-0.000 000 000 742 224 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 224(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
-0.000 000 000 742 224(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:

|-0.000 000 000 742 224| = 0.000 000 000 742 224


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: 0.000 000 000 742 224.

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) 0.000 000 000 742 224 × 2 = 0 + 0.000 000 001 484 448;
  • 2) 0.000 000 001 484 448 × 2 = 0 + 0.000 000 002 968 896;
  • 3) 0.000 000 002 968 896 × 2 = 0 + 0.000 000 005 937 792;
  • 4) 0.000 000 005 937 792 × 2 = 0 + 0.000 000 011 875 584;
  • 5) 0.000 000 011 875 584 × 2 = 0 + 0.000 000 023 751 168;
  • 6) 0.000 000 023 751 168 × 2 = 0 + 0.000 000 047 502 336;
  • 7) 0.000 000 047 502 336 × 2 = 0 + 0.000 000 095 004 672;
  • 8) 0.000 000 095 004 672 × 2 = 0 + 0.000 000 190 009 344;
  • 9) 0.000 000 190 009 344 × 2 = 0 + 0.000 000 380 018 688;
  • 10) 0.000 000 380 018 688 × 2 = 0 + 0.000 000 760 037 376;
  • 11) 0.000 000 760 037 376 × 2 = 0 + 0.000 001 520 074 752;
  • 12) 0.000 001 520 074 752 × 2 = 0 + 0.000 003 040 149 504;
  • 13) 0.000 003 040 149 504 × 2 = 0 + 0.000 006 080 299 008;
  • 14) 0.000 006 080 299 008 × 2 = 0 + 0.000 012 160 598 016;
  • 15) 0.000 012 160 598 016 × 2 = 0 + 0.000 024 321 196 032;
  • 16) 0.000 024 321 196 032 × 2 = 0 + 0.000 048 642 392 064;
  • 17) 0.000 048 642 392 064 × 2 = 0 + 0.000 097 284 784 128;
  • 18) 0.000 097 284 784 128 × 2 = 0 + 0.000 194 569 568 256;
  • 19) 0.000 194 569 568 256 × 2 = 0 + 0.000 389 139 136 512;
  • 20) 0.000 389 139 136 512 × 2 = 0 + 0.000 778 278 273 024;
  • 21) 0.000 778 278 273 024 × 2 = 0 + 0.001 556 556 546 048;
  • 22) 0.001 556 556 546 048 × 2 = 0 + 0.003 113 113 092 096;
  • 23) 0.003 113 113 092 096 × 2 = 0 + 0.006 226 226 184 192;
  • 24) 0.006 226 226 184 192 × 2 = 0 + 0.012 452 452 368 384;
  • 25) 0.012 452 452 368 384 × 2 = 0 + 0.024 904 904 736 768;
  • 26) 0.024 904 904 736 768 × 2 = 0 + 0.049 809 809 473 536;
  • 27) 0.049 809 809 473 536 × 2 = 0 + 0.099 619 618 947 072;
  • 28) 0.099 619 618 947 072 × 2 = 0 + 0.199 239 237 894 144;
  • 29) 0.199 239 237 894 144 × 2 = 0 + 0.398 478 475 788 288;
  • 30) 0.398 478 475 788 288 × 2 = 0 + 0.796 956 951 576 576;
  • 31) 0.796 956 951 576 576 × 2 = 1 + 0.593 913 903 153 152;
  • 32) 0.593 913 903 153 152 × 2 = 1 + 0.187 827 806 306 304;
  • 33) 0.187 827 806 306 304 × 2 = 0 + 0.375 655 612 612 608;
  • 34) 0.375 655 612 612 608 × 2 = 0 + 0.751 311 225 225 216;
  • 35) 0.751 311 225 225 216 × 2 = 1 + 0.502 622 450 450 432;
  • 36) 0.502 622 450 450 432 × 2 = 1 + 0.005 244 900 900 864;
  • 37) 0.005 244 900 900 864 × 2 = 0 + 0.010 489 801 801 728;
  • 38) 0.010 489 801 801 728 × 2 = 0 + 0.020 979 603 603 456;
  • 39) 0.020 979 603 603 456 × 2 = 0 + 0.041 959 207 206 912;
  • 40) 0.041 959 207 206 912 × 2 = 0 + 0.083 918 414 413 824;
  • 41) 0.083 918 414 413 824 × 2 = 0 + 0.167 836 828 827 648;
  • 42) 0.167 836 828 827 648 × 2 = 0 + 0.335 673 657 655 296;
  • 43) 0.335 673 657 655 296 × 2 = 0 + 0.671 347 315 310 592;
  • 44) 0.671 347 315 310 592 × 2 = 1 + 0.342 694 630 621 184;
  • 45) 0.342 694 630 621 184 × 2 = 0 + 0.685 389 261 242 368;
  • 46) 0.685 389 261 242 368 × 2 = 1 + 0.370 778 522 484 736;
  • 47) 0.370 778 522 484 736 × 2 = 0 + 0.741 557 044 969 472;
  • 48) 0.741 557 044 969 472 × 2 = 1 + 0.483 114 089 938 944;
  • 49) 0.483 114 089 938 944 × 2 = 0 + 0.966 228 179 877 888;
  • 50) 0.966 228 179 877 888 × 2 = 1 + 0.932 456 359 755 776;
  • 51) 0.932 456 359 755 776 × 2 = 1 + 0.864 912 719 511 552;
  • 52) 0.864 912 719 511 552 × 2 = 1 + 0.729 825 439 023 104;
  • 53) 0.729 825 439 023 104 × 2 = 1 + 0.459 650 878 046 208;
  • 54) 0.459 650 878 046 208 × 2 = 0 + 0.919 301 756 092 416;

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:


0.000 000 000 742 224(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0101 0111 10(2)

6. Positive number before normalization:

0.000 000 000 742 224(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0101 0111 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:


0.000 000 000 742 224(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0101 0111 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0001 0101 0111 10(2) × 20 =

1.1001 1000 0000 1010 1011 110(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 1010 1011 110


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 0101 0101 1110 =

100 1100 0000 0101 0101 1110


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 0101 0101 1110


Decimal number -0.000 000 000 742 224 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0101 0101 1110

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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