-0.000 000 000 742 146 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 146 5(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
-0.000 000 000 742 146 5(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:

|-0.000 000 000 742 146 5| = 0.000 000 000 742 146 5


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: 0.000 000 000 742 146 5.

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) 0.000 000 000 742 146 5 × 2 = 0 + 0.000 000 001 484 293;
  • 2) 0.000 000 001 484 293 × 2 = 0 + 0.000 000 002 968 586;
  • 3) 0.000 000 002 968 586 × 2 = 0 + 0.000 000 005 937 172;
  • 4) 0.000 000 005 937 172 × 2 = 0 + 0.000 000 011 874 344;
  • 5) 0.000 000 011 874 344 × 2 = 0 + 0.000 000 023 748 688;
  • 6) 0.000 000 023 748 688 × 2 = 0 + 0.000 000 047 497 376;
  • 7) 0.000 000 047 497 376 × 2 = 0 + 0.000 000 094 994 752;
  • 8) 0.000 000 094 994 752 × 2 = 0 + 0.000 000 189 989 504;
  • 9) 0.000 000 189 989 504 × 2 = 0 + 0.000 000 379 979 008;
  • 10) 0.000 000 379 979 008 × 2 = 0 + 0.000 000 759 958 016;
  • 11) 0.000 000 759 958 016 × 2 = 0 + 0.000 001 519 916 032;
  • 12) 0.000 001 519 916 032 × 2 = 0 + 0.000 003 039 832 064;
  • 13) 0.000 003 039 832 064 × 2 = 0 + 0.000 006 079 664 128;
  • 14) 0.000 006 079 664 128 × 2 = 0 + 0.000 012 159 328 256;
  • 15) 0.000 012 159 328 256 × 2 = 0 + 0.000 024 318 656 512;
  • 16) 0.000 024 318 656 512 × 2 = 0 + 0.000 048 637 313 024;
  • 17) 0.000 048 637 313 024 × 2 = 0 + 0.000 097 274 626 048;
  • 18) 0.000 097 274 626 048 × 2 = 0 + 0.000 194 549 252 096;
  • 19) 0.000 194 549 252 096 × 2 = 0 + 0.000 389 098 504 192;
  • 20) 0.000 389 098 504 192 × 2 = 0 + 0.000 778 197 008 384;
  • 21) 0.000 778 197 008 384 × 2 = 0 + 0.001 556 394 016 768;
  • 22) 0.001 556 394 016 768 × 2 = 0 + 0.003 112 788 033 536;
  • 23) 0.003 112 788 033 536 × 2 = 0 + 0.006 225 576 067 072;
  • 24) 0.006 225 576 067 072 × 2 = 0 + 0.012 451 152 134 144;
  • 25) 0.012 451 152 134 144 × 2 = 0 + 0.024 902 304 268 288;
  • 26) 0.024 902 304 268 288 × 2 = 0 + 0.049 804 608 536 576;
  • 27) 0.049 804 608 536 576 × 2 = 0 + 0.099 609 217 073 152;
  • 28) 0.099 609 217 073 152 × 2 = 0 + 0.199 218 434 146 304;
  • 29) 0.199 218 434 146 304 × 2 = 0 + 0.398 436 868 292 608;
  • 30) 0.398 436 868 292 608 × 2 = 0 + 0.796 873 736 585 216;
  • 31) 0.796 873 736 585 216 × 2 = 1 + 0.593 747 473 170 432;
  • 32) 0.593 747 473 170 432 × 2 = 1 + 0.187 494 946 340 864;
  • 33) 0.187 494 946 340 864 × 2 = 0 + 0.374 989 892 681 728;
  • 34) 0.374 989 892 681 728 × 2 = 0 + 0.749 979 785 363 456;
  • 35) 0.749 979 785 363 456 × 2 = 1 + 0.499 959 570 726 912;
  • 36) 0.499 959 570 726 912 × 2 = 0 + 0.999 919 141 453 824;
  • 37) 0.999 919 141 453 824 × 2 = 1 + 0.999 838 282 907 648;
  • 38) 0.999 838 282 907 648 × 2 = 1 + 0.999 676 565 815 296;
  • 39) 0.999 676 565 815 296 × 2 = 1 + 0.999 353 131 630 592;
  • 40) 0.999 353 131 630 592 × 2 = 1 + 0.998 706 263 261 184;
  • 41) 0.998 706 263 261 184 × 2 = 1 + 0.997 412 526 522 368;
  • 42) 0.997 412 526 522 368 × 2 = 1 + 0.994 825 053 044 736;
  • 43) 0.994 825 053 044 736 × 2 = 1 + 0.989 650 106 089 472;
  • 44) 0.989 650 106 089 472 × 2 = 1 + 0.979 300 212 178 944;
  • 45) 0.979 300 212 178 944 × 2 = 1 + 0.958 600 424 357 888;
  • 46) 0.958 600 424 357 888 × 2 = 1 + 0.917 200 848 715 776;
  • 47) 0.917 200 848 715 776 × 2 = 1 + 0.834 401 697 431 552;
  • 48) 0.834 401 697 431 552 × 2 = 1 + 0.668 803 394 863 104;
  • 49) 0.668 803 394 863 104 × 2 = 1 + 0.337 606 789 726 208;
  • 50) 0.337 606 789 726 208 × 2 = 0 + 0.675 213 579 452 416;
  • 51) 0.675 213 579 452 416 × 2 = 1 + 0.350 427 158 904 832;
  • 52) 0.350 427 158 904 832 × 2 = 0 + 0.700 854 317 809 664;
  • 53) 0.700 854 317 809 664 × 2 = 1 + 0.401 708 635 619 328;
  • 54) 0.401 708 635 619 328 × 2 = 0 + 0.803 417 271 238 656;

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:


0.000 000 000 742 146 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 10(2)

6. Positive number before normalization:

0.000 000 000 742 146 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 10(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:


0.000 000 000 742 146 5(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1010 10(2) × 20 =

1.1001 0111 1111 1111 1101 010(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 1101 010


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 1110 1010 =

100 1011 1111 1111 1110 1010


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 1110 1010


Decimal number -0.000 000 000 742 146 5 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1110 1010

Share

» Convert decimal -0.000 000 000 742 143 5 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: Your greatest rival, your most powerful ally. NotebookLM YouTube Video of 7.54 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