 × 2 = 1 + 0.999 737 038 954 823 68;
  • 39) 0.999 737 038 954 823 68 × 2 = 1 + 0.999 474 077 909 647 36;
  • 40) 0.999 474 077 909 647 36 × 2 = 1 + 0.998 948 155 819 294 72;
  • 41) 0.998 948 155 819 294 72 × 2 = 1 + 0.997 896 311 638 589 44;
  • 42) 0.997 896 311 638 589 44 × 2 = 1 + 0.995 792 623 277 178 88;
  • 43) 0.995 792 623 277 178 88 × 2 = 1 + 0.991 585 246 554 357 76;
  • 44) 0.991 585 246 554 357 76 × 2 = 1 + 0.983 170 493 108 715 52;
  • 45) 0.983 170 493 108 715 52 × 2 = 1 + 0.966 340 986 217 431 04;
  • 46) 0.966 340 986 217 431 04 × 2 = 1 + 0.932 681 972 434 862 08;
  • 47) 0.932 681 972 434 862 08 × 2 = 1 + 0.865 363 944 869 724 16;
  • 48) 0.865 363 944 869 724 16 × 2 = 1 + 0.730 727 889 739 448 32;
  • 49) 0.730 727 889 739 448 32 × 2 = 1 + 0.461 455 779 478 896 64;
  • 50) 0.461 455 779 478 896 64 × 2 = 0 + 0.922 911 558 957 793 28;
  • 51) 0.922 911 558 957 793 28 × 2 = 1 + 0.845 823 117 915 586 56;
  • 52) 0.845 823 117 915 586 56 × 2 = 1 + 0.691 646 235 831 173 12;
  • 53) 0.691 646 235 831 173 12 × 2 = 1 + 0.383 292 471 662 346 24;
  • 54) 0.383 292 471 662 346 24 × 2 = 0 + 0.766 584 943 324 692 48;

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:


0.000 000 000 742 146 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2)

6. Positive number before normalization:

0.000 000 000 742 146 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(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:


0.000 000 000 742 146 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1011 10(2) × 20 =

1.1001 0111 1111 1111 1101 110(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 1101 110


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 1110 1110 =

100 1011 1111 1111 1110 1110


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 1110 1110


Decimal number -0.000 000 000 742 146 72 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 1110

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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