-0.000 000 000 742 148 61 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 148 61(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
-0.000 000 000 742 148 61(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:

|-0.000 000 000 742 148 61| = 0.000 000 000 742 148 61


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: 0.000 000 000 742 148 61.

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) 0.000 000 000 742 148 61 × 2 = 0 + 0.000 000 001 484 297 22;
  • 2) 0.000 000 001 484 297 22 × 2 = 0 + 0.000 000 002 968 594 44;
  • 3) 0.000 000 002 968 594 44 × 2 = 0 + 0.000 000 005 937 188 88;
  • 4) 0.000 000 005 937 188 88 × 2 = 0 + 0.000 000 011 874 377 76;
  • 5) 0.000 000 011 874 377 76 × 2 = 0 + 0.000 000 023 748 755 52;
  • 6) 0.000 000 023 748 755 52 × 2 = 0 + 0.000 000 047 497 511 04;
  • 7) 0.000 000 047 497 511 04 × 2 = 0 + 0.000 000 094 995 022 08;
  • 8) 0.000 000 094 995 022 08 × 2 = 0 + 0.000 000 189 990 044 16;
  • 9) 0.000 000 189 990 044 16 × 2 = 0 + 0.000 000 379 980 088 32;
  • 10) 0.000 000 379 980 088 32 × 2 = 0 + 0.000 000 759 960 176 64;
  • 11) 0.000 000 759 960 176 64 × 2 = 0 + 0.000 001 519 920 353 28;
  • 12) 0.000 001 519 920 353 28 × 2 = 0 + 0.000 003 039 840 706 56;
  • 13) 0.000 003 039 840 706 56 × 2 = 0 + 0.000 006 079 681 413 12;
  • 14) 0.000 006 079 681 413 12 × 2 = 0 + 0.000 012 159 362 826 24;
  • 15) 0.000 012 159 362 826 24 × 2 = 0 + 0.000 024 318 725 652 48;
  • 16) 0.000 024 318 725 652 48 × 2 = 0 + 0.000 048 637 451 304 96;
  • 17) 0.000 048 637 451 304 96 × 2 = 0 + 0.000 097 274 902 609 92;
  • 18) 0.000 097 274 902 609 92 × 2 = 0 + 0.000 194 549 805 219 84;
  • 19) 0.000 194 549 805 219 84 × 2 = 0 + 0.000 389 099 610 439 68;
  • 20) 0.000 389 099 610 439 68 × 2 = 0 + 0.000 778 199 220 879 36;
  • 21) 0.000 778 199 220 879 36 × 2 = 0 + 0.001 556 398 441 758 72;
  • 22) 0.001 556 398 441 758 72 × 2 = 0 + 0.003 112 796 883 517 44;
  • 23) 0.003 112 796 883 517 44 × 2 = 0 + 0.006 225 593 767 034 88;
  • 24) 0.006 225 593 767 034 88 × 2 = 0 + 0.012 451 187 534 069 76;
  • 25) 0.012 451 187 534 069 76 × 2 = 0 + 0.024 902 375 068 139 52;
  • 26) 0.024 902 375 068 139 52 × 2 = 0 + 0.049 804 750 136 279 04;
  • 27) 0.049 804 750 136 279 04 × 2 = 0 + 0.099 609 500 272 558 08;
  • 28) 0.099 609 500 272 558 08 × 2 = 0 + 0.199 219 000 545 116 16;
  • 29) 0.199 219 000 545 116 16 × 2 = 0 + 0.398 438 001 090 232 32;
  • 30) 0.398 438 001 090 232 32 × 2 = 0 + 0.796 876 002 180 464 64;
  • 31) 0.796 876 002 180 464 64 × 2 = 1 + 0.593 752 004 360 929 28;
  • 32) 0.593 752 004 360 929 28 × 2 = 1 + 0.187 504 008 721 858 56;
  • 33) 0.187 504 008 721 858 56 × 2 = 0 + 0.375 008 017 443 717 12;
  • 34) 0.375 008 017 443 717 12 × 2 = 0 + 0.750 016 034 887 434 24;
  • 35) 0.750 016 034 887 434 24 × 2 = 1 + 0.500 032 069 774 868 48;
  • 36) 0.500 032 069 774 868 48 × 2 = 1 + 0.000 064 139 549 736 96;
  • 37) 0.000 064 139 549 736 96 × 2 = 0 + 0.000 128 279 099 473 92;
  • 38) 0.000 128 279 099 473 92 × 2 = 0 + 0.000 256 558 198 947 84;
  • 39) 0.000 256 558 198 947 84 × 2 = 0 + 0.000 513 116 397 895 68;
  • 40) 0.000 513 116 397 895 68 × 2 = 0 + 0.001 026 232 795 791 36;
  • 41) 0.001 026 232 795 791 36 × 2 = 0 + 0.002 052 465 591 582 72;
  • 42) 0.002 052 465 591 582 72 × 2 = 0 + 0.004 104 931 183 165 44;
  • 43) 0.004 104 931 183 165 44 × 2 = 0 + 0.008 209 862 366 330 88;
  • 44) 0.008 209 862 366 330 88 × 2 = 0 + 0.016 419 724 732 661 76;
  • 45) 0.016 419 724 732 661 76 × 2 = 0 + 0.032 839 449 465 323 52;
  • 46) 0.032 839 449 465 323 52 × 2 = 0 + 0.065 678 898 930 647 04;
  • 47) 0.065 678 898 930 647 04 × 2 = 0 + 0.131 357 797 861 294 08;
  • 48) 0.131 357 797 861 294 08 × 2 = 0 + 0.262 715 595 722 588 16;
  • 49) 0.262 715 595 722 588 16 × 2 = 0 + 0.525 431 191 445 176 32;
  • 50) 0.525 431 191 445 176 32 × 2 = 1 + 0.050 862 382 890 352 64;
  • 51) 0.050 862 382 890 352 64 × 2 = 0 + 0.101 724 765 780 705 28;
  • 52) 0.101 724 765 780 705 28 × 2 = 0 + 0.203 449 531 561 410 56;
  • 53) 0.203 449 531 561 410 56 × 2 = 0 + 0.406 899 063 122 821 12;
  • 54) 0.406 899 063 122 821 12 × 2 = 0 + 0.813 798 126 245 642 24;

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:


0.000 000 000 742 148 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0100 00(2)

6. Positive number before normalization:

0.000 000 000 742 148 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0100 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:


0.000 000 000 742 148 61(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0100 00(2) × 20 =

1.1001 1000 0000 0000 0010 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 0010 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 0001 0000 =

100 1100 0000 0000 0001 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 0001 0000


Decimal number -0.000 000 000 742 148 61 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0001 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