-0.000 000 000 742 147 815 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 815(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
-0.000 000 000 742 147 815(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:

|-0.000 000 000 742 147 815| = 0.000 000 000 742 147 815


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: 0.000 000 000 742 147 815.

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) 0.000 000 000 742 147 815 × 2 = 0 + 0.000 000 001 484 295 63;
  • 2) 0.000 000 001 484 295 63 × 2 = 0 + 0.000 000 002 968 591 26;
  • 3) 0.000 000 002 968 591 26 × 2 = 0 + 0.000 000 005 937 182 52;
  • 4) 0.000 000 005 937 182 52 × 2 = 0 + 0.000 000 011 874 365 04;
  • 5) 0.000 000 011 874 365 04 × 2 = 0 + 0.000 000 023 748 730 08;
  • 6) 0.000 000 023 748 730 08 × 2 = 0 + 0.000 000 047 497 460 16;
  • 7) 0.000 000 047 497 460 16 × 2 = 0 + 0.000 000 094 994 920 32;
  • 8) 0.000 000 094 994 920 32 × 2 = 0 + 0.000 000 189 989 840 64;
  • 9) 0.000 000 189 989 840 64 × 2 = 0 + 0.000 000 379 979 681 28;
  • 10) 0.000 000 379 979 681 28 × 2 = 0 + 0.000 000 759 959 362 56;
  • 11) 0.000 000 759 959 362 56 × 2 = 0 + 0.000 001 519 918 725 12;
  • 12) 0.000 001 519 918 725 12 × 2 = 0 + 0.000 003 039 837 450 24;
  • 13) 0.000 003 039 837 450 24 × 2 = 0 + 0.000 006 079 674 900 48;
  • 14) 0.000 006 079 674 900 48 × 2 = 0 + 0.000 012 159 349 800 96;
  • 15) 0.000 012 159 349 800 96 × 2 = 0 + 0.000 024 318 699 601 92;
  • 16) 0.000 024 318 699 601 92 × 2 = 0 + 0.000 048 637 399 203 84;
  • 17) 0.000 048 637 399 203 84 × 2 = 0 + 0.000 097 274 798 407 68;
  • 18) 0.000 097 274 798 407 68 × 2 = 0 + 0.000 194 549 596 815 36;
  • 19) 0.000 194 549 596 815 36 × 2 = 0 + 0.000 389 099 193 630 72;
  • 20) 0.000 389 099 193 630 72 × 2 = 0 + 0.000 778 198 387 261 44;
  • 21) 0.000 778 198 387 261 44 × 2 = 0 + 0.001 556 396 774 522 88;
  • 22) 0.001 556 396 774 522 88 × 2 = 0 + 0.003 112 793 549 045 76;
  • 23) 0.003 112 793 549 045 76 × 2 = 0 + 0.006 225 587 098 091 52;
  • 24) 0.006 225 587 098 091 52 × 2 = 0 + 0.012 451 174 196 183 04;
  • 25) 0.012 451 174 196 183 04 × 2 = 0 + 0.024 902 348 392 366 08;
  • 26) 0.024 902 348 392 366 08 × 2 = 0 + 0.049 804 696 784 732 16;
  • 27) 0.049 804 696 784 732 16 × 2 = 0 + 0.099 609 393 569 464 32;
  • 28) 0.099 609 393 569 464 32 × 2 = 0 + 0.199 218 787 138 928 64;
  • 29) 0.199 218 787 138 928 64 × 2 = 0 + 0.398 437 574 277 857 28;
  • 30) 0.398 437 574 277 857 28 × 2 = 0 + 0.796 875 148 555 714 56;
  • 31) 0.796 875 148 555 714 56 × 2 = 1 + 0.593 750 297 111 429 12;
  • 32) 0.593 750 297 111 429 12 × 2 = 1 + 0.187 500 594 222 858 24;
  • 33) 0.187 500 594 222 858 24 × 2 = 0 + 0.375 001 188 445 716 48;
  • 34) 0.375 001 188 445 716 48 × 2 = 0 + 0.750 002 376 891 432 96;
  • 35) 0.750 002 376 891 432 96 × 2 = 1 + 0.500 004 753 782 865 92;
  • 36) 0.500 004 753 782 865 92 × 2 = 1 + 0.000 009 507 565 731 84;
  • 37) 0.000 009 507 565 731 84 × 2 = 0 + 0.000 019 015 131 463 68;
  • 38) 0.000 019 015 131 463 68 × 2 = 0 + 0.000 038 030 262 927 36;
  • 39) 0.000 038 030 262 927 36 × 2 = 0 + 0.000 076 060 525 854 72;
  • 40) 0.000 076 060 525 854 72 × 2 = 0 + 0.000 152 121 051 709 44;
  • 41) 0.000 152 121 051 709 44 × 2 = 0 + 0.000 304 242 103 418 88;
  • 42) 0.000 304 242 103 418 88 × 2 = 0 + 0.000 608 484 206 837 76;
  • 43) 0.000 608 484 206 837 76 × 2 = 0 + 0.001 216 968 413 675 52;
  • 44) 0.001 216 968 413 675 52 × 2 = 0 + 0.002 433 936 827 351 04;
  • 45) 0.002 433 936 827 351 04 × 2 = 0 + 0.004 867 873 654 702 08;
  • 46) 0.004 867 873 654 702 08 × 2 = 0 + 0.009 735 747 309 404 16;
  • 47) 0.009 735 747 309 404 16 × 2 = 0 + 0.019 471 494 618 808 32;
  • 48) 0.019 471 494 618 808 32 × 2 = 0 + 0.038 942 989 237 616 64;
  • 49) 0.038 942 989 237 616 64 × 2 = 0 + 0.077 885 978 475 233 28;
  • 50) 0.077 885 978 475 233 28 × 2 = 0 + 0.155 771 956 950 466 56;
  • 51) 0.155 771 956 950 466 56 × 2 = 0 + 0.311 543 913 900 933 12;
  • 52) 0.311 543 913 900 933 12 × 2 = 0 + 0.623 087 827 801 866 24;
  • 53) 0.623 087 827 801 866 24 × 2 = 1 + 0.246 175 655 603 732 48;
  • 54) 0.246 175 655 603 732 48 × 2 = 0 + 0.492 351 311 207 464 96;

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:


0.000 000 000 742 147 815(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2)

6. Positive number before normalization:

0.000 000 000 742 147 815(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 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:


0.000 000 000 742 147 815(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 10(2) × 20 =

1.1001 1000 0000 0000 0000 010(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 010


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 0010 =

100 1100 0000 0000 0000 0010


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 0010


Decimal number -0.000 000 000 742 147 815 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0010

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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