-0.000 000 000 742 147 676 632 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 632 6(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
-0.000 000 000 742 147 676 632 6(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:

|-0.000 000 000 742 147 676 632 6| = 0.000 000 000 742 147 676 632 6


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: 0.000 000 000 742 147 676 632 6.

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) 0.000 000 000 742 147 676 632 6 × 2 = 0 + 0.000 000 001 484 295 353 265 2;
  • 2) 0.000 000 001 484 295 353 265 2 × 2 = 0 + 0.000 000 002 968 590 706 530 4;
  • 3) 0.000 000 002 968 590 706 530 4 × 2 = 0 + 0.000 000 005 937 181 413 060 8;
  • 4) 0.000 000 005 937 181 413 060 8 × 2 = 0 + 0.000 000 011 874 362 826 121 6;
  • 5) 0.000 000 011 874 362 826 121 6 × 2 = 0 + 0.000 000 023 748 725 652 243 2;
  • 6) 0.000 000 023 748 725 652 243 2 × 2 = 0 + 0.000 000 047 497 451 304 486 4;
  • 7) 0.000 000 047 497 451 304 486 4 × 2 = 0 + 0.000 000 094 994 902 608 972 8;
  • 8) 0.000 000 094 994 902 608 972 8 × 2 = 0 + 0.000 000 189 989 805 217 945 6;
  • 9) 0.000 000 189 989 805 217 945 6 × 2 = 0 + 0.000 000 379 979 610 435 891 2;
  • 10) 0.000 000 379 979 610 435 891 2 × 2 = 0 + 0.000 000 759 959 220 871 782 4;
  • 11) 0.000 000 759 959 220 871 782 4 × 2 = 0 + 0.000 001 519 918 441 743 564 8;
  • 12) 0.000 001 519 918 441 743 564 8 × 2 = 0 + 0.000 003 039 836 883 487 129 6;
  • 13) 0.000 003 039 836 883 487 129 6 × 2 = 0 + 0.000 006 079 673 766 974 259 2;
  • 14) 0.000 006 079 673 766 974 259 2 × 2 = 0 + 0.000 012 159 347 533 948 518 4;
  • 15) 0.000 012 159 347 533 948 518 4 × 2 = 0 + 0.000 024 318 695 067 897 036 8;
  • 16) 0.000 024 318 695 067 897 036 8 × 2 = 0 + 0.000 048 637 390 135 794 073 6;
  • 17) 0.000 048 637 390 135 794 073 6 × 2 = 0 + 0.000 097 274 780 271 588 147 2;
  • 18) 0.000 097 274 780 271 588 147 2 × 2 = 0 + 0.000 194 549 560 543 176 294 4;
  • 19) 0.000 194 549 560 543 176 294 4 × 2 = 0 + 0.000 389 099 121 086 352 588 8;
  • 20) 0.000 389 099 121 086 352 588 8 × 2 = 0 + 0.000 778 198 242 172 705 177 6;
  • 21) 0.000 778 198 242 172 705 177 6 × 2 = 0 + 0.001 556 396 484 345 410 355 2;
  • 22) 0.001 556 396 484 345 410 355 2 × 2 = 0 + 0.003 112 792 968 690 820 710 4;
  • 23) 0.003 112 792 968 690 820 710 4 × 2 = 0 + 0.006 225 585 937 381 641 420 8;
  • 24) 0.006 225 585 937 381 641 420 8 × 2 = 0 + 0.012 451 171 874 763 282 841 6;
  • 25) 0.012 451 171 874 763 282 841 6 × 2 = 0 + 0.024 902 343 749 526 565 683 2;
  • 26) 0.024 902 343 749 526 565 683 2 × 2 = 0 + 0.049 804 687 499 053 131 366 4;
  • 27) 0.049 804 687 499 053 131 366 4 × 2 = 0 + 0.099 609 374 998 106 262 732 8;
  • 28) 0.099 609 374 998 106 262 732 8 × 2 = 0 + 0.199 218 749 996 212 525 465 6;
  • 29) 0.199 218 749 996 212 525 465 6 × 2 = 0 + 0.398 437 499 992 425 050 931 2;
  • 30) 0.398 437 499 992 425 050 931 2 × 2 = 0 + 0.796 874 999 984 850 101 862 4;
  • 31) 0.796 874 999 984 850 101 862 4 × 2 = 1 + 0.593 749 999 969 700 203 724 8;
  • 32) 0.593 749 999 969 700 203 724 8 × 2 = 1 + 0.187 499 999 939 400 407 449 6;
  • 33) 0.187 499 999 939 400 407 449 6 × 2 = 0 + 0.374 999 999 878 800 814 899 2;
  • 34) 0.374 999 999 878 800 814 899 2 × 2 = 0 + 0.749 999 999 757 601 629 798 4;
  • 35) 0.749 999 999 757 601 629 798 4 × 2 = 1 + 0.499 999 999 515 203 259 596 8;
  • 36) 0.499 999 999 515 203 259 596 8 × 2 = 0 + 0.999 999 999 030 406 519 193 6;
  • 37) 0.999 999 999 030 406 519 193 6 × 2 = 1 + 0.999 999 998 060 813 038 387 2;
  • 38) 0.999 999 998 060 813 038 387 2 × 2 = 1 + 0.999 999 996 121 626 076 774 4;
  • 39) 0.999 999 996 121 626 076 774 4 × 2 = 1 + 0.999 999 992 243 252 153 548 8;
  • 40) 0.999 999 992 243 252 153 548 8 × 2 = 1 + 0.999 999 984 486 504 307 097 6;
  • 41) 0.999 999 984 486 504 307 097 6 × 2 = 1 + 0.999 999 968 973 008 614 195 2;
  • 42) 0.999 999 968 973 008 614 195 2 × 2 = 1 + 0.999 999 937 946 017 228 390 4;
  • 43) 0.999 999 937 946 017 228 390 4 × 2 = 1 + 0.999 999 875 892 034 456 780 8;
  • 44) 0.999 999 875 892 034 456 780 8 × 2 = 1 + 0.999 999 751 784 068 913 561 6;
  • 45) 0.999 999 751 784 068 913 561 6 × 2 = 1 + 0.999 999 503 568 137 827 123 2;
  • 46) 0.999 999 503 568 137 827 123 2 × 2 = 1 + 0.999 999 007 136 275 654 246 4;
  • 47) 0.999 999 007 136 275 654 246 4 × 2 = 1 + 0.999 998 014 272 551 308 492 8;
  • 48) 0.999 998 014 272 551 308 492 8 × 2 = 1 + 0.999 996 028 545 102 616 985 6;
  • 49) 0.999 996 028 545 102 616 985 6 × 2 = 1 + 0.999 992 057 090 205 233 971 2;
  • 50) 0.999 992 057 090 205 233 971 2 × 2 = 1 + 0.999 984 114 180 410 467 942 4;
  • 51) 0.999 984 114 180 410 467 942 4 × 2 = 1 + 0.999 968 228 360 820 935 884 8;
  • 52) 0.999 968 228 360 820 935 884 8 × 2 = 1 + 0.999 936 456 721 641 871 769 6;
  • 53) 0.999 936 456 721 641 871 769 6 × 2 = 1 + 0.999 872 913 443 283 743 539 2;
  • 54) 0.999 872 913 443 283 743 539 2 × 2 = 1 + 0.999 745 826 886 567 487 078 4;

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:


0.000 000 000 742 147 676 632 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 632 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(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:


0.000 000 000 742 147 676 632 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(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 111


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 1111 =

100 1011 1111 1111 1111 1111


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 1111


Decimal number -0.000 000 000 742 147 676 632 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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