-0.000 000 000 742 147 676 643 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 643 8(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
-0.000 000 000 742 147 676 643 8(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:

|-0.000 000 000 742 147 676 643 8| = 0.000 000 000 742 147 676 643 8


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: 0.000 000 000 742 147 676 643 8.

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) 0.000 000 000 742 147 676 643 8 × 2 = 0 + 0.000 000 001 484 295 353 287 6;
  • 2) 0.000 000 001 484 295 353 287 6 × 2 = 0 + 0.000 000 002 968 590 706 575 2;
  • 3) 0.000 000 002 968 590 706 575 2 × 2 = 0 + 0.000 000 005 937 181 413 150 4;
  • 4) 0.000 000 005 937 181 413 150 4 × 2 = 0 + 0.000 000 011 874 362 826 300 8;
  • 5) 0.000 000 011 874 362 826 300 8 × 2 = 0 + 0.000 000 023 748 725 652 601 6;
  • 6) 0.000 000 023 748 725 652 601 6 × 2 = 0 + 0.000 000 047 497 451 305 203 2;
  • 7) 0.000 000 047 497 451 305 203 2 × 2 = 0 + 0.000 000 094 994 902 610 406 4;
  • 8) 0.000 000 094 994 902 610 406 4 × 2 = 0 + 0.000 000 189 989 805 220 812 8;
  • 9) 0.000 000 189 989 805 220 812 8 × 2 = 0 + 0.000 000 379 979 610 441 625 6;
  • 10) 0.000 000 379 979 610 441 625 6 × 2 = 0 + 0.000 000 759 959 220 883 251 2;
  • 11) 0.000 000 759 959 220 883 251 2 × 2 = 0 + 0.000 001 519 918 441 766 502 4;
  • 12) 0.000 001 519 918 441 766 502 4 × 2 = 0 + 0.000 003 039 836 883 533 004 8;
  • 13) 0.000 003 039 836 883 533 004 8 × 2 = 0 + 0.000 006 079 673 767 066 009 6;
  • 14) 0.000 006 079 673 767 066 009 6 × 2 = 0 + 0.000 012 159 347 534 132 019 2;
  • 15) 0.000 012 159 347 534 132 019 2 × 2 = 0 + 0.000 024 318 695 068 264 038 4;
  • 16) 0.000 024 318 695 068 264 038 4 × 2 = 0 + 0.000 048 637 390 136 528 076 8;
  • 17) 0.000 048 637 390 136 528 076 8 × 2 = 0 + 0.000 097 274 780 273 056 153 6;
  • 18) 0.000 097 274 780 273 056 153 6 × 2 = 0 + 0.000 194 549 560 546 112 307 2;
  • 19) 0.000 194 549 560 546 112 307 2 × 2 = 0 + 0.000 389 099 121 092 224 614 4;
  • 20) 0.000 389 099 121 092 224 614 4 × 2 = 0 + 0.000 778 198 242 184 449 228 8;
  • 21) 0.000 778 198 242 184 449 228 8 × 2 = 0 + 0.001 556 396 484 368 898 457 6;
  • 22) 0.001 556 396 484 368 898 457 6 × 2 = 0 + 0.003 112 792 968 737 796 915 2;
  • 23) 0.003 112 792 968 737 796 915 2 × 2 = 0 + 0.006 225 585 937 475 593 830 4;
  • 24) 0.006 225 585 937 475 593 830 4 × 2 = 0 + 0.012 451 171 874 951 187 660 8;
  • 25) 0.012 451 171 874 951 187 660 8 × 2 = 0 + 0.024 902 343 749 902 375 321 6;
  • 26) 0.024 902 343 749 902 375 321 6 × 2 = 0 + 0.049 804 687 499 804 750 643 2;
  • 27) 0.049 804 687 499 804 750 643 2 × 2 = 0 + 0.099 609 374 999 609 501 286 4;
  • 28) 0.099 609 374 999 609 501 286 4 × 2 = 0 + 0.199 218 749 999 219 002 572 8;
  • 29) 0.199 218 749 999 219 002 572 8 × 2 = 0 + 0.398 437 499 998 438 005 145 6;
  • 30) 0.398 437 499 998 438 005 145 6 × 2 = 0 + 0.796 874 999 996 876 010 291 2;
  • 31) 0.796 874 999 996 876 010 291 2 × 2 = 1 + 0.593 749 999 993 752 020 582 4;
  • 32) 0.593 749 999 993 752 020 582 4 × 2 = 1 + 0.187 499 999 987 504 041 164 8;
  • 33) 0.187 499 999 987 504 041 164 8 × 2 = 0 + 0.374 999 999 975 008 082 329 6;
  • 34) 0.374 999 999 975 008 082 329 6 × 2 = 0 + 0.749 999 999 950 016 164 659 2;
  • 35) 0.749 999 999 950 016 164 659 2 × 2 = 1 + 0.499 999 999 900 032 329 318 4;
  • 36) 0.499 999 999 900 032 329 318 4 × 2 = 0 + 0.999 999 999 800 064 658 636 8;
  • 37) 0.999 999 999 800 064 658 636 8 × 2 = 1 + 0.999 999 999 600 129 317 273 6;
  • 38) 0.999 999 999 600 129 317 273 6 × 2 = 1 + 0.999 999 999 200 258 634 547 2;
  • 39) 0.999 999 999 200 258 634 547 2 × 2 = 1 + 0.999 999 998 400 517 269 094 4;
  • 40) 0.999 999 998 400 517 269 094 4 × 2 = 1 + 0.999 999 996 801 034 538 188 8;
  • 41) 0.999 999 996 801 034 538 188 8 × 2 = 1 + 0.999 999 993 602 069 076 377 6;
  • 42) 0.999 999 993 602 069 076 377 6 × 2 = 1 + 0.999 999 987 204 138 152 755 2;
  • 43) 0.999 999 987 204 138 152 755 2 × 2 = 1 + 0.999 999 974 408 276 305 510 4;
  • 44) 0.999 999 974 408 276 305 510 4 × 2 = 1 + 0.999 999 948 816 552 611 020 8;
  • 45) 0.999 999 948 816 552 611 020 8 × 2 = 1 + 0.999 999 897 633 105 222 041 6;
  • 46) 0.999 999 897 633 105 222 041 6 × 2 = 1 + 0.999 999 795 266 210 444 083 2;
  • 47) 0.999 999 795 266 210 444 083 2 × 2 = 1 + 0.999 999 590 532 420 888 166 4;
  • 48) 0.999 999 590 532 420 888 166 4 × 2 = 1 + 0.999 999 181 064 841 776 332 8;
  • 49) 0.999 999 181 064 841 776 332 8 × 2 = 1 + 0.999 998 362 129 683 552 665 6;
  • 50) 0.999 998 362 129 683 552 665 6 × 2 = 1 + 0.999 996 724 259 367 105 331 2;
  • 51) 0.999 996 724 259 367 105 331 2 × 2 = 1 + 0.999 993 448 518 734 210 662 4;
  • 52) 0.999 993 448 518 734 210 662 4 × 2 = 1 + 0.999 986 897 037 468 421 324 8;
  • 53) 0.999 986 897 037 468 421 324 8 × 2 = 1 + 0.999 973 794 074 936 842 649 6;
  • 54) 0.999 973 794 074 936 842 649 6 × 2 = 1 + 0.999 947 588 149 873 685 299 2;

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:


0.000 000 000 742 147 676 643 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 643 8(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:


0.000 000 000 742 147 676 643 8(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 -0.000 000 000 742 147 676 643 8 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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