-0.000 000 000 742 147 676 653 7 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 653 7(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
-0.000 000 000 742 147 676 653 7(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:

|-0.000 000 000 742 147 676 653 7| = 0.000 000 000 742 147 676 653 7


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: 0.000 000 000 742 147 676 653 7.

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) 0.000 000 000 742 147 676 653 7 × 2 = 0 + 0.000 000 001 484 295 353 307 4;
  • 2) 0.000 000 001 484 295 353 307 4 × 2 = 0 + 0.000 000 002 968 590 706 614 8;
  • 3) 0.000 000 002 968 590 706 614 8 × 2 = 0 + 0.000 000 005 937 181 413 229 6;
  • 4) 0.000 000 005 937 181 413 229 6 × 2 = 0 + 0.000 000 011 874 362 826 459 2;
  • 5) 0.000 000 011 874 362 826 459 2 × 2 = 0 + 0.000 000 023 748 725 652 918 4;
  • 6) 0.000 000 023 748 725 652 918 4 × 2 = 0 + 0.000 000 047 497 451 305 836 8;
  • 7) 0.000 000 047 497 451 305 836 8 × 2 = 0 + 0.000 000 094 994 902 611 673 6;
  • 8) 0.000 000 094 994 902 611 673 6 × 2 = 0 + 0.000 000 189 989 805 223 347 2;
  • 9) 0.000 000 189 989 805 223 347 2 × 2 = 0 + 0.000 000 379 979 610 446 694 4;
  • 10) 0.000 000 379 979 610 446 694 4 × 2 = 0 + 0.000 000 759 959 220 893 388 8;
  • 11) 0.000 000 759 959 220 893 388 8 × 2 = 0 + 0.000 001 519 918 441 786 777 6;
  • 12) 0.000 001 519 918 441 786 777 6 × 2 = 0 + 0.000 003 039 836 883 573 555 2;
  • 13) 0.000 003 039 836 883 573 555 2 × 2 = 0 + 0.000 006 079 673 767 147 110 4;
  • 14) 0.000 006 079 673 767 147 110 4 × 2 = 0 + 0.000 012 159 347 534 294 220 8;
  • 15) 0.000 012 159 347 534 294 220 8 × 2 = 0 + 0.000 024 318 695 068 588 441 6;
  • 16) 0.000 024 318 695 068 588 441 6 × 2 = 0 + 0.000 048 637 390 137 176 883 2;
  • 17) 0.000 048 637 390 137 176 883 2 × 2 = 0 + 0.000 097 274 780 274 353 766 4;
  • 18) 0.000 097 274 780 274 353 766 4 × 2 = 0 + 0.000 194 549 560 548 707 532 8;
  • 19) 0.000 194 549 560 548 707 532 8 × 2 = 0 + 0.000 389 099 121 097 415 065 6;
  • 20) 0.000 389 099 121 097 415 065 6 × 2 = 0 + 0.000 778 198 242 194 830 131 2;
  • 21) 0.000 778 198 242 194 830 131 2 × 2 = 0 + 0.001 556 396 484 389 660 262 4;
  • 22) 0.001 556 396 484 389 660 262 4 × 2 = 0 + 0.003 112 792 968 779 320 524 8;
  • 23) 0.003 112 792 968 779 320 524 8 × 2 = 0 + 0.006 225 585 937 558 641 049 6;
  • 24) 0.006 225 585 937 558 641 049 6 × 2 = 0 + 0.012 451 171 875 117 282 099 2;
  • 25) 0.012 451 171 875 117 282 099 2 × 2 = 0 + 0.024 902 343 750 234 564 198 4;
  • 26) 0.024 902 343 750 234 564 198 4 × 2 = 0 + 0.049 804 687 500 469 128 396 8;
  • 27) 0.049 804 687 500 469 128 396 8 × 2 = 0 + 0.099 609 375 000 938 256 793 6;
  • 28) 0.099 609 375 000 938 256 793 6 × 2 = 0 + 0.199 218 750 001 876 513 587 2;
  • 29) 0.199 218 750 001 876 513 587 2 × 2 = 0 + 0.398 437 500 003 753 027 174 4;
  • 30) 0.398 437 500 003 753 027 174 4 × 2 = 0 + 0.796 875 000 007 506 054 348 8;
  • 31) 0.796 875 000 007 506 054 348 8 × 2 = 1 + 0.593 750 000 015 012 108 697 6;
  • 32) 0.593 750 000 015 012 108 697 6 × 2 = 1 + 0.187 500 000 030 024 217 395 2;
  • 33) 0.187 500 000 030 024 217 395 2 × 2 = 0 + 0.375 000 000 060 048 434 790 4;
  • 34) 0.375 000 000 060 048 434 790 4 × 2 = 0 + 0.750 000 000 120 096 869 580 8;
  • 35) 0.750 000 000 120 096 869 580 8 × 2 = 1 + 0.500 000 000 240 193 739 161 6;
  • 36) 0.500 000 000 240 193 739 161 6 × 2 = 1 + 0.000 000 000 480 387 478 323 2;
  • 37) 0.000 000 000 480 387 478 323 2 × 2 = 0 + 0.000 000 000 960 774 956 646 4;
  • 38) 0.000 000 000 960 774 956 646 4 × 2 = 0 + 0.000 000 001 921 549 913 292 8;
  • 39) 0.000 000 001 921 549 913 292 8 × 2 = 0 + 0.000 000 003 843 099 826 585 6;
  • 40) 0.000 000 003 843 099 826 585 6 × 2 = 0 + 0.000 000 007 686 199 653 171 2;
  • 41) 0.000 000 007 686 199 653 171 2 × 2 = 0 + 0.000 000 015 372 399 306 342 4;
  • 42) 0.000 000 015 372 399 306 342 4 × 2 = 0 + 0.000 000 030 744 798 612 684 8;
  • 43) 0.000 000 030 744 798 612 684 8 × 2 = 0 + 0.000 000 061 489 597 225 369 6;
  • 44) 0.000 000 061 489 597 225 369 6 × 2 = 0 + 0.000 000 122 979 194 450 739 2;
  • 45) 0.000 000 122 979 194 450 739 2 × 2 = 0 + 0.000 000 245 958 388 901 478 4;
  • 46) 0.000 000 245 958 388 901 478 4 × 2 = 0 + 0.000 000 491 916 777 802 956 8;
  • 47) 0.000 000 491 916 777 802 956 8 × 2 = 0 + 0.000 000 983 833 555 605 913 6;
  • 48) 0.000 000 983 833 555 605 913 6 × 2 = 0 + 0.000 001 967 667 111 211 827 2;
  • 49) 0.000 001 967 667 111 211 827 2 × 2 = 0 + 0.000 003 935 334 222 423 654 4;
  • 50) 0.000 003 935 334 222 423 654 4 × 2 = 0 + 0.000 007 870 668 444 847 308 8;
  • 51) 0.000 007 870 668 444 847 308 8 × 2 = 0 + 0.000 015 741 336 889 694 617 6;
  • 52) 0.000 015 741 336 889 694 617 6 × 2 = 0 + 0.000 031 482 673 779 389 235 2;
  • 53) 0.000 031 482 673 779 389 235 2 × 2 = 0 + 0.000 062 965 347 558 778 470 4;
  • 54) 0.000 062 965 347 558 778 470 4 × 2 = 0 + 0.000 125 930 695 117 556 940 8;

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:


0.000 000 000 742 147 676 653 7(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 653 7(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:


0.000 000 000 742 147 676 653 7(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 -0.000 000 000 742 147 676 653 7 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