-0.000 000 000 742 147 72 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

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

|-0.000 000 000 742 147 72| = 0.000 000 000 742 147 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: 0.000 000 000 742 147 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) 0.000 000 000 742 147 72 × 2 = 0 + 0.000 000 001 484 295 44;
  • 2) 0.000 000 001 484 295 44 × 2 = 0 + 0.000 000 002 968 590 88;
  • 3) 0.000 000 002 968 590 88 × 2 = 0 + 0.000 000 005 937 181 76;
  • 4) 0.000 000 005 937 181 76 × 2 = 0 + 0.000 000 011 874 363 52;
  • 5) 0.000 000 011 874 363 52 × 2 = 0 + 0.000 000 023 748 727 04;
  • 6) 0.000 000 023 748 727 04 × 2 = 0 + 0.000 000 047 497 454 08;
  • 7) 0.000 000 047 497 454 08 × 2 = 0 + 0.000 000 094 994 908 16;
  • 8) 0.000 000 094 994 908 16 × 2 = 0 + 0.000 000 189 989 816 32;
  • 9) 0.000 000 189 989 816 32 × 2 = 0 + 0.000 000 379 979 632 64;
  • 10) 0.000 000 379 979 632 64 × 2 = 0 + 0.000 000 759 959 265 28;
  • 11) 0.000 000 759 959 265 28 × 2 = 0 + 0.000 001 519 918 530 56;
  • 12) 0.000 001 519 918 530 56 × 2 = 0 + 0.000 003 039 837 061 12;
  • 13) 0.000 003 039 837 061 12 × 2 = 0 + 0.000 006 079 674 122 24;
  • 14) 0.000 006 079 674 122 24 × 2 = 0 + 0.000 012 159 348 244 48;
  • 15) 0.000 012 159 348 244 48 × 2 = 0 + 0.000 024 318 696 488 96;
  • 16) 0.000 024 318 696 488 96 × 2 = 0 + 0.000 048 637 392 977 92;
  • 17) 0.000 048 637 392 977 92 × 2 = 0 + 0.000 097 274 785 955 84;
  • 18) 0.000 097 274 785 955 84 × 2 = 0 + 0.000 194 549 571 911 68;
  • 19) 0.000 194 549 571 911 68 × 2 = 0 + 0.000 389 099 143 823 36;
  • 20) 0.000 389 099 143 823 36 × 2 = 0 + 0.000 778 198 287 646 72;
  • 21) 0.000 778 198 287 646 72 × 2 = 0 + 0.001 556 396 575 293 44;
  • 22) 0.001 556 396 575 293 44 × 2 = 0 + 0.003 112 793 150 586 88;
  • 23) 0.003 112 793 150 586 88 × 2 = 0 + 0.006 225 586 301 173 76;
  • 24) 0.006 225 586 301 173 76 × 2 = 0 + 0.012 451 172 602 347 52;
  • 25) 0.012 451 172 602 347 52 × 2 = 0 + 0.024 902 345 204 695 04;
  • 26) 0.024 902 345 204 695 04 × 2 = 0 + 0.049 804 690 409 390 08;
  • 27) 0.049 804 690 409 390 08 × 2 = 0 + 0.099 609 380 818 780 16;
  • 28) 0.099 609 380 818 780 16 × 2 = 0 + 0.199 218 761 637 560 32;
  • 29) 0.199 218 761 637 560 32 × 2 = 0 + 0.398 437 523 275 120 64;
  • 30) 0.398 437 523 275 120 64 × 2 = 0 + 0.796 875 046 550 241 28;
  • 31) 0.796 875 046 550 241 28 × 2 = 1 + 0.593 750 093 100 482 56;
  • 32) 0.593 750 093 100 482 56 × 2 = 1 + 0.187 500 186 200 965 12;
  • 33) 0.187 500 186 200 965 12 × 2 = 0 + 0.375 000 372 401 930 24;
  • 34) 0.375 000 372 401 930 24 × 2 = 0 + 0.750 000 744 803 860 48;
  • 35) 0.750 000 744 803 860 48 × 2 = 1 + 0.500 001 489 607 720 96;
  • 36) 0.500 001 489 607 720 96 × 2 = 1 + 0.000 002 979 215 441 92;
  • 37) 0.000 002 979 215 441 92 × 2 = 0 + 0.000 005 958 430 883 84;
  • 38) 0.000 005 958 430 883 84 × 2 = 0 + 0.000 011 916 861 767 68;
  • 39) 0.000 011 916 861 767 68 × 2 = 0 + 0.000 023 833 723 535 36;
  • 40) 0.000 023 833 723 535 36 × 2 = 0 + 0.000 047 667 447 070 72;
  • 41) 0.000 047 667 447 070 72 × 2 = 0 + 0.000 095 334 894 141 44;
  • 42) 0.000 095 334 894 141 44 × 2 = 0 + 0.000 190 669 788 282 88;
  • 43) 0.000 190 669 788 282 88 × 2 = 0 + 0.000 381 339 576 565 76;
  • 44) 0.000 381 339 576 565 76 × 2 = 0 + 0.000 762 679 153 131 52;
  • 45) 0.000 762 679 153 131 52 × 2 = 0 + 0.001 525 358 306 263 04;
  • 46) 0.001 525 358 306 263 04 × 2 = 0 + 0.003 050 716 612 526 08;
  • 47) 0.003 050 716 612 526 08 × 2 = 0 + 0.006 101 433 225 052 16;
  • 48) 0.006 101 433 225 052 16 × 2 = 0 + 0.012 202 866 450 104 32;
  • 49) 0.012 202 866 450 104 32 × 2 = 0 + 0.024 405 732 900 208 64;
  • 50) 0.024 405 732 900 208 64 × 2 = 0 + 0.048 811 465 800 417 28;
  • 51) 0.048 811 465 800 417 28 × 2 = 0 + 0.097 622 931 600 834 56;
  • 52) 0.097 622 931 600 834 56 × 2 = 0 + 0.195 245 863 201 669 12;
  • 53) 0.195 245 863 201 669 12 × 2 = 0 + 0.390 491 726 403 338 24;
  • 54) 0.390 491 726 403 338 24 × 2 = 0 + 0.780 983 452 806 676 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:


0.000 000 000 742 147 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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 72(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 72 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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