-0.000 000 000 742 147 673 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 673 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
-0.000 000 000 742 147 673 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:

|-0.000 000 000 742 147 673 6| = 0.000 000 000 742 147 673 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: 0.000 000 000 742 147 673 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) 0.000 000 000 742 147 673 6 × 2 = 0 + 0.000 000 001 484 295 347 2;
  • 2) 0.000 000 001 484 295 347 2 × 2 = 0 + 0.000 000 002 968 590 694 4;
  • 3) 0.000 000 002 968 590 694 4 × 2 = 0 + 0.000 000 005 937 181 388 8;
  • 4) 0.000 000 005 937 181 388 8 × 2 = 0 + 0.000 000 011 874 362 777 6;
  • 5) 0.000 000 011 874 362 777 6 × 2 = 0 + 0.000 000 023 748 725 555 2;
  • 6) 0.000 000 023 748 725 555 2 × 2 = 0 + 0.000 000 047 497 451 110 4;
  • 7) 0.000 000 047 497 451 110 4 × 2 = 0 + 0.000 000 094 994 902 220 8;
  • 8) 0.000 000 094 994 902 220 8 × 2 = 0 + 0.000 000 189 989 804 441 6;
  • 9) 0.000 000 189 989 804 441 6 × 2 = 0 + 0.000 000 379 979 608 883 2;
  • 10) 0.000 000 379 979 608 883 2 × 2 = 0 + 0.000 000 759 959 217 766 4;
  • 11) 0.000 000 759 959 217 766 4 × 2 = 0 + 0.000 001 519 918 435 532 8;
  • 12) 0.000 001 519 918 435 532 8 × 2 = 0 + 0.000 003 039 836 871 065 6;
  • 13) 0.000 003 039 836 871 065 6 × 2 = 0 + 0.000 006 079 673 742 131 2;
  • 14) 0.000 006 079 673 742 131 2 × 2 = 0 + 0.000 012 159 347 484 262 4;
  • 15) 0.000 012 159 347 484 262 4 × 2 = 0 + 0.000 024 318 694 968 524 8;
  • 16) 0.000 024 318 694 968 524 8 × 2 = 0 + 0.000 048 637 389 937 049 6;
  • 17) 0.000 048 637 389 937 049 6 × 2 = 0 + 0.000 097 274 779 874 099 2;
  • 18) 0.000 097 274 779 874 099 2 × 2 = 0 + 0.000 194 549 559 748 198 4;
  • 19) 0.000 194 549 559 748 198 4 × 2 = 0 + 0.000 389 099 119 496 396 8;
  • 20) 0.000 389 099 119 496 396 8 × 2 = 0 + 0.000 778 198 238 992 793 6;
  • 21) 0.000 778 198 238 992 793 6 × 2 = 0 + 0.001 556 396 477 985 587 2;
  • 22) 0.001 556 396 477 985 587 2 × 2 = 0 + 0.003 112 792 955 971 174 4;
  • 23) 0.003 112 792 955 971 174 4 × 2 = 0 + 0.006 225 585 911 942 348 8;
  • 24) 0.006 225 585 911 942 348 8 × 2 = 0 + 0.012 451 171 823 884 697 6;
  • 25) 0.012 451 171 823 884 697 6 × 2 = 0 + 0.024 902 343 647 769 395 2;
  • 26) 0.024 902 343 647 769 395 2 × 2 = 0 + 0.049 804 687 295 538 790 4;
  • 27) 0.049 804 687 295 538 790 4 × 2 = 0 + 0.099 609 374 591 077 580 8;
  • 28) 0.099 609 374 591 077 580 8 × 2 = 0 + 0.199 218 749 182 155 161 6;
  • 29) 0.199 218 749 182 155 161 6 × 2 = 0 + 0.398 437 498 364 310 323 2;
  • 30) 0.398 437 498 364 310 323 2 × 2 = 0 + 0.796 874 996 728 620 646 4;
  • 31) 0.796 874 996 728 620 646 4 × 2 = 1 + 0.593 749 993 457 241 292 8;
  • 32) 0.593 749 993 457 241 292 8 × 2 = 1 + 0.187 499 986 914 482 585 6;
  • 33) 0.187 499 986 914 482 585 6 × 2 = 0 + 0.374 999 973 828 965 171 2;
  • 34) 0.374 999 973 828 965 171 2 × 2 = 0 + 0.749 999 947 657 930 342 4;
  • 35) 0.749 999 947 657 930 342 4 × 2 = 1 + 0.499 999 895 315 860 684 8;
  • 36) 0.499 999 895 315 860 684 8 × 2 = 0 + 0.999 999 790 631 721 369 6;
  • 37) 0.999 999 790 631 721 369 6 × 2 = 1 + 0.999 999 581 263 442 739 2;
  • 38) 0.999 999 581 263 442 739 2 × 2 = 1 + 0.999 999 162 526 885 478 4;
  • 39) 0.999 999 162 526 885 478 4 × 2 = 1 + 0.999 998 325 053 770 956 8;
  • 40) 0.999 998 325 053 770 956 8 × 2 = 1 + 0.999 996 650 107 541 913 6;
  • 41) 0.999 996 650 107 541 913 6 × 2 = 1 + 0.999 993 300 215 083 827 2;
  • 42) 0.999 993 300 215 083 827 2 × 2 = 1 + 0.999 986 600 430 167 654 4;
  • 43) 0.999 986 600 430 167 654 4 × 2 = 1 + 0.999 973 200 860 335 308 8;
  • 44) 0.999 973 200 860 335 308 8 × 2 = 1 + 0.999 946 401 720 670 617 6;
  • 45) 0.999 946 401 720 670 617 6 × 2 = 1 + 0.999 892 803 441 341 235 2;
  • 46) 0.999 892 803 441 341 235 2 × 2 = 1 + 0.999 785 606 882 682 470 4;
  • 47) 0.999 785 606 882 682 470 4 × 2 = 1 + 0.999 571 213 765 364 940 8;
  • 48) 0.999 571 213 765 364 940 8 × 2 = 1 + 0.999 142 427 530 729 881 6;
  • 49) 0.999 142 427 530 729 881 6 × 2 = 1 + 0.998 284 855 061 459 763 2;
  • 50) 0.998 284 855 061 459 763 2 × 2 = 1 + 0.996 569 710 122 919 526 4;
  • 51) 0.996 569 710 122 919 526 4 × 2 = 1 + 0.993 139 420 245 839 052 8;
  • 52) 0.993 139 420 245 839 052 8 × 2 = 1 + 0.986 278 840 491 678 105 6;
  • 53) 0.986 278 840 491 678 105 6 × 2 = 1 + 0.972 557 680 983 356 211 2;
  • 54) 0.972 557 680 983 356 211 2 × 2 = 1 + 0.945 115 361 966 712 422 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:


0.000 000 000 742 147 673 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 673 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:


0.000 000 000 742 147 673 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 -0.000 000 000 742 147 673 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 672 7 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Hidden Requirement in Every Single Job Description. NotebookLM YouTube Video of 8.15 minutes

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