-0.000 000 000 742 147 535 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 535(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
-0.000 000 000 742 147 535(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:

|-0.000 000 000 742 147 535| = 0.000 000 000 742 147 535


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: 0.000 000 000 742 147 535.

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) 0.000 000 000 742 147 535 × 2 = 0 + 0.000 000 001 484 295 07;
  • 2) 0.000 000 001 484 295 07 × 2 = 0 + 0.000 000 002 968 590 14;
  • 3) 0.000 000 002 968 590 14 × 2 = 0 + 0.000 000 005 937 180 28;
  • 4) 0.000 000 005 937 180 28 × 2 = 0 + 0.000 000 011 874 360 56;
  • 5) 0.000 000 011 874 360 56 × 2 = 0 + 0.000 000 023 748 721 12;
  • 6) 0.000 000 023 748 721 12 × 2 = 0 + 0.000 000 047 497 442 24;
  • 7) 0.000 000 047 497 442 24 × 2 = 0 + 0.000 000 094 994 884 48;
  • 8) 0.000 000 094 994 884 48 × 2 = 0 + 0.000 000 189 989 768 96;
  • 9) 0.000 000 189 989 768 96 × 2 = 0 + 0.000 000 379 979 537 92;
  • 10) 0.000 000 379 979 537 92 × 2 = 0 + 0.000 000 759 959 075 84;
  • 11) 0.000 000 759 959 075 84 × 2 = 0 + 0.000 001 519 918 151 68;
  • 12) 0.000 001 519 918 151 68 × 2 = 0 + 0.000 003 039 836 303 36;
  • 13) 0.000 003 039 836 303 36 × 2 = 0 + 0.000 006 079 672 606 72;
  • 14) 0.000 006 079 672 606 72 × 2 = 0 + 0.000 012 159 345 213 44;
  • 15) 0.000 012 159 345 213 44 × 2 = 0 + 0.000 024 318 690 426 88;
  • 16) 0.000 024 318 690 426 88 × 2 = 0 + 0.000 048 637 380 853 76;
  • 17) 0.000 048 637 380 853 76 × 2 = 0 + 0.000 097 274 761 707 52;
  • 18) 0.000 097 274 761 707 52 × 2 = 0 + 0.000 194 549 523 415 04;
  • 19) 0.000 194 549 523 415 04 × 2 = 0 + 0.000 389 099 046 830 08;
  • 20) 0.000 389 099 046 830 08 × 2 = 0 + 0.000 778 198 093 660 16;
  • 21) 0.000 778 198 093 660 16 × 2 = 0 + 0.001 556 396 187 320 32;
  • 22) 0.001 556 396 187 320 32 × 2 = 0 + 0.003 112 792 374 640 64;
  • 23) 0.003 112 792 374 640 64 × 2 = 0 + 0.006 225 584 749 281 28;
  • 24) 0.006 225 584 749 281 28 × 2 = 0 + 0.012 451 169 498 562 56;
  • 25) 0.012 451 169 498 562 56 × 2 = 0 + 0.024 902 338 997 125 12;
  • 26) 0.024 902 338 997 125 12 × 2 = 0 + 0.049 804 677 994 250 24;
  • 27) 0.049 804 677 994 250 24 × 2 = 0 + 0.099 609 355 988 500 48;
  • 28) 0.099 609 355 988 500 48 × 2 = 0 + 0.199 218 711 977 000 96;
  • 29) 0.199 218 711 977 000 96 × 2 = 0 + 0.398 437 423 954 001 92;
  • 30) 0.398 437 423 954 001 92 × 2 = 0 + 0.796 874 847 908 003 84;
  • 31) 0.796 874 847 908 003 84 × 2 = 1 + 0.593 749 695 816 007 68;
  • 32) 0.593 749 695 816 007 68 × 2 = 1 + 0.187 499 391 632 015 36;
  • 33) 0.187 499 391 632 015 36 × 2 = 0 + 0.374 998 783 264 030 72;
  • 34) 0.374 998 783 264 030 72 × 2 = 0 + 0.749 997 566 528 061 44;
  • 35) 0.749 997 566 528 061 44 × 2 = 1 + 0.499 995 133 056 122 88;
  • 36) 0.499 995 133 056 122 88 × 2 = 0 + 0.999 990 266 112 245 76;
  • 37) 0.999 990 266 112 245 76 × 2 = 1 + 0.999 980 532 224 491 52;
  • 38) 0.999 980 532 224 491 52 × 2 = 1 + 0.999 961 064 448 983 04;
  • 39) 0.999 961 064 448 983 04 × 2 = 1 + 0.999 922 128 897 966 08;
  • 40) 0.999 922 128 897 966 08 × 2 = 1 + 0.999 844 257 795 932 16;
  • 41) 0.999 844 257 795 932 16 × 2 = 1 + 0.999 688 515 591 864 32;
  • 42) 0.999 688 515 591 864 32 × 2 = 1 + 0.999 377 031 183 728 64;
  • 43) 0.999 377 031 183 728 64 × 2 = 1 + 0.998 754 062 367 457 28;
  • 44) 0.998 754 062 367 457 28 × 2 = 1 + 0.997 508 124 734 914 56;
  • 45) 0.997 508 124 734 914 56 × 2 = 1 + 0.995 016 249 469 829 12;
  • 46) 0.995 016 249 469 829 12 × 2 = 1 + 0.990 032 498 939 658 24;
  • 47) 0.990 032 498 939 658 24 × 2 = 1 + 0.980 064 997 879 316 48;
  • 48) 0.980 064 997 879 316 48 × 2 = 1 + 0.960 129 995 758 632 96;
  • 49) 0.960 129 995 758 632 96 × 2 = 1 + 0.920 259 991 517 265 92;
  • 50) 0.920 259 991 517 265 92 × 2 = 1 + 0.840 519 983 034 531 84;
  • 51) 0.840 519 983 034 531 84 × 2 = 1 + 0.681 039 966 069 063 68;
  • 52) 0.681 039 966 069 063 68 × 2 = 1 + 0.362 079 932 138 127 36;
  • 53) 0.362 079 932 138 127 36 × 2 = 0 + 0.724 159 864 276 254 72;
  • 54) 0.724 159 864 276 254 72 × 2 = 1 + 0.448 319 728 552 509 44;

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:


0.000 000 000 742 147 535(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2)

6. Positive number before normalization:

0.000 000 000 742 147 535(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(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:


0.000 000 000 742 147 535(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 01(2) × 20 =

1.1001 0111 1111 1111 1111 101(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 101


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 1101 =

100 1011 1111 1111 1111 1101


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 1101


Decimal number -0.000 000 000 742 147 535 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1101

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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