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

Convert decimal -0.000 000 000 742 147(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(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| = 0.000 000 000 742 147


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.

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 × 2 = 0 + 0.000 000 001 484 294;
  • 2) 0.000 000 001 484 294 × 2 = 0 + 0.000 000 002 968 588;
  • 3) 0.000 000 002 968 588 × 2 = 0 + 0.000 000 005 937 176;
  • 4) 0.000 000 005 937 176 × 2 = 0 + 0.000 000 011 874 352;
  • 5) 0.000 000 011 874 352 × 2 = 0 + 0.000 000 023 748 704;
  • 6) 0.000 000 023 748 704 × 2 = 0 + 0.000 000 047 497 408;
  • 7) 0.000 000 047 497 408 × 2 = 0 + 0.000 000 094 994 816;
  • 8) 0.000 000 094 994 816 × 2 = 0 + 0.000 000 189 989 632;
  • 9) 0.000 000 189 989 632 × 2 = 0 + 0.000 000 379 979 264;
  • 10) 0.000 000 379 979 264 × 2 = 0 + 0.000 000 759 958 528;
  • 11) 0.000 000 759 958 528 × 2 = 0 + 0.000 001 519 917 056;
  • 12) 0.000 001 519 917 056 × 2 = 0 + 0.000 003 039 834 112;
  • 13) 0.000 003 039 834 112 × 2 = 0 + 0.000 006 079 668 224;
  • 14) 0.000 006 079 668 224 × 2 = 0 + 0.000 012 159 336 448;
  • 15) 0.000 012 159 336 448 × 2 = 0 + 0.000 024 318 672 896;
  • 16) 0.000 024 318 672 896 × 2 = 0 + 0.000 048 637 345 792;
  • 17) 0.000 048 637 345 792 × 2 = 0 + 0.000 097 274 691 584;
  • 18) 0.000 097 274 691 584 × 2 = 0 + 0.000 194 549 383 168;
  • 19) 0.000 194 549 383 168 × 2 = 0 + 0.000 389 098 766 336;
  • 20) 0.000 389 098 766 336 × 2 = 0 + 0.000 778 197 532 672;
  • 21) 0.000 778 197 532 672 × 2 = 0 + 0.001 556 395 065 344;
  • 22) 0.001 556 395 065 344 × 2 = 0 + 0.003 112 790 130 688;
  • 23) 0.003 112 790 130 688 × 2 = 0 + 0.006 225 580 261 376;
  • 24) 0.006 225 580 261 376 × 2 = 0 + 0.012 451 160 522 752;
  • 25) 0.012 451 160 522 752 × 2 = 0 + 0.024 902 321 045 504;
  • 26) 0.024 902 321 045 504 × 2 = 0 + 0.049 804 642 091 008;
  • 27) 0.049 804 642 091 008 × 2 = 0 + 0.099 609 284 182 016;
  • 28) 0.099 609 284 182 016 × 2 = 0 + 0.199 218 568 364 032;
  • 29) 0.199 218 568 364 032 × 2 = 0 + 0.398 437 136 728 064;
  • 30) 0.398 437 136 728 064 × 2 = 0 + 0.796 874 273 456 128;
  • 31) 0.796 874 273 456 128 × 2 = 1 + 0.593 748 546 912 256;
  • 32) 0.593 748 546 912 256 × 2 = 1 + 0.187 497 093 824 512;
  • 33) 0.187 497 093 824 512 × 2 = 0 + 0.374 994 187 649 024;
  • 34) 0.374 994 187 649 024 × 2 = 0 + 0.749 988 375 298 048;
  • 35) 0.749 988 375 298 048 × 2 = 1 + 0.499 976 750 596 096;
  • 36) 0.499 976 750 596 096 × 2 = 0 + 0.999 953 501 192 192;
  • 37) 0.999 953 501 192 192 × 2 = 1 + 0.999 907 002 384 384;
  • 38) 0.999 907 002 384 384 × 2 = 1 + 0.999 814 004 768 768;
  • 39) 0.999 814 004 768 768 × 2 = 1 + 0.999 628 009 537 536;
  • 40) 0.999 628 009 537 536 × 2 = 1 + 0.999 256 019 075 072;
  • 41) 0.999 256 019 075 072 × 2 = 1 + 0.998 512 038 150 144;
  • 42) 0.998 512 038 150 144 × 2 = 1 + 0.997 024 076 300 288;
  • 43) 0.997 024 076 300 288 × 2 = 1 + 0.994 048 152 600 576;
  • 44) 0.994 048 152 600 576 × 2 = 1 + 0.988 096 305 201 152;
  • 45) 0.988 096 305 201 152 × 2 = 1 + 0.976 192 610 402 304;
  • 46) 0.976 192 610 402 304 × 2 = 1 + 0.952 385 220 804 608;
  • 47) 0.952 385 220 804 608 × 2 = 1 + 0.904 770 441 609 216;
  • 48) 0.904 770 441 609 216 × 2 = 1 + 0.809 540 883 218 432;
  • 49) 0.809 540 883 218 432 × 2 = 1 + 0.619 081 766 436 864;
  • 50) 0.619 081 766 436 864 × 2 = 1 + 0.238 163 532 873 728;
  • 51) 0.238 163 532 873 728 × 2 = 0 + 0.476 327 065 747 456;
  • 52) 0.476 327 065 747 456 × 2 = 0 + 0.952 654 131 494 912;
  • 53) 0.952 654 131 494 912 × 2 = 1 + 0.905 308 262 989 824;
  • 54) 0.905 308 262 989 824 × 2 = 1 + 0.810 616 525 979 648;

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(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 11(2)

6. Positive number before normalization:

0.000 000 000 742 147(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 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(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1100 11(2) × 20 =

1.1001 0111 1111 1111 1110 011(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 1110 011


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 0011 =

100 1011 1111 1111 1111 0011


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 0011


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

1 - 0110 0000 - 100 1011 1111 1111 1111 0011

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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