-0.000 000 000 742 161 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 161(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
-0.000 000 000 742 161(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:

|-0.000 000 000 742 161| = 0.000 000 000 742 161


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: 0.000 000 000 742 161.

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) 0.000 000 000 742 161 × 2 = 0 + 0.000 000 001 484 322;
  • 2) 0.000 000 001 484 322 × 2 = 0 + 0.000 000 002 968 644;
  • 3) 0.000 000 002 968 644 × 2 = 0 + 0.000 000 005 937 288;
  • 4) 0.000 000 005 937 288 × 2 = 0 + 0.000 000 011 874 576;
  • 5) 0.000 000 011 874 576 × 2 = 0 + 0.000 000 023 749 152;
  • 6) 0.000 000 023 749 152 × 2 = 0 + 0.000 000 047 498 304;
  • 7) 0.000 000 047 498 304 × 2 = 0 + 0.000 000 094 996 608;
  • 8) 0.000 000 094 996 608 × 2 = 0 + 0.000 000 189 993 216;
  • 9) 0.000 000 189 993 216 × 2 = 0 + 0.000 000 379 986 432;
  • 10) 0.000 000 379 986 432 × 2 = 0 + 0.000 000 759 972 864;
  • 11) 0.000 000 759 972 864 × 2 = 0 + 0.000 001 519 945 728;
  • 12) 0.000 001 519 945 728 × 2 = 0 + 0.000 003 039 891 456;
  • 13) 0.000 003 039 891 456 × 2 = 0 + 0.000 006 079 782 912;
  • 14) 0.000 006 079 782 912 × 2 = 0 + 0.000 012 159 565 824;
  • 15) 0.000 012 159 565 824 × 2 = 0 + 0.000 024 319 131 648;
  • 16) 0.000 024 319 131 648 × 2 = 0 + 0.000 048 638 263 296;
  • 17) 0.000 048 638 263 296 × 2 = 0 + 0.000 097 276 526 592;
  • 18) 0.000 097 276 526 592 × 2 = 0 + 0.000 194 553 053 184;
  • 19) 0.000 194 553 053 184 × 2 = 0 + 0.000 389 106 106 368;
  • 20) 0.000 389 106 106 368 × 2 = 0 + 0.000 778 212 212 736;
  • 21) 0.000 778 212 212 736 × 2 = 0 + 0.001 556 424 425 472;
  • 22) 0.001 556 424 425 472 × 2 = 0 + 0.003 112 848 850 944;
  • 23) 0.003 112 848 850 944 × 2 = 0 + 0.006 225 697 701 888;
  • 24) 0.006 225 697 701 888 × 2 = 0 + 0.012 451 395 403 776;
  • 25) 0.012 451 395 403 776 × 2 = 0 + 0.024 902 790 807 552;
  • 26) 0.024 902 790 807 552 × 2 = 0 + 0.049 805 581 615 104;
  • 27) 0.049 805 581 615 104 × 2 = 0 + 0.099 611 163 230 208;
  • 28) 0.099 611 163 230 208 × 2 = 0 + 0.199 222 326 460 416;
  • 29) 0.199 222 326 460 416 × 2 = 0 + 0.398 444 652 920 832;
  • 30) 0.398 444 652 920 832 × 2 = 0 + 0.796 889 305 841 664;
  • 31) 0.796 889 305 841 664 × 2 = 1 + 0.593 778 611 683 328;
  • 32) 0.593 778 611 683 328 × 2 = 1 + 0.187 557 223 366 656;
  • 33) 0.187 557 223 366 656 × 2 = 0 + 0.375 114 446 733 312;
  • 34) 0.375 114 446 733 312 × 2 = 0 + 0.750 228 893 466 624;
  • 35) 0.750 228 893 466 624 × 2 = 1 + 0.500 457 786 933 248;
  • 36) 0.500 457 786 933 248 × 2 = 1 + 0.000 915 573 866 496;
  • 37) 0.000 915 573 866 496 × 2 = 0 + 0.001 831 147 732 992;
  • 38) 0.001 831 147 732 992 × 2 = 0 + 0.003 662 295 465 984;
  • 39) 0.003 662 295 465 984 × 2 = 0 + 0.007 324 590 931 968;
  • 40) 0.007 324 590 931 968 × 2 = 0 + 0.014 649 181 863 936;
  • 41) 0.014 649 181 863 936 × 2 = 0 + 0.029 298 363 727 872;
  • 42) 0.029 298 363 727 872 × 2 = 0 + 0.058 596 727 455 744;
  • 43) 0.058 596 727 455 744 × 2 = 0 + 0.117 193 454 911 488;
  • 44) 0.117 193 454 911 488 × 2 = 0 + 0.234 386 909 822 976;
  • 45) 0.234 386 909 822 976 × 2 = 0 + 0.468 773 819 645 952;
  • 46) 0.468 773 819 645 952 × 2 = 0 + 0.937 547 639 291 904;
  • 47) 0.937 547 639 291 904 × 2 = 1 + 0.875 095 278 583 808;
  • 48) 0.875 095 278 583 808 × 2 = 1 + 0.750 190 557 167 616;
  • 49) 0.750 190 557 167 616 × 2 = 1 + 0.500 381 114 335 232;
  • 50) 0.500 381 114 335 232 × 2 = 1 + 0.000 762 228 670 464;
  • 51) 0.000 762 228 670 464 × 2 = 0 + 0.001 524 457 340 928;
  • 52) 0.001 524 457 340 928 × 2 = 0 + 0.003 048 914 681 856;
  • 53) 0.003 048 914 681 856 × 2 = 0 + 0.006 097 829 363 712;
  • 54) 0.006 097 829 363 712 × 2 = 0 + 0.012 195 658 727 424;

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:


0.000 000 000 742 161(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0011 1100 00(2)

6. Positive number before normalization:

0.000 000 000 742 161(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0011 1100 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:


0.000 000 000 742 161(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0011 1100 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0011 1100 00(2) × 20 =

1.1001 1000 0000 0001 1110 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 0001 1110 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 1111 0000 =

100 1100 0000 0000 1111 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 1111 0000


Decimal number -0.000 000 000 742 161 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 1111 0000

Share

» Convert decimal -0.000 000 000 742 068 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: Reputation: Bridge Between Company's Actions and Market Value. NotebookLM YouTube Video of 6.55 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